SETS AND PROBABILITY. Chapter Sets. 7.1 Exercises Î {2,5, 7,9,10} The number 3 is not an element of the set, so the statement is false.

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1 Chapter 7 SETS AND PROBABILITY 7. Sets Your Turn There are three states whose names begin with O, Ohio and Oklahoma. Thus { x x is a state that begins with the letter O} {Ohio, Oklahoma, Oregon}. Your Turn {, 4, 6} Í {6,, 4} is true because each element of the first set is an element of the second set. (In this example the sets are in fact equal.) Your Turn 3 A set of k distinct elements has k subsets, so since there are four seasons, { x x is a season of the year} 4 has 6 subsets. Your Turn 4 U {0,,,...,0} A {3, 6,9} B {, 4, 6,8} Then A {0,,,4,5,7,8,0}. The elements common to A and B are,4, and 8. Thus A Ç B {, 4,8}. Your Turn 5 U {0,,,...,} A {,3,5,7,9,} B {3, 6,9,} C {,, 3, 4, 5} To find A È ( B Ç C ) begin with the expression in parentheses and find C, which includes all the elements of the universal set that are not in C. C {6,7,8,9,0,,} The elements in C that are also in B are 6, 9, and, so B Ç C {6, 9,}. Now we list the elements of A and include any elements of B Ç C that are not already listed: A È ( B Ç C ) {,3,5,6,7,9,,} 7. Exercises. 3 Î {,5, 7,9,0} The number 3 is not an element of the set, so the statement is false.. 6 Î {-, 6,9,5} Since 6 is an element of the given set, the statement is true Ï {,,5,8} Since 9 is not an element of the set, the statement is true Ï {7,6,5,4} Since 3 is not an element of the given set, the statement is true. 5. {,5,8,9} {,5,9,8} The sets contain exactly the same elements, so they are equal. The statement is true. 6. {3,7,,4} {3,7,,4,0} Two sets are equal only if they contain exactly the same elements. Since 0 is an element of the second set but not the first, the statement is false. 7. {All whole numbers greater than 7 and less than 0} {8,9} Since 8 and 9 are the only such numbers, the statement is true. 8. { x x is an odd integer, 6 x 8} {7,9,,5,7} The number 3 should be included in the set so the statement is false Î 0/ The empty set has no elements. The statement is false. 0. / 0 Î {/ 0} Since {0} / has 0/ as a member, this statement is true. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley. 385

2 386 Chapter 7 SETS AND PROBABILITY In Exercises, and A {, 4, 6,8,0,}, B {, 4,8,0}, C {4,8,}, D {,0}, E {6}, U {,4,6,8,0,,4}.. Since every element of A is also an element of U, A is a subset of U, written A Í U.. Since every element of E is also an element of A, E is a subset of A, written E Í A. 3. A contains elements that do not belong to E, namely, 4, 8, 0, and, so A is not a subset of E, written A Í E. 4. Since 0 is an element of B but is not an element of C, B is not a subset of C, written B Í C. 5. The empty set is a subset of every set, so / 0 Í A. 6. Since 0 is an element of {0, }, but is not an element of D, {0, } Í D. 7. Every element of D is also an element of B, so D is a subset of B, D Í B. 8. Since, 6, and 0 are elements of A, but are not elements of C, A Í C. 9. Since every element of A is also an element of U, and A ¹ U, A Ì U. Since every element of E is also an element of A, and E ¹ A, E Ì A. Since every element of A is not also an element of E, A Ë E. Since every element of B is not also an element of C, B Ë C. Since 0 is a subset of every set, and 0 ¹ A, 0 Ì A. Since every element of {0, } is not also an element of D, {0, } Ë D. Since every element of D is not also an element of B, and D ¹ B, D Ì B. Since every element of A is not also an element of C, A Ë C. 4. Since B has 4 elements, it has or 6 subsets. There are exactly 6 subsets of B. 3. A set with n distinct elements has n subsets, and C has n 3 elements. Therefore, there 3 are exactly 8 subsets of C. 4. Since D has elements, it has or 4 subsets. There are exactly 4 subsets of D. 5. Since {7,9} is the set of elements belonging to both sets, which is the intersection of the two sets, we write {5,7,9,9} Ç {7,9,,5} {7,9}. 6. {8,,5} Ç {8,,9,0} {8,} {8,} is the set of all elements belonging to both of the first two sets, so it is the intersection of those sets. 7. Since {,,5,7,9} is the set of elements belonging to one or the other (or both) of the listed sets, it is their union. {,,7} È {,5,9} {,,5,7,9} 8. Since {6,,4,6,9} is the set of elements belonging to one or the other (or both) of the listed sets, it is their union. {6,,4,6} È {6,4,9} {6,,4,6,9} 9. Since 0/ contains no elements, there are no elements belonging to both sets. Thus, the intersection is the empty set, and we write {3,5,9,0} Ç / 0 / {3,5,9,0} È / 0 {3,5,9,0} The empty set contains no elements, so the union of any set with the empty set will result in an answer set that is identical to the original set. (On the other hand, {3,5,9,0} Ç / 0 / 0.) 3. {,,4} is the set of elements belonging to both sets, and {,,4} is also the set of elements in the first set or in the second set or possibly both. Thus, {,,4} Ç {,,4} {,,4} and {,,4} È {,,4} {,,4} are both true statements.. A set with n distinct elements has n subsets. A has n 6 elements, so there are exactly 6 64 subsets of A. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

3 Section {0,0} is the set of elements belonging to both of the two sets on the left, and it is also the set of elements belonging to one or the other of these two sets, or to both. Thus {0,0} Ç {0, 0} {0,0} and {0,0} È {0, 0} {0,0} are both true statements. In Exercises 35 44, and U {,,3,4,5,6,7,8,9}, X {, 4, 6,8}, Y {,3, 4,5, 6}, Z {,,3,8,9}. 35. X Ç Y, the intersection of X and Y, is the set of elements belonging to both X and Y. Thus, X Ç Y {, 4, 6,8} Ç{,3, 4, 5, 6} {, 4, 6}. 36. X È Y, the union of X and Y, is the set of all elements belonging to X or Y or both. Thus, X È Y {,3,4,5,6,8}. 37. X, the complement of X, consists of those elements of U that are not in X. Thus, X {,3,5,7,9}. 38. Y, the complement of Y, is the set of all elements of U that do not belong to Y. Thus, Y {, 7, 8,9}. 39. From Exercise 37, X {,3,5,7,9}; from Exercise 38, Y {, 7, 8,9}. There are no elements common to both X and Y so X Ç Y {, 7, 9}. 40. X Ç Z {,3,5,7,9} Ç{,,3,8,9} {,3,9} 4. First find X È Z. X È Z {, 4, 6,8} È{,,3,8,9} {,, 3, 4,6, 8,9} Now find Y Ç ( X È Z). Y Ç ( X È Z) {,3, 4,5, 6} Ç{,,3, 4, 6,8,9} {,3, 4, 6} 4. From Exercise 37, X {,3,5,7,9}; from Exercise 38, Y {, 7, 8,9}. X Ç( Y È Z) {,3,5,7,9} Ç ({,7,8,9} È{,,3,8,9}) {,3,5,7,9} Ç{,,3,7,8,9} {,3,7,9} 43. U {,,3,4,5,6,7,8,9} and Z {,,3,8,9}, so Z {4,5,6,7}. From Exercise 38, Y {, 7, 8,9}. ( X Ç Y ) È ( Z Ç Y ) ({,4,6,8} Ç{,7,8,9}) È ({4,5,6,7} Ç{,7,8,9}) {8} È{7} {7,8} 44. From Exercise 37, X {,3,5,7,9}. ( X Ç Y) È ( X Ç Z) ({,4,6,8} Ç {,3,4,5,6}) È ({,3,5,7,9} Ç {,,3,8,9}) {,4,6} È {,3,9} {,,3,4,6,9} 45. ( A Ç B) È ( A Ç B ) ({3,6,9} Ç{,4,6,8}) È ({3,6,9} Ç{0,,3,5,7,9,0}) {6} È{3,9} {3,6,9} A 47. M consists of all students in U who are not in M, so M consists of all students in this school not taking this course. 48. M È N is the set of all students in this school taking this course or taking accounting. 49. N Ç P is the set of all students in this school taking both accounting and zoology. 50. N Ç P is the set of all students in this school not taking accounting and not taking zoology. 5. A {, 4, 6,8,0,}, B {, 4, 8,0}, C {4,8,}, D {,0}, E {6}, U {,4,6,8,0,,4} A pair of sets is disjoint if the two sets have no elements in common. The pairs of these sets that are disjoint are B and E, C and E, D and E, and C and D. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

4 388 Chapter 7 SETS AND PROBABILITY 5. U X Y {,,3,4,5,6,7,8,9} {, 4, 6,8} {,3, 4,5, 6} Z {,,3,8,9} Since each pair of sets has at least one element in common (for example, the element ), none of the pairs of sets are disjoint. 53. B is the set of all stocks on the list with a closing price below $60 or above $70. B {AT&T, Coca-Cola, FedEx, Disney} 54. A Ç B is the set of all stocks on the list with a high price greater than $50 and a closing price between $60 and $70. A Ç B {McDonald s, PepsiCo} 55. ( A Ç B) is the set of all stocks on the list that do not have both a high price greater than $50 and a closing price between $60 and $70. ( A Ç B) {AT&T, Coca-Cola, FedEx, Disney} 56. ( C È D) is the set of all stocks on the list that do not have either a positive price change or a low price less than $ A {,, 3,{3},{, 4,7}} (a) (b) {3} (c) {} (d) 4 Î A is true. Î is true. ( C È D) 0 Î is false. ({} Í ) Î is false. (4 Î {, 4, 7}) (e) {{ 3}} Ì A is true. (f ) {,4,7} Î A is true. (g) {, 4, 7} Í A is false. ({, 4,7} Î A) 58. B {,,,{},{, abc d e f}} (a) a (b) {, bcd, } (c) {} d (d) {} d (e) {, e f} Î B is true. (f) {,{, a e f}} (g) {, e f} Ì B is false. ( d Ï B) B is true. Í B is false. ({ d} Î B) Î B is true. Ì B is true. Ì B is false. ({ e, f} Î B) For Exercises 59 through 6 refer to this abbreviated version of the table: Vanguard 500 (V ) Fidelity New Millennium Fund (F ) Janus Perkins Large Cap Value (J ) Templeton Large Cap Value Fund (T) Exxon Pfizer Exxon IBM Apple Cisco GE GE GE Wal-Mart Wal-Mart HP IBM Apple AT&T Home Depot JPMorgan JPMorgan JPMorgan Aflac 59. V Ç J {Exxon, Apple, GE, IBM, JPMorgan} Ç{Exxon, GE, Wal-Mart, AT&T, JPMorgan} {Exxon, GE, JPMorgan} 60. V Ç ( F È T) {Exxon, Apple, GE, IBM, JPMorgan} Ç ({Pfizer, Cisco, Wal-Mart, Apple, JPMorgan} È {IBM, GE, HP, Home Depot, Aflac}) {Exxon, Apple, GE, IBM, JPMorgan} Ç {Pfizer, Cisco, Wal-Mart, Apple, JPMorgan, IBM, GE, HP, Home Depot, Aflac} {Apple, GE, IBM, JPMorgan} 6. J È F {Exxon, GE, Wal-Mart, AT&T, JPMorgan} È {Pfizer, Cisco,Wal-Mart, Apple, JPMorgan} {Exxon, GE, Wal-Mart, AT&T, JPMorgan, Pfizer, Cisco, Apple} ( J È F) {IBM, HP, Home Depot, Aflac} 6. J Ç T is the set of elements neither in J nor in T. This is the set of elements in the first two columns of the table that are not in the last two columns. J Ç T {Pfizer, Cisco, Apple} 63. The number of subsets of a set with k elements is k, so the number of possible sets of customers 9 (including the empty set) is U {, sdcgimh,,,,, } and O {, imhg,, }, so O { s, d, c}. 65. U {, sdcgimh,,,,, } and N {, sdcg,, },so N {, i m, h}. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

5 Section N Ç O {, s d, c, g} Ç {, i m, h, g} { g} 67. N È O {, s d, c, g} È{, i m, h, g} {, s dcgimh,,,,, } U 68. N Ç O {, s d, c, g} Ç {, s d, c} {, s d,} c 69. The number of subsets of a set with 5 elements (50 states plus the District of Columbia) is 5 5» The total number of subsets is 6. The number of subsets with no elements is, and the number of subsets with one element is 4. Therefore, the number of subsets with at least two elements is 6 - ( + 4) or. For Exercises 7 through 76 refer to this table: Network The Discovery Channel Subscribers (millions) 4 Launch Content Nonfiction, nature, science TNT Movies, sports, original programming USA Network Sports, family entertainment TLC Original programming, family entertainment TBS Movies, sports, original programming 7. F {USA, TLC, TBS} 7. G {TNT, USA, TBS} 73. H {Discovery, TNT} 74. F Ç H {USA, TLC, TBS} Ç {Discovery, TNT} ; the set of networks launched before 985 that also have more than 97.6 million viewers. 75. G È H {TNT, USA, TBS} È{Discovery, TNT} {TNT, USA, TBS, Discovery}; the set of networks that feature sports or that have more than 97.6 million viewers. 76. G {Discovery, TLC}; the set of networks that do not feature sports. 77. Joe should always first choose the complement of what Dorothy chose. This will leave only two sets to choose from, and Joe will get the last choice. 78. (a) A È ( B Ç C) B Ç C is the set of states with both a population over 4,000,000 and an area greater than 40,000 square miles. Therefore, ( B Ç C) is the set of states with not both a population over 4,000,000 and an area greater than 40,000 square miles. As a result, A È ( B Ç C) is the set of states whose name contains the letter e or which are not both over 4,000,000 in population and over 40,000 square miles in area. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley. (b) ( B Ç C ) ({ Alabama, Colorado, Florida, Indiana, Kentucky, New Jersey} Ç {Alabama, Alaska, Colorado, Florida, Kentucky, Nebraska}) {Alabama, Colorado, Florida, Kentucky} {Alaska, Hawaii, Indiana, Maine, Nebraska, New Jersey} A È ( B Ç C ) { Kentucky, Maine, Nebraska, New Jersey) È {Alaska, Hawaii, Indiana, Maine, Nebraska, New Jersey} { Alaska, Hawaii, Indiana, Kentucky, Maine, Nebraska, New Jersey} 79. (a) (A È B ) Ç C (b) A È B is the set of states whose name contains the letter e or which have a population over 4,000,000. Therefore, ( A È B ) is the set of states which are not among those whose name contains the letter e or which have a population over 4,000,000. As a result, ( A È B) Ç C is the set of states which are not among those whose name contains the letter e or which have a population over 4,000,000 and which also have an area over 40,000 square miles. ( AÈ B ) {(Kentucky, Maine, Nebraska, New Jersey} È {Alabama, Colorado, Florida, Indiana, Kentucky, New Jersey}) {Alabama, Colorado, Florida, Indiana, Kentucky, Maine, Nebraska, New Jersey} {Alaska, Hawaii} ( AÈ B) Ç C {Alaska, Hawai} Ç{Alabama, Alaska, Colorado, Florida, Kentucky, Nebraska} {Alaska}

6 390 Chapter 7 SETS AND PROBABILITY 7. Applications of Venn Diagrams Your Turn A È B is the set of elements in A or not in B or both in A and not in B. U M 4 3 W 6 T U A B Since nt ( ) 46, the number who ate only at Taco Bell is Your Turn 4 Your Turn A Ç( B È C) First find A. Let T represent the set of those texting and M represent the set of those listening to music. The number in the lounge is nt ( È M) nt ( ) + nm ( ) - nt ( Ç M). We know that nt ( ) 5, nm ( ), and nt ( Ç M) 8. Then nt ( È M) U A B 7. Exercises C. B Ç A is the set of all elements in B and not in A. Then find B È C. U U A B A B Then intersect these regions. C B A'. A È B is the set of all elements in A or not in B, or both. A B U C Your Turn 3 Start with nm ( Ç TÇ W) 3 and label the corresponding region with 3. Since nm ( Ç T) 7, there are an additional 4 elements in M Ç T but not in M Ç T Ç W. Label the corresponding region with 4. Since nt ( Ç W) 9, there are an additional 6 elements in T Ç W but not in M Ç T Ç W. Label the corresponding region with 6. The Venn diagram now looks like this: 3. A È B is the set of all elements that do not belong to A or that do belong to B, or both. A A' B B U Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

7 Section A Ç B is the set of all elements not in A and not in B U, so the entire region is shaded. 5. B È ( A Ç B ) First find A Ç B, the set of elements not in A and not in B. 9. Three sets divide the universal set into at most 8 regions. (Examples of this situation will be seen in Exercises 7.). ( A Ç B) Ç C First form the intersection of A with B. For the union, we want those elements in B or ( A Ç B ), or both. A B U A Ç B Now form the intersection of A Ç B with C. The result will be the set of all elements that belong to all three sets. B' (A' B' ) 6. ( A Ç B) È B First find A Ç B, the set of elements in A and in B. Now combine this region with B, the set of all elements not in B. For the union, we want those elements in A Ç B or B, or both.. ( A Ç C ) È B First find A Ç C, the region in A and not in C. 7. U is the empty set 0. U A B U' 0 Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

8 39 Chapter 7 SETS AND PROBABILITY For the union, we want the region in ( A Ç C ) or in B, or both. 3. A Ç ( B È C ) Then find B Ç C, the region where B and C overlap. C is the set of all elements in U that are not elements of C. Now intersect these regions. Now form the union of C with B. Finally, find the intersection of this region with A. 5. ( A Ç B ) Ç C A Ç B is the part of the universal set not in A and not in B: A B U C A Ç B 4. A Ç ( B Ç C) First find A, the region not in A. C is the part of the universal set not in C: U A B C C Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

9 Section Now intersect the shaded regions in these two diagrams: A B U Now form the union of these two regions. U A B C C 6. ( A Ç B ) Ç C (A' B') C' First find A Ç B, the region in A and not in B. 8. A Ç ( B È C) First find A. (A B') C' Now intersect this region with C. Then find B È C, the region not in B or in C, or both. 7. ( A Ç B ) È C First find A Ç B, the region in A and not in B: Now intersect these regions. 9. ( A È B ) Ç C First find A È B, the region in A or B or both. C is the region of the universal set not in C: C Intersect this with C. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

10 394 Chapter 7 SETS AND PROBABILITY A C B U 4. na ( È B) na ( ) + nb ( ) - na ( Ç B) nb ( ) nb ( ) nb ( ) A È ( B Ç C ) First find A : (A B' ) C 5. nu ( ) 4 na ( ) 6 na ( Ç B) nb' ( ) 0 Now find B Ç C : First put in A Ç B. Since na ( ) 6, and are in A Ç B, there must be 4 elements in A that are not in A Ç B. n( B ) 0, so there are 0 not in B. We already have 4 not in B (but in A), so there must be another 6 outside B and outside A. So far we have accounted for 3, and nu ( ) 4, so 9 must be in B but not in any region yet identified. Thus na ( Ç B) 9. U A B B Ç C Now find the union of these two regions: U A B C A' (B' C'). na ( È B) na ( ) + nb ( ) - na ( Ç B) na ( È B) na ( ) + nb ( ) - na ( Ç B) na ( Ç B) na ( Ç B) na ( Ç B) 3. na ( È B) na ( ) + nb ( ) - na ( Ç B) na ( ) na ( ) na ( ) 6. na ( ) 8 nb ( ) na ( È B) 3 na ( ) 9 na ( È B) na ( ) + nb ( ) - na ( Ç B) na ( Ç B) na ( Ç B) na ( Ç B) 8 To fill in the regions, start with A Ç B. n( A) 8 and na ( Ç B) 8, so na ( Ç B ) 0. nb ( ) and na ( Ç B) 8, so nb ( Ç A ) 4. Since na ( ) 9,4 of which are accounted for in B Ç A, 5 elements remain in A Ç B. A 8 B 0 4 U 5 Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

11 Section na ( È B) 4 na ( Ç B) 6 na ( ) na ( È B ) 5 Start with na ( Ç B) 6. Since na ( ), there must be 5 more in A not in B. na ( È B) 4; we already have, so 3 more must be in B not yet counted. A È B consists of all the region not in A Ç B, where we have 6. So far are in this region, so another must be outside both A and B. Also, na ( Ç C), so - 9 are in A Ç C but not in A Ç B Ç C. Now we turn our attention to na ( ) 8. So far we have in A; there must be another 8-6 in A not yet counted. Similarly, nb ( ) 34; we have so far, and more must be put in B. For C, nc ( ) 5; we have counted so far. Then there must be 8 more in C not yet counted. The count now stands at 56, and nu ( ) 59, so 3 must be outside the three sets. A 6 B 5 3 U A B 9 6 C 8 U 3 8. na ( ) 3 nb ( ) 5 na ( È B ) 46 na ( Ç B) nb ( ) 5 and na ( Ç B), so nb ( Ç A ) 3. Since na ( ) 3, of which 3 elements are accounted for, 8 elements are in A Ç B. na ( È B ) na ( ) + nb ( ) - na ( Ç B ) nb ( ) nb ( ) 33 nb ( ) 8 elements are in A Ç B, so the rest are in A Ç B, and na ( Ç B ) 5. A B na ( ) 8 nb ( ) 34 nc ( ) 5 na ( Ç B) 4 nb ( Ç C) 5 na ( Ç C) na ( Ç BÇ C) 9 nu ( ) 59 U na ( ) 54 na ( Ç B) na ( È B) 85 na ( Ç BÇ C) 4 na ( Ç C) 5 nb ( Ç C) 6 nc ( ) 44 nb ( ) 63 Start with A Ç B Ç C. We have na ( Ç C) 5, of which 4 elements are in A Ç B Ç C, so na ( Ç B Ç C). nb ( Ç C) 6, of which 4 elements are in A Ç B Ç C, so nb ( Ç CÇ A ). nc ( ) 44, so 7 elements are in C Ç A Ç B. na ( Ç B), so 8 elements are in A Ç B Ç C. na ( ) 54, so elements are in A Ç B Ç C. Now use the union rule. na ( È B) na ( ) + nb ( ) - na ( Ç B) nb ( ) - 53 nb ( ) This leaves 9 elements in B Ç A Ç C. nb ( ) 63, of which are accounted for, leaving 4 elements in A ÇB Ç C. We start with na ( Ç BÇ C) 9. If na ( Ç B) 4, an additional 5 are in A Ç B but not in A Ç B Ç C. Similarly, nb ( Ç C) 5, so are in B Ç C but not in A Ç B Ç C. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

12 396 Chapter 7 SETS AND PROBABILITY 3. na ( Ç B) 6 na ( Ç BÇ C) 4 na ( Ç C) 7 nb ( Ç C) 4 na ( Ç C ) nb ( Ç C ) 8 n( C) 5 n( A Ç B Ç C ) 5 Start with na ( Ç B) 6 and na ( Ç BÇ C) 4 to get 6-4 in that portion of A Ç B outside of C. From nb ( Ç C) 4, there are elements in that portion of B Ç C outside of A. Use na ( Ç C) 7 to get elements in that portion of A Ç C outside of B. Since na ( Ç C ), there are - 9 elements in that part of A outside of B and C. Use nb ( Ç C ) 8 to get 8-6 elements in that part of B outside of A and C. Since nc ( ) 5, there are elements in C outside of A and B. Finally, 5 must be outside all three sets, since na ( Ç B Ç C ) A B C 8 U 33. ( A È B) A Ç B For ( A È B ), first find A È B. Now find ( A È B ), the region outside A È B na ( ) 3 na ( Ç BÇ C) 4 na ( Ç C) 6 na ( Ç B ) 6 nb ( Ç C) 6 nb ( Ç C ) nb ( È C) na ( Ç B Ç C ) 5 For A Ç B, first find A and B individually. Start with the regions A Ç B Ç C and A Ç B Ç C. nb ( Ç C) 6, leaving elements in B Ç C Ç A. na ( Ç C) 6, leaving elements in A Ç C Ç B. na ( Ç B ) 6, leaving 4 elements in A Ç B Ç C. na ( ) 3, leaving 3 elements in A Ç B Ç C. nb ( Ç C ), leaving 8 elements in B Ç C Ç A. Now use the union rule. nb ( È C) nb ( ) + nc ( ) - nb ( Ç C) 7 + nc ( ) - 6 nc ( ) This leaves 3 elements in C Ç A Ç B. Then A Ç B is the region where A and B overlap, which is the entire region outside A È B (the same result as in the second diagram). Therefore, ( A È B) A Ç B. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

13 Section ( A Ç B ) is the complement of the intersection of A and B; hence it contains all elements not in A Ç B. A È B is the union of the complements of A and B; hence it contains any element that is either not in A or not in B. A is the set of all elements in U that are not in A. Then A Ç ( B È C) is the region where the above two diagram overlap. B is the set of all elements in U that are not in B. Next find A Ç B and A Ç C individually. Form the union of A and B. The Venn diagrams show that as claimed. ( A Ç B) A È B, 35. A Ç ( B È C) ( A Ç B) È ( A Ç C) First find A and B È C individually. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

14 398 Chapter 7 SETS AND PROBABILITY Then ( A Ç B) È ( A Ç C) is the union of the above two diagrams. The Venn diagram for A Ç ( B È C) is identical to the Venn diagram for ( A Ç B) È ( A Ç C), so conclude that A Ç ( B È C) ( A Ç B) È ( A Ç C). 36. A È ( B Ç C) contains all the elements in A or in both B and C. Comparing the Venn diagrams, we see that as claimed. A È ( B Ç C) ( A È B) Ç ( A È C), 37. Prove ( A È B) Ç ( A È C) contains the intersection A È B and A È C. na ( È BÈ C) na ( ) + nb ( ) + nc ( ) - na ( Ç B) - na ( Ç C) - nb ( Ç C) + na ( Ç BÇ C) na ( È BÈ C) na [ È ( BÈ C)] na ( ) + nb ( È C) - na [ Ç ( BÈ C)] na ( ) + nb ( ) + nc ( ) - nb ( Ç C) - n[( A Ç B) È ( A Ç C)] na ( ) + nb ( ) + nc ( ) - nb ( Ç C) -{( naç B) + na ( Ç C) - n[( A Ç B) Ç ( A Ç C)]} na ( ) + nb ( ) + nc ( ) - nb ( Ç C) - na ( Ç B) - na ( Ç C) + na ( Ç BÇ C) Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

15 Section Let M be the set of those who use a microwave oven, E be the set of those who use an electric range, and G be the set of those who use a gas range. We are given the following information. nu ( ) 40 nm ( ) 58 ne ( ) 63 ng ( ) 58 nm ( Ç E) 9 nm ( Ç G) 7 ng ( Ç E) 4 nm ( Ç GÇ E) nm ( Ç G Ç E ) Since nm ( Ç GÇ E), there is element in the region where the three sets overlap. Since nm ( Ç E) 9, there are 9-8 elements in M Ç E but not in M Ç G Ç E. Since nm ( Ç G) 7, there are 7-6 elements in M Ç G but not in M Ç G Ç E. Since ng ( Ç E) 4, there are 4-3 elements in G Ç E but not in M Ç G Ç E. Now consider nm ( ) 58. So far we have in M; there must be another in M not yet counted. Similarly, ne ( ) 63; we have counted so far. There must be 63-4 more in E not yet counted. Also, ng ( ) 58; we have counted so far. There must be more in G not yet counted. Lastly, nm ( Ç G Ç E ) indicates that there are elements outside of all three sets. 39. Let A be the set of trucks that carried early peaches, B be the set of trucks that carried late peaches, and C be the set of trucks that carried extra late peaches. We are given the following information. na ( ) 34 nb ( ) 6 nc ( ) 50 na ( Ç B) 5 nb ( Ç C) 30 na ( Ç C) 8 na ( Ç BÇ C) 6 na ( Ç B Ç C ) 9 Start with A Ç B Ç C. We know that na ( Ç BÇ C) 6. Since na ( Ç B) 5, the number in A Ç B but not in C is Since nb ( Ç C) 30, the number in B Ç C but not in A is Since na ( Ç C) 8, the number in A Ç C but not in B is 8-6. Since na ( ) 34, the number in A but not in B or C is 34 - ( ) 7. Since nb ( ) 6, the number in B but not in A or C is 6 - ( ). Since nc ( ) 50, the number in C but not in A or B is 50 - ( ) 8. Since na ( Ç B Ç C ) 9, the number outside A È B È C is 9. Note that the numbers in the Venn diagram add up to 4 even though nu ( ) 40. Jeff has made some error, and he should definitely be reassigned. (a) From the Venn diagram, trucks carried only late peaches. (b) From the Venn diagram, 8 trucks carried only extra late peaches. (c) From the Venn diagram, trucks carried only one type of peach. (d) From the Venn diagram, trucks went out during the week. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

16 400 Chapter 7 SETS AND PROBABILITY 40. (a) ny ( Ç R) 40 since 40 is the number in the table where the Y row and the R column meet. (b) nm ( Ç D) 30 since 30 is the number in the table where the M row and the D column meet. (c) nd ( Ç Y) 5 and nm ( ) 80 since that is the total in the M row. nm ( Ç ( DÇ Y)) 0 since no person can simultaneously have an age in the range 5 and have an age in the range By the union rule for sets, nm ( È ( DÇ Y)) nm ( ) + nd ( Ç Y) - nm ( Ç ( DÇ Y)) (d) Y Ç( D È N) consists of all people in the D column or in the N column who are at the same time not in the Y row. Therefore, ny ( Ç( DÈ N)) (e) nn ( ) 45 no ( ) 70 no ( ) no ( Ç N) By the union rule, no ( È N) no ( ) + nn ( ) - no ( Ç N) (f) M Ç ( R Ç N ) consists of all people who are not in the R column and not in the N column and who are at the same time not in the M row. Therefore, nm ( Ç ( R Ç N )) (d) no [ È ( S È A)] no ( ) + ns ( È A) - no [ Ç ( SÈ A)] (3 + 33) + (5 + 4) - ( ) 00 (e) Since is M È O is the entire set, ( M È O ) Ç B B. Therefore, n[( M È O ) Ç B] n( B) 7. (f) Y Ç ( S È B) is the set of all bank customers who are of age 8 9 and who invest in stocks or bonds. 4. We start with the innermost region, which is the number of professors who invested in stocks and bonds and certificates of deposit (CDs). Since this is our unknown, place an x in this region. If 80 invested in stocks and bonds, then 80 - x invested in only stocks and bonds. If 83 invested in bonds and CDs, then 83 - x invested in only bonds and CDs. If 85 invested in stocks and CDs, then 85 - x invested in only stocks and CDs. If invested in stocks, then the number who invested in only stocks is: -[(80 - x) + (85 - x) + x] x If 98 invested in bonds, then the number who invested in only bonds is; 98 -[(80 - x) + (83 - x) + x] x If 00 invested in CDs, then the number who invested in only CDs is: 00 -[(85 - x) + (83 - x) + x] x There are 9 who did not invest in any of the three, so place a 9 outside all three circles. Now, the sum of all the regions is 50, so (g) M È ( D Ç Y) consists of all people age 5 who drink diet cola or anyone age (a) ny ( Ç B) since is the number in the table where the Y row and the B column meet. (b) nm ( È A) nm ( ) + na ( ) - nm ( Ç A) (c) ny [ Ç ( S È B)] ( x - 54) + (80 - x) + ( x - 65) + x + (85 - x) + (83 - x) + ( x - 68) x x professors invested in stocks and bonds and certificates of deposits. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

17 Section Let T be the set of all tall pea plants, G be the set of plants with green peas, and S be the set of plants with smooth peas. We are given the following information. nu ( ) 50 nt ( ) ng ( ) 5 ns ( ) 39 nt ( Ç G) 9 ng ( Ç S) 0 nt ( Ç GÇ S) 6 nt ( Ç G Ç S ) 4 Start by filling in the Venn Diagram with the numbers for the last two regions, T Ç G Ç S and T Ç G Ç S, as shown below. With nt ( Ç G) 9, this leaves nt ( Ç GÇ S ) With ng ( Ç S) 0, this leaves nt ( ÇGÇ S) Since ng ( ) 5, n( T Ç G Ç S ) With no other regions that we can calculate, denote by x the number in T Ç G Ç S. Then nt ( Ç G Ç S ) x 3 - x, and nt ( Ç G Ç S) x 9 - x, as shown. Summing the values for all eight regions, (3 - x) x (9 - x) x 50 x (a) nt ( Ç S) (b) nt ( Ç G Ç S ) 3 - x 3 - (c) nt ( Ç GÇ S) (a) The blood has the A antigen but is Rh negative and has no B antigen. This blood type is A-negative. (b) Both A and B antigens are present and the blood is Rh negative. This blood type is AB-negative. (c) Only the B antigen is present. This blood type is B-negative. (d) Both A and Rh antigens are present. This is blood type is A-positive. (e) All antigens are present. This blood type is AB-positive. (f ) Both B and Rh antigens are present. This blood type is B-positive. (g) Only the Rh antigen is present. This blood type is O-positive. (h) No antigens at all are present. This blood type is O-negative. 45. First fill in the Venn diagram, starting with the region common to all three sets. (a) The total of these numbers in the diagram is 54. (b) had only one antigen. (c) had exactly two antigens. (d) A person with O-positive blood has only the Rh antigen, so this number is 7. (e) A person with AB-positive blood has all three antigens, so this number is 5. (f ) A person with B-negative blood has only the B antigen, so this number is 3. (g) A person with O-negative blood has none of the antigens. There are such people. (h) A person with A-positive blood has the A and Rh antigens, but not the B-antigen. The number is. 46. Extend the table to include totals for each row and column. W B I A Totals F,055, 4, ,35,4,3 M,0,38 48, ,38,0,94 Totals,077,549 89,97 4,037 44,707,46,64 (a) n(f ) is the total for the first row in the table. ( ),4,3. nf Thus, there are,4,3 people in the set F. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

18 40 Chapter 7 SETS AND PROBABILITY (b) nf ( Ç ( IÈ A)) is the sum of the entries in the first row in the I and A columns. Therefore, nf ( Ç ( IÈ A)) nf ( Ç I) + nf ( Ç A) 7,73. Therefore, there are 7,73 people in the set F Ç ( I È A). (c) nm ( È B) nm ( ) + nb ( ) - nm ( Ç B),0, ,97-48,60,343,3. There are,343,3 people in the set M È B. (d) W È I È A is the universe, since each person is either not white, or not American Indian, or not Asian or Pacific Islander. Thus, there are, 46,64 people in the set W' È I È A. (e) Females who are either American Indian or Pacific Islander. 47. Extend the table to include totals for each row and each column. H F Total A B C D Total (a) na ( Ç F) is the entry in the table that is in both row A and column F. Thus, there are 34 players in the set A Ç F. (b) Since all players in the set C are either in set H or set F, C Ç ( H È F) C. Thus, nc ( Ç ( H È F)) nc ( ) 6, the total for row C. There are 6 players in the set C Ç ( H È F). (c) nd ( È F) nd ( ) + nf ( ) - nd ( Ç F) (d) B Ç C is the set of players who are both not in B and not in C. Thus, B Ç C A È D, and since A and D are disjoint, na ( È D) na ( ) + nd ( ) There are 48 players in the set B Ç C. 48. Use the following table: A B C D Totals O 4,3, ,505 E 59,40 5,965 4,5,749 65,340 M Totals 74,4 64,984 5,09,95 03,375 (a) na ( È B) is the total number of female personnel serving in the Army or the Air Force: na ( È B) na ( ) + nb ( ) 39,395. (b) E È ( C È D) includes all enlisted personnel, and all personnel in the Navy and Marines. In order not to count any person twice, we take ne ( ) and add nc ( Ç E ) and nd ( Ç E ). This gives us 65,340 + ( ) + (0 + 0) 75,346. There are 75,346 people in the set E È ( C È D). (c) O Ç M E, since the enlisted are the only group who are both not officers and not cadets or midshipmen. Thus no ( Ç M ) ne ( ) 65,340. There are 65,340 people in the set O Ç M. 49. Reading directly from the table, na ( Ç B) 0.6. Thus, there are 0.6 million people in the set A Ç B. 50. ng ( È B) ng ( ) + nb ( ) - ng ( Ç B) Thus, there are.7 million people in the set G È B. 5. ng ( È ( CÇ H)) ng ( ) + nc ( Ç H) There are 85.4 million people in the set G È ( C Ç H). 5. The only intersection of the set F and the set B È H is the set F Ç B. Therefore, nf ( Ç ( BÈ H)) nf ( Ç B) There are 37.6 million people in the set F Ç ( B È H). 53. nh ( È D) nh ( ) + nd ( ) - nh ( Ç D) There are 7.0 million people in the set H È D. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

19 Section First of all, A Ç C B È D È E since the only people not in set A and not in the set C are the people in set B or set D or set E. Second G F È H, since the only people not in the set G are either in the set F or the set H. Thus, the set G Ç ( A Ç C ) consists of people in either F or H and also in B, D, or E. Therefore, ng ( Ç ( A Ç C )) nf ( Ç B) + nf ( Ç D) + nf ( Ç E) + nh ( Ç B) + nh ( Ç D) + nh ( Ç E) There are 60. million people in the set G Ç ( A Ç C ). For Exercises 55 through 58, use the following table, where the numbers are in thousands and the table has been extended to include row and column totals. W B H A Totals N 54,05 3,547, ,9 M 06, , ,946 I, ,83 D 3, ,778 Totals 95,736 9,454 3,677,43 69, N Ç ( B È H) is the set of Blacks or Hispanics who never married. These people are located in the first row of the table, in the B and H columns, so nn ( Ç ( BÈ H)) 3,547 +, 0 5,568; since the table values are in thousands, this set contains 5,568,000 people. 56. ( M È I) Ç A is the set of married or widowed people who are also Asian/Pacific Islanders. They are found in the M and I rows of the table in the A column, so nm ( È I) Ç A) ,48; since the table values are in thousands, this set contains 7,48,000 people. 57. ( D ÈW) Ç A is the set of Whites or Divorced/ separated people who are not Asian/Pacific Islanders. The number of these people is found by adding the total of the W column to the total of the D row and then subtracting the number of Divorced/separated Whites (so we don t count them twice) and subtracting the number of Divorced/separated Asians/Pacific Islanders (whom we don t want to count). n(( D ÈW) Ç A ) 95, ,778-3, ,803; since the table values are in thousands, this set contains 03,803,000 people. 58. M Ç ( B È A) is the set of unmarried people who are either Blacks or Asian/Pacific Islanders. Their number can be found by adding the totals of the B and A columns and subtracting the number of married Blacks and the number of married Asian/Pacific Islanders. nm ( Ç ( BÈ A)) 9, 454 +, , 567; since the table values are in thousands, this set contains 4,567,000 people. 59. Let W be the set of women, C be the set of those who speak Cantonese, and F be the set of those who set off firecrackers. We are given the following information. nw ( ) 0 nc ( ) 50 nf ( ) 70 nw' ( Ç C) 08 nw' ( Ç F' ) 00 nw ( Ç C' Ç F) 8 nw' ( Ç C' Ç F' ) 78 nw ( Ç CÇ F) 30 Note that nw ( Ç CÇ F ) nw ( Ç F ) - nw ( Ç C Ç F ) Furthermore, nw ( Ç CÇ F) nw ( Ç C) - nw ( Ç CÇ F ) We now have nw ( Ç CÇ F ) nc ( ) - nw ( Ç C Ç F) - nw ( Ç C Ç F) - nw ( Ç C Ç F ) With all of the overlaps of W, C, and F determined, we can now compute nw ( Ç C Ç F ) 60 and nw ( Ç C Ç F) 36. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

20 404 Chapter 7 SETS AND PROBABILITY (a) Adding up the disjoint components, we find the total attendance to be (b) nc ( ) 34 - nc ( ) (c) nw ( Ç F ) (d) nw ( Ç CÇ F) Let F be the set of people who brought food, C be the set of those who brought costumes, and K be the set of those who brought crafts. We are given the following information. nu ( ) 75 nf ( Ç CÇ K) 5 nf ( Ç K) 5 nf ( ) 4 nk ( ) 35 nk ( Ç C ) 4 nf ( Ç C Ç K ) 0 nc ( Ç K ) 8 Fill in regions in the Venn diagram in this order: Put 5 in the region F Ç C Ç K. Since nf ( Ç K) 5, we can put in F Ç C Ç K. Since nk ( Ç C ) 4, we can now put in K Ç F Ç C. Since nk ( ) 35, we can then put in C Ç K Ç F. Now we have run out of numbers that we can compute directly, so we put the unknown x into F Ç C Ç K. Since nf ( ) 4 we can now put x or 7 - x into F Ç C Ç K. Since nc ( Ç K ) 8, we can put 8 - x in C Ç F Ç K. Finally, since 0 families brought none of the three items, we can put a 0 outside the circles. The Venn diagram now looks like this: To find x, we note that since there were 75 families and 0 brought none of the three items, the total of the seven inner regions must be x x x implies that 70 - x 65 or x 5. This allows us to fill in all the missing regions. We can now read the answers to the questions from the completed diagram. (a) nc ( Ç F) (b) nc ( ) (c) nf ( Ç C ) + 0 (d) nk ( ) (e) nf ( È C) Let F be the set of fat chickens (so F is the set of thin chickens), R be the set of red chickens (so R is the set of brown chickens), and M be the set of male chickens, or roosters (so M is the set of female chickens, or hens). We are given the following information. nf ( Ç RÇ M) 9 nf ( Ç R Ç M ) 3 nr ( Ç M) 5 nf ( Ç R) nr ( Ç M ) 7 nf ( ) 56 nm ( ) 4 nm ( ) 48 First, note that nm ( ) + nm ( ) nu ( ) 89, the total number of chickens. Since nr ( Ç M) 5, nf ( Ç RÇ M) Since nf ( Ç R), nf ( Ç RÇ M ) Since nr ( Ç M ) 7, nf ( Ç RÇ M ) 7-5. Since nm ( ) 48, nf ( Ç R Ç M ) 48 - ( ) 8. Since nf ( ) 56, nf ( Ç R Ç M) 56 - ( ) 7. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

21 Section And, finally, since nm ( ) 4, nf ( Ç R Ç M) 4 - ( ) 9. F: worker is white, so F : worker is not white. Thus, E Ç F is the event that the worker is 0 or over and is not white. Your Turn 4 The sample space S is S {,,3,4,5,6}. The event H that the die shows a number less than 5 is H {,, 3, 4}. The probability of H is nh ( ) 4 PH ( ). ns ( ) 6 3 (a) nu ( ) (b) nr ( ) nr ( Ç M) + nr ( Ç M ) (c) nf ( Ç M) nf ( Ç RÇ M) + nf ( Ç R Ç M) (d) nf ( Ç M ) nf ( ) - nf ( Ç M) (e) nf ( Ç R ) nf ( Ç R Ç M) + nf ( Ç R Ç M ) (f) nf ( Ç R) nf ( Ç RÇ M) + nf ( Ç RÇ M ) Introduction to Probability Your Turn The sample space of equally likely outcomes is {hh, ht, th, tt}. Your Turn The sample space for tossing two coins is {hh, ht, th, tt}. The event E: the coins show exactly one head is E { ht, th}. Your Turn 3 E: worker is under 0, so E : worker is 0 or over. Your Turn 5 In a standard 5-card deck there are 4 jacks and 4 kings, so P (jack or king) Exercises The sample space is the set of the twelve months, {January, February, March,..., December}. 4. List the days in April. S {,,3,4,...,9,30} 5. The possible number of points earned could be any whole number from 0 to 80. The sample space is the set {0,,,3,...,79,80}. 6. List the possible number of hours of TV watching in a day. S {0,,, 3,..., 3, 4} 7. The possible decisions are to go ahead with a new oil shale plant or to cancel it. The sample space is the set {go ahead, cancel}. 8. Let u up and d down. There are possible 3-day outcomes, so 3 8 S { uuu, uud, udu, duu, ddu, dud, udd, ddd}. 9. Let h heads and t tails for the coin; the die can display 6 different numbers. There are possible outcomes in the sample space, which is the set {( h,), ( h,), ( h,3), ( h,4), ( h,5), ( h,6), ( t,),( t,),( t,3),( t,4),( t,5),( t,6)}. 0. There are possible outcomes. S {(,),(,),(,3),(,4),(,5), (,),(,),(,3),(,4),(,5), (3,),(3, ),(3,3),(3, 4),(3,5), (4,),(4,),(4,3),(4,4),(4,5), (5,),(5, ),(5,3),(5, 4),(5,5)} Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

22 406 Chapter 7 SETS AND PROBABILITY 3. Use the first letter of each name. The sample space is the set S {AB, AC, AD, AE, BC, BD, BE,CD,CE, DE}. ns ( ) 0. Assuming the committee is selected at random, the outcomes are equally likely. (a) One of the committee members must be Chinn. This event is {AC, BC, CD, CE}. (b) Alam, Bartolini, and Chinn may be on any committee; Dickson and Ellsberg may not be on the same committee. This event is {AB,AC,AD,AE,BC,BD,BE,CD,CE}. (c) Both Alam and Chinn are on the committee. This event is {AC}. 4. S {(CA, CO, NJ), (CA, CO, NY), (CA, CO, UT) (CA, NJ, NY), (CA, NJ, UT), (CA, NY, UT), (CO, NJ, NY), (CO, NJ, UT), (CO, NY, UT), (NJ, NY, UT)} Ns () 0 Assuming the states are chosen at random, the outcomes are equally likely. (a) Of the states listed, the ones that border an ocean are CA, NJ, and NY. Therefore, the event all three states border an ocean is the set {(CA, NJ, NY)}. (b) The states that border an ocean are CA, NJ, and NY. Therefore, the event exactly two of the three states border an ocean is the set {(CA, CO, NJ), (CA, CO, NY), (CA, NJ, UT), (CA, NY, UT), (CO, NJ, NY), (NJ, NY, UT)}. (c) The states that lie west of the Mississippi River are CA, CO, and UT. Therefore, the event exactly one of the three states is west of the Mississippi River is the set {(CA, NJ, NY), (CO, NJ, NY), (NJ, NY, UT)}. 5. Each outcome consists of two of the numbers,, 3, 4, and 5, without regard for order. For example, let (, 5) represent the outcome that the slips of paper marked with and 5 are drawn. There are ten equally likely outcomes in this sample space, which is S {(,),(,3),(,4),(,5),(,3), (,4), (,5), (3,4), (3,5), (4,5)}. (a) Both numbers in the outcome pair are even. This event is {(, 4)}, which is called a simple event since it consists of only one outcome. (b) One number in the pair is even and the other number is odd. This event is {(, ), (,4),(,3), (,5), (3, 4),(4,5)}. (c) Each slip of paper has a different number written on it, so it is not possible to draw two slips marked with the same number. This event is / 0, which is called an impossible event since it contains no outcomes. 6. Let w wrong; c correct. S { www, wwc, wcw, cww, ccw, cwc, wcc, ccc} ns ( ) 8. The problem states the outcomes are equally likely. (a) The student gets three answers wrong. This event is written {www}. (b) The student gets exactly two answers correct. Since either the first, second, or third answer can be wrong, this can happen in three ways. The event is written {ccw, cwc, wcc}. (c) The student gets only the first answer correct. The second and third answers must be wrong. This event is written {cww}. 7. S { HH, THH, HTH, TTHH, THTH, HTTH, TTTH, TTHT, THTT, HTTT, TTTT} ns ( ). The outcomes are not equally likely. (a) The coin is tossed four times. This event is written {TTHH, THTH, HTTH, TTTH, TTHT, THTT, HTTT, TTTT}. (b) Exactly two heads are tossed. This event is written {HH, THH, HTH, TTHH, THTH, HTTH}. (c) No heads are tossed. This event is written {TTTT}. 8. There are 4 possibilities for the first choice and 5 for the second choice. The sample space is S {(,),(,),(,3),(,4),(,5), (,),(, ),(,3),(, 4),(,5), (3,),(3, ),(3,3),(3, 4),(3,5), (4,),(4, ),(4,3),(4, 4),(4,5)}. ns ( ) 0. Yes, the outcomes are equally likely. (a) The first choice must be or 4; the second can range from to 5: {(,), (, ), (,3), (, 4), (,5), (4,),(4,),(4,3),(4,4),(4,5)}. (b) The first choice can range from to 4; the second must be or 4: {(,),(,4),(,),(,4),(3,),(3,4),(4,),(4,4)}. (c) The choices must add up to 5: {(,4),(,3),(3,),(4,)}. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

23 Section (d) It is not possible for the sum to be : / 0. For Exercises 9 4, use the sample space S {,,3,4,5,6}. 9. Getting a is the event E {}, so ne ( ) and ns ( ) 6. If all the outcomes in a sample space S are equally likely, then the probability of an event E is In this problem, ne ( ) PE ( ). ns ( ) ne ( ) PE ( ). ns ( ) 6 0. Let O be the event getting an odd number. O {,3,5} 3 PO ( ) 6. Getting a number less than 5 is the event E {,,3,4}, so ne ( ) 4. 4 PE ( ) Let A be the event getting a number greater than. A {3, 4,5, 6} 4 PA ( ) Getting a 3 or a 4 is the event E {3,4}, so ne ( ). PE ( ) Let B be the event getting any number except 3. B {,, 4, 5,6} 5 PB ( ) 6 For Exercises 5 34, the sample space contains all 5 cards in the deck, so ns ( ) Let E be the event a 9 is drawn. There are four 9 s in the deck, so ne ( ) 4. ne ( ) 4 P(9) P( E) ns ( ) Let B be the event drawing a black card. There are 6 black cards in the deck, 3 spades and 3 clubs. nb ( ) 6 6 PB ( ) 5 7. Let F be the event a black 9 is drawn. There are two black 9 s in the deck, so nf ( ). nf ( ) P(black 9) P( F) ns ( ) Let H be the event a heart is drawn. There are 3 hearts in the deck. nh ( ) 3 3 PH ( ) Let G be the event a 9 of hearts is drawn. There is only one 9 of hearts in a deck of 5 cards, so ng ( ). ng ( ) P(9 of hearts) P( G) ns ( ) Let F be the event drawing a face card. The face cards are the jack, queen, and king of each of the four suits. nf ( ) 3 PF ( ) Let H be the event a or a queen is drawn. There are four s and four queens in the deck, so nh ( ) 8. nh ( ) 8 P( or queen) P( H) ns ( ) Let R be the event drawing a black 7 or a red 8. There are two black 7 s and two red 8 s. nr ( ) 4 4 PR ( ) Let E be the event a red card or a ten is drawn. There are 6 red cards and 4 tens in the deck. But tens are red cards and are counted twice. Use the result from the previous section. ne ( ) n(red cards) + n(tens) - n(red tens) Now calculate the probability of E. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

24 408 Chapter 7 SETS AND PROBABILITY P(red cards or ten) ne ( ) ns ( ) Let E be the event a spade or a king is drawn. There are 3 spades and 4 kings in the deck. But the king of spades is counted twice. Use the result from the previous section. ne ( ) n(spades) + n(kings) - n(king of spades) Now calculate the probability of E. ne ( ) P(spade or king) ns ( ) For Exercises 35 40, the sample space consists of all the marbles in the jar. There are marbles, so ns ( ) of the marbles are white, so 3 P (white) There are 4 orange marbles. 4 P (orange) of the marbles are yellow, so 5 P (yellow) There are 8 black marbles. 8 P (black) of the marbles are not black, so 3 P (not black) There are 9 marbles which are orange or yellow. 9 P (orange or yellow) 0 4. It is possible to establish an exact probability for this event, so it is not an empirical probability. 4. This is empirical; only a survey could determine the probability. 43. It is not possible to establish an exact probability for this event, so this an empirical probability. 44. This is not empirical; a formula can compute the probability exactly. 45. It is not possible to establish an exact probability for this event, so this is an empirical probability. 46. This is empirical, based on experience and conditions rather than probability theory. 47. The gambler s claim is a mathematical fact, so this is not an empirical probability. 48. This is empirical, based on experience rather than probability theory. 49. The outcomes are not equally likely. 50. Let W l be the event win on the first draw and W be the event win on the second draw. First of all, PW ( ), a simple choice from 3 three slips of paper. To determine P(W ), the first slip of paper is drawn (from the three available) and thrown away. There are two slips left which may be chosen in any way. So PW ( ). Altogether, the 3 6 probability of winning using this strategy is PW ( ) + PW ( ) E: worker is female F: worker has worked less than 5 yr G: worker contributes to a voluntary retirement plan (a) E occurs when E does not, so E is the event worker is male. (b) E Ç F occurs when both E and F occur, so E Ç F is the event worker is female and has worked less than 5 yr. (c) E È G is the event worker is female or does not contribute to a voluntary retirement plan. (d) F occurs when F does not, so F is the event worker has worked 5 yr or more. (e) F È G occurs when F or G occurs or both, so F È G is the event worker has worked less than 5 yr or contributes to a voluntary retirement plan. (f) F Ç G occurs when F does not and G does not, so F Ç G is the event worker has worked 5 yr or more and does not contribute to a voluntary retirement plan. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

25 Section (a) (b) 03.7 P (Federal government) P (Industry) (c) P (Academic institutions) (a) From the solution to Exercise 4 in Section 7. we know that 80 investors made all three types of investments, so the probability that a randomly chosen professor invested in stocks and bonds and certificates of deposit is (a) (b) From the solution to Exercise 4 in Section 7. we find that or 5 professors invested only in bonds, so the probability that a randomly chosen professor invested only in bonds is (b) civilian laborer in 7.9 P æ ö ç is 55 or older çè ø 54.3 civilian laborer in 39.8 P æ ö ç is 55 or older çè ø E: person smokes F: person has a family history of heart disease G: person is overweight (a) G : person is not overweight. (b) F Ç G: person has a family history of heart disease and is overweight. (c) E È G : person smokes or is not overweight. 56. E: person smokes 57. (a) F: person has a family history of heart disease G: person is overweight (a) E È F occurs when E or F or both occur, so E È F is the event person smokes or has a family history of heart disease, or both. (b) E Ç F occurs when E does not occur and F does occur, so E Ç F is the event person does not smoke and has a family history of heart disease. (c) F È G is the event person does not have a family history of heart disease or is not overweight, or both. 65,65 P (heart disease) 0.540,44,059 (b),75,838 P (cancer or heart disease),44, (c) P(not accident and not diabetes mellitus),44,059-7,075-70,905,44,059,36,079,44, The total population for 00 is 3,74, and the total for 050 is 393,93. (a) 5,65 P (Hispanic in 00) 3,74» 0.63 (b) 96,508 P (Hispanic in 050) 393,93» (c) P (Black in 00) 4,538 3,74» ,555 (d) P (Black in 050) 393,93» (a) 7 P(served 0-9 years) (b) (c) (d) 988 P (Corps)» ,950,803 P (lost in battle)» , P (I Corps lost in battle)» , P(I Corps not lost in battle), » , P(II Corps not lost in battle), » 0.650,347 P(III Corps not lost in battle) 0,675-4» ,675 P(V Corps not lost in battle) 0,907-87» ,907 Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

26 40 Chapter 7 SETS AND PROBABILITY (e) P(VI Corps not lost in battle) 3,596-4» ,596 P(XI Corps not lost in battle) » P(XII Corps not lost in battle) » P( Cavalry not lost in battle),85-60» ,85 P(Artillery not lost in battle) 376-4» VI Corps had the highest probability of not being lost in battle P (I Corps loss)» , 4369 P (II Corps loss )» ,347 4 P (III Corps loss )» , P (V Corps loss )» ,907 (d) (e) P(I Corps not lost in battle) 0, » ,706 P(II Corps not lost in battle) 0, » ,666 P(III Corps not lost in battle), » ,083 P(Cavalry not lost in battle) 66-86» The Cavalry had the highest probability of not being lost in battle. 766 P (I Corps loss)» , P (II Corps loss)» , P (III Corps loss)» 0.366, P (Cavalry loss)» I Corps had the highest probability of loss. 6. (a) (b) (c) 4 P (VI Corps loss )» , P (XI Corps loss )» P (XII Corps loss )» P (Cavalry loss )» 0.055,85 4 P (Artillery loss )» I Corps had the highest probability of loss.,083 P (III Corps)» ,076,557 P (lost in battle)» , P (ICorps lost in battle)» , (a) (b) (c) 0 4 P (brought costumes and food) 75 5 P(brought crafts, but neither food 4 nor costumes) 75 P (brought food or costumes) 63. There were 34 in attendance. (a) (b) (c) P (speaks Cantonese) P (does not speaks Cantonese) P (woman who did not light firecracker) Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

27 Section Basic Concepts of Probability Your Turn Let A stand for the event ace and C stand for the event club. PA ( È C) PA ( ) + PC ( ) - PA ( Ç C) Your Turn Let E stand for the event eight and B stand for the event both dice show the same number. From Figure 8 we see that E contains the 5 events 6-, 5-3, 4-4, 3-5, and -6. B contains the 6 events -, -, 3-3, 4-4, 5-5, and 6-6. Only the even 4-4 belongs to both E and B. The sample space contains 36 equally likely events. PE ( È B) PE ( ) + PB ( ) - PE ( Ç B) Your Turn 3 The complement of the event sum < is sum or sum. Figure 8 shows that sum or sum contains the 3 events 5-6, 6-5, and 6-6. P(sum < ) - P(sum or sum ) Your Turn 4 Let E be the event snow tomorrow. Since PE ( ) 3, 0 PE ( ) - PE ( ) 7. The odds in favor of snow 0 tomorrow are PE ( ) 3/0 3. PE ( ) 7/0 7 We can write these odds as 3 to 7 or 3:7. Your Turn 5 Let E be the event package delivered on time. The odds in favor of E are 7 to 3, so PE ( ) The probability that the package will not be 0 delivered on time is 3 0. Your Turn 6 If the odds against the horse winning are 7 to 3, 7 7 P P(loses), so (wins) Exercises 3. A person can own a dog and own an MP3 player at the same time. No, these events are not mutually exclusive. 4. A person can be from Texas and be a business major at the same time. No, these events are not mutually exclusive. 5. A person can be retired and be over 70 years old at the same time. No, these events are not mutually exclusive. 6. A person cannot be a teenager and be 70 years old at the same time. Yes, these events are mutually exclusive. 7. A person cannot be one of the ten tallest people in the United States and be under 4 feet tall at the same time. Yes, these events are mutually exclusive. 8. A person can be male and be a nurse at the same time. No, these events are not mutually exclusive. 9. When two dice are rolled, there are 36 equally likely outcomes. (a) Of the 36 ordered pairs, there is only one for which the sum is, namely {(,)}. Thus, P (sum is ). 36 (b) {(,3),(,),(3,)} comprise the ways of getting a sum of 4. Thus, 3 P (sum is 4). 36 (c) {(,4),(,3),(3,),(4,)}comprise the ways of getting a sum of 5. Thus, 4 P (sum is 5) (d) {(,5),(,4),(3,3),(4,),(5,)} comprise the ways of getting a sum of 6. Thus, 5 P (sum is 6). 36 Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

28 4 Chapter 7 SETS AND PROBABILITY 0. When the two dice are rolled, there are 36 equally likely outcomes. Let 5-3 represent the outcome the first die shows a 5 and the second die shows a 3, and so on. (a) Rolling a sum of 8 occurs when the outcome is -6, 3-5, 4-4, 5-3, or 6-. Therefore, since there are five such outcomes, the probability of this event is 5 P (sum is 8). 36 (b) A sum of 9 occurs when the outcome is 3-6, 4-5, 5-4, or 6-3, so 4 P (sum is 9) (c) A sum of 0 occurs when the outcome is 4-6, 5-5, or 6-4, so 3 P (sum is 0). 36 (d) A sum of 3 does not occur in any of the 36 outcomes, so 0 P (sum is 3) Again, when two dice are rolled there are 36 equally likely outcomes. (a) Here, the event is the union of four mutually exclusive events, namely, the sum is 9, the sum is 0, the sum is, and the sum is. Hence, P(sum is 9 or more) P(sum is 9) + P(sum is 0) + P(sum is ) + (sum is ) (b) P(sum is less than 7) P() + P(3) + P(4) + P(5) + P(6) (c) P(sum is between 5 and 8) P(sum is 6) + P(sum is 7) Again, when two dice are rolled, there are 36 equally likely outcomes. (a) P(sum is not more than 5) P(sum is ) + P(sum is ) + P(sum is 3) + P(sum is 4) + P(sum is 5) (b) P(sum is not less than 8) P(8) + P(9) + P(0) + P() + P() (c) P(sum is between 3 and 7) P(4) + P(5) + P(6) P(first die is 3 or sum is 8) P(first die is 3) + P(sum is 8) - P(first die is 3 and sum is 8) Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

29 Section P(second die is 5 or the sum is 0) P(second die is 5) + P(sum is 0) - P(second die is 5 and sum is 0) (a) The events E, 9 is drawn, and F, 0 is drawn, are mutually exclusive, so PE ( Ç F) 0. Using the union rule, P(9 or 0) P(9) + P(0) (b) P(red or 3) P(red) + P(3) - P(red and 3) (c) Since these events are mutually exclusive, P(9 or black 0) P(9) + P(black 0) (d) (e) P (heart or black ) P(face card or diamond) P(face card) + P(diamond) - P(face card and diamond) (a) Less than a 4 would be an ace, a, or a 3. There are a total of aces, s, and 3 s in a deck of 5, so 3 P (ace or or 3). 5 3 (b) There are 3 diamonds plus three 7 s in other suits, so 6 4 P (diamond or 7). 5 3 Alternatively, using the union rule for probability, P(diamond ) + P (7) - P(7 of diamonds) (c) There are 6 black cards plus red aces, so 8 7 P (black or ace). 5 3 (d) P(heart or jack) P(heart) + P(jack) - P(heart and jack) (e) There are 6 red cards plus 6 black face cards, so 3 8 P (red or face card) (a) Since these events are mutually exclusive, P(brother or uncle) P(brother) + P(uncle) (b) Since these events are mutually exclusive, P(brother or cousin) P(brother) + P(cousin) (c) Since these events are mutually exclusive, P(brother or mother) P(brother) + P(mother) Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

30 44 Chapter 7 SETS AND PROBABILITY 8. (a) There are 3 uncles plus 5 cousins out of 3, so 8 P (uncle or cousin). 3 (b) There are 3 uncles, brothers, and 5 cousins, for a total of 0 out of 3, so 0 P (male or cousin). 3 (c) There are aunts, 5 cousins, and mother, for a total of 5 out of 0, so 8 P (female or cousin) (a) There are 5 possible numbers on the first slip drawn, and for each of these, 4 possible numbers on the second, so the sample space contains ordered pairs. Two of these ordered pairs have a sum of 9: (4, 5) and (5, 4). Thus, P (sum is 9). 0 0 (b) The outcomes for which the sum is 5 or less are (, ), (, 3), (, 4), (, ), (, 3), (3, ), (3, ), and (4,). Thus, 8 P (sum is 5 or less). 0 5 (c) Let A be the event the first number is and B the event the sum is 6. Use the union rule. PA ( È B) PA ( ) + PB ( ) - PA ( Ç B) (a) The sample space for this experiment is listed in part (b) below. The only outcomes in which both numbers are even are (, 4) and (4, ), so P (even). 0 0 (b) The sample space is {(,),(,3),(,4),(,5),(,),(,3),(,4),(,5),(3,),(3,), (3,4), (3,5), (4,), (4,),(4,3), (4,5), (5,), (5,), (5,3), (5,4)}. The 8 underlined pairs are the outcomes in which one number is even or greater than 3, so 8 9 P (even or > 3). 0 0 (c) The sum is 5 in the outcomes (, 4), (, 3), (3, ), and (4, l). The second draw is in the outcomes (, ), (3, ), (4, ), and (5, ). There are 7 distinct outcomes out of 0, so 7 P (sum is 5 or second number is ). 0. Since PE ( Ç F) 0.6, the overlapping region E Ç F is assigned the probability 0.6 in the diagram. Since PE ( ) 0.6 and PE ( Ç F) 0.6, the region in E but not F is given the label 0.0. Similarly, the remaining regions are labeled. (a) PE ( È F) Consequently, the part of U outside E È F receives the label (b) PE ( Ç F) P(in Fbut not in E) 0.5 (c) The region E Ç F is that part of E which is not in F. Thus, PE ( Ç F ) 0.0. (d) PE ( È F ) PE ( ) + PF ( ) - PE ( Ç F ) PZ ( ) 0.4, PY ( ) 0.35, PZ ( È Y) 0.59 Begin by using the union rule for probability. PZ ( È Y) PZ ( ) + PY ( ) - PZ ( Ç Y) PZ ( Ç Y) PZ ( Ç Y) PZ ( Ç Y) 0.8 PZ ( Ç Y) This gives the first value to be labeled in the Venn diagram. Then the part of Z outside Y must contain , and the part of Y outside Z must contain Observe that , which agrees with the given information that PZ ( È Y) The part of U outside both Y and Z must contain Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

31 Section This Venn diagram may now be used to find the following probabilities. (a) Z Ç Y is the event presented by the part of the Venn diagram that is outside Z and outside Y. PZ ( Ç Y ) 0.4 (b) Z È Y is everything outside Z or outside Y or both, which is all of U except Z Ç Y. PZ ( Ç Y ) (c) Z È Y is everything outside Z or inside Y or both. PZ ( È Y) (d) Z Ç Y is everything inside Z and outside Y. PZ ( Ç Y ) (a) The sample space is where the first number in each pair is the number that appears on A and the second the number that appears on B. B beats A in 0 of 36 possible outcomes. Thus, 0 5 PB ( beats A ) (b) The sample space is where the first number in each pair is the number that appears on B and the second the number that appears on C. C beats B in 0 of 36 possible outcomes. Thus, 0 5 PC ( beats B ) (c) The sample space is where the first number in each pair is the number that appears on A and the second the number that appears on C. A beats C in 0 of 36 possible outcomes. Thus, 0 5 PA ( beats C ) The statement is not correct. Assume two dice are rolled. P(you win) P(first die is greater than second) P(other player wins) - P(you win) Let E be the event a 3 is rolled. 5 PE ( ) and PE ( ). 6 6 The odds in favor of rolling a 3 are PE ( ) 6 5, PE ( ) 5 6 which is written to Let E be the event 4, 5, or 6 is rolled. Here E {4,5, 6}, so PE ( ) 3, and 6 PE ( ). The odds in favor of rolling 4, 5 or 6 are PE ( ). PE ( ) which is written to. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

32 46 Chapter 7 SETS AND PROBABILITY 9. Let E be the event a, 3, 4, or 5 is rolled. Here PE ( ) 4 and PE ( ). The odds in favor of E are PE ( ) 3, PE ( ) 3 which is written to. 30. Let F be the event some number less than 6 is rolled. Here F {,, 3, 4, 5}, so PF ( ) 5 6 and PF ( ). The odds in favor of rolling a 6 number less than are 5 PF ( ) 6 5, PF ( ) 6 which is written 5 to. 3. (a) Yellow: There are 3 ways to win and 5 ways to lose. The odds in favor of drawing yellow are 3 to 5, or to 5. (b) Blue: There are ways to win and 7 ways to lose; the odds in favor of drawing blue are to 7. (c) White: There are 4 ways to win and 4 ways to lose; the odds in favor of drawing white are 4 to 4, or to 7. (d) Not white: Since the odds in favor of white are to 7, the odds in favor of not white are 7 to. 3. (a) From Figure 8 we see that a sum of 3 includes the outcomes - and -; there are 36 - or 34 remaining outcomes. Thus the odds in favor of a sum of 3 are to 34 or to 7. (b) From Figure 8 we see that a sum of 7 or includes the 8 outcomes -6, -5, 3-4, 4-3, 5-, 6-, 5-6, and 6-5; there are 36-8 or 8 remaining outcomes. Thus the odds in favor of a sum of 7 or are 8 to 8 or to 7. (c) A sum less than 5 must be, 3 or 4, which includes the 6 outcomes -, -, -, -3, -, and 3-; there are remaining outcomes. Thus the odds in favor of a sum less than 5 are 6 to 30 or to 5. (d) Not a sum of 6 is everything except the 5 outcomes -5, -4, 3-3, 4-, and 5-. So outcomes favor the event and the odds in favor of not a sum of 6 are 3 to Each of the probabilities is between 0 and and the sum of all the probabilities is , so this assignment is possible. 36. The probability assignment is possible because the probability of each outcome is a number between 0 and, and the sum of the probabilities of all the outcomes is The sum of the probabilities <, so this assignment is not possible. 38. The probability assignment is not possible. All of the probabilities are between 0 and, but the sum of the probabilities is which is greater than. 39. This assignment is not possible because one of the probabilities is , which is not between 0 and. A probability cannot be negative. 40. The probability assignment is not possible. One of the probabilities is negative instead of being between 0 and, and the sum of the probabilities is not. 4. The answers that are given are theoretical. Using the Monte Carlo method with at least 50 repetitions on a graphing calculator should give values close to these. (a) (b) The answers that are given are theoretical. Using the Monte Carlo method with at least 50 repetitions on a graphing calculator should give values close to these. (a) (b) Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

33 Section The answers that are given are theoretical. Using the Monte Carlo method with at least 00 repetitions should give values close to these. (a) (b) The answers that are given are theoretical. Using the Monte Carlo method with at least 50 repetitions on a graphing calculator should give values close to these. (a) (b) First, notice ( B Ç A ) È ( A È B ) U. As a result, PB ( Ç A ) + ( AÈ B ) PU ( ) PB ( Ç A ) - PA ( È B ) From the Venn diagram, note that A È B A È ( B Ç A ). Therefore, PA ( È B) PA ( ) + PB ( Ç A ) 0.7 PA ( ) + 0. PA ( ) 0.6 The correct answer is choice d. 47. Let C be the event the calculator has a good case, and let B be the event the calculator has good batteries. PC ( Ç B) - P[( C Ç B) ] - PC ( È B ) -[ PC ( ) + PB ( ) - PC ( Ç B )] - ( ) 0.84 Thus, the probability that the calculator has a good case and good batteries is (a) P(no profit) - P(Profit) (b) The odds against profit are or 3 to (a) P($500 or more) - P(less than $500) - ( ) (b) P (less than $000) (c) P ($500 to $999) (d) P ($3000 or more) (a) P (sales or service) (b) P (not in construction) (c) The probability of a worker being in production is , so the probability of a worker not being in production is Thus the odds in favor of a worker being in production are » which we could write as 7 to 883. Since 6 : these odds are approximately 5. (a) The probability of Female and 6 to 4 years old is the entry in the first row in the Female column, which is (b) 6 to 54 years old includes the first two rows of the table, so adding the totals for these two rows we find that the probability is (c) For Male or 5 to 54, we add the totals of the second row and the Male column and then subtract the value for Male and 5 to 54 in order not to count it twice. Thus the probability is (d) For Female or 6 to 4 we add the totals of the first row and the Female column and then subtract the value for Female and 6 to 4 in order not to count it twice. Thus the probability is Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

34 48 Chapter 7 SETS AND PROBABILITY 5. Let S be the event the person is short, and let O be the event the person is overweight. From the Venn diagram, x x x 0.3. The probability that a person is (a) overweight is ; (b) short, but not overweight is 0.0; (c) tall (not short) and overweight is PC ( ) 0.039, PM ( Ç C) 0.035, PM ( È C) 0.49 Place the given information in a Venn diagram by starting with in the intersection of the regions for M and C. Since PC ( ) 0.039, goes inside region C, but outside the intersection of C and M. Thus, Since PC ( Ç M ) PM ( È C) 0.49, goes inside region M, but outside the intersection of C and M. Thus, PM ( Ç C ) The labeled regions have probability Since the entire region of the Venn diagram must have probability, the region outside M and C, or M Ç C, has probability (a) PC ( ) - PC ( ) (b) PM ( ) (c) PM ( ) - PM ( ) (d) PM ( Ç C ) (e) PC ( Ç M ) (f) PC ( È M ) PC ( ) + PM ( ) - PC ( Ç M ) (a) Since red is dominant, the event plant has red flowers (b) 3 { RR, RW, WR}; P(red). 4 P(white) - P(red) 55. (a) Now red is no longer dominant, and RW or WR results in pink, so 4 P(red) P( RR). 4 (b) P(pink) P( RW) + P( WR) (c) P(white) P( WW) (a) P(no more than 4 good toes) (b) P (5 toes) Let L be the event visit results in lab work and R be the event visit results in referral to specialist. We are given the probability a visit results in neither is 35%, so P(( L È R ) ) Since PL ( È R) - P(( LÈ R ) ), we have PL ( È R) We are also given ( ) 0.40 PL and ( ) PR Using the union rule for probability, Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

35 Section PL ( È R) PL ( ) + PR ( ) - PL ( Ç R) PL ( Ç R) PL ( Ç R) PL ( Ç R) 0.05 The correct answer choice is a. 58. Let T be the event patient visits a physical therapist and C be the event patient visits a chiropractor. If % of patients visit both a physical therapist and a chiropractor, then PT ( Ç C) 0.. If % of patients visit neither, then P(( T È C ) ) 0.. This means PT ( È C) - P(( T È C )) Let x P( T). If the probability that a patient visits a chiropractor exceeds by 0.4 the probability that a patient visits a physical therapist, then PC ( ) PT ( ) x Using the union rule for probability, we have PT ( È C) PT ( ) + PC ( ) - PT ( Ç C) 0.88 x + ( x + 0.4) x x x 0.48 The correct answer choice is d. 59. Let x P( A Ç B), y P( B Ç C), z P( A Ç C), and w P(( A È B È C) ). If an employee must choose exactly two or none of the supplementary coverages A, B, and C, then PA ( Ç BÇ C) 0 and the probabilities of the region representing a single choice of coverages A, B, or C are also 0. We can represent the choices and probabilities with the following Venn diagram. x + y + z + w x + z 4 x + y 3 5 y + z Using a graphing calculator or computer program, the solution to the system is x /, y /4, z /6, w /. Since the probability that a randomly chosen employee will choose no supplementary coverage is w, the correct answer choice is c. 60. Gore: Daschle: Kerry: Dodd: Lieberman: Biden: Leahy: Feingold: Edwards: Gephardt: Find the sum of the probabilities » Since the sum of the probabilities of all possible outcomes cannot be greater than, there is some thing wrong with the assignment of odds. The information given leads to the following system of equations. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

36 40 Chapter 7 SETS AND PROBABILITY 6. Since 55 of the workers were women, were men. Since 3 of the women earned more than $40,000, of them earned $40,000 or less. Since 6 of the men earned $40,000 or less, earned more than $40,000. These data for the 30 workers can be summarized in the following table. Men Women $40,000 or less 6 5 Over $40, (a) P(a woman earning $40,000 or less) (b) P(a man earning more than $40,000) (c) P(a man or is earning more than $40,000) (d) P(a woman or is earning $40,000 or less) Let R be the event the person buys rock music, and let T be the event the person is a teenager. The probability that a person (a) is a teenager who buys nonrock music is 0.06; 6 00 (b) buys rock music or is a teenager is ; (c) is not a teenager is ; 00 (d) is not a teenager, but buys rock music, Let A be the set of refugees who came to escape abject poverty and B be the set of refugees who came to escape political oppression. Then PA ( ) 0.80, PB ( ) 0.90, and PA ( Ç B) PA ( È B) PA ( ) + PB ( ) - PA ( Ç B) PA ( Ç B ) - PA ( Ç B) - 0 The probability that a refugee in the camp was neither poor nor seeking political asylum is There are 8 people, so ns ( ) 8. (a) P(Chinese) P(Chinese man) + P(Chinese woman) (b) P(Korean or woman) P(Korean) + P(woman) - P(Korean woman) (c) P(man or Vietnamese) P(man) + P(Vietnamese) - P(Vietnamese man) (d) These events are mutually exclusive, so P(Chinese or Vietnamese) P(Chinese) + P(Vietnamese) (e) P (Korean woman) 8 9 Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

37 Section The odds of winning are 3 to ; this means there are 3 ways to win and ways to lose, out of a total of ways altogether. Hence, the probability of losing is (a) We divide each number in the given table by 03,375, which produces the following table of probabilities: O E M A B C D (b) PA ( ) (c) This is the sum of the entries in the O row and the C and D columns: PO ( Ç ( CÈ D)) (d) This is the sum of the entries in the A and B columns. We have already computed P(A) in part (b). PA ( È B) PA ( ) + PB ( ) (e) We can find PC ( È D) by subtracting the probability we found in (d) from : PC ( È D) To this we want to add the first two values in the E row to form the union with E: PE ( È ( CÈ D) (a) P(somewhat or extremely intolerant of Facists) P(somewhat intolerant of Facists) + P(extremely intolerant of Facists) (b) P(completely tolerant of Communists) P(no intolerance at all of Communists) (a) P(at least some intolerance of Fascists) P(not very much) + P(somewhat) + P(extremely intolerant of Facists) (b) P(at least some intolerance of Communists) P(not very much) + P(somewhat) + P(extremely intolerant of Communists) (a) There are possible judging combinations with Sasha Cohen finishing in first place. The probability of this outcome is, therefore, 9/0 3/55. (b) There are 67 possible judging combinations with Irina Slutskaya finishing in second place. The probability of this outcome is, therefore, 67/0. (c) There are possible judging combinations with Shizuka Arahawa finishing in third place. The probability of this outcome is, therefore, 59/ The odds are as follows or 3 to or 3 to or 6 to or to or to The probabilities are as follows: + 95,99,999 95,00, , , Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

38 4 Chapter 7 SETS AND PROBABILITY 7.5 Conditional Probability; Independent Events Your Turn Reduce the sample space to B and then find na ( Ç B ) and nb ( ). PA ( Ç B ) na ( Ç B ) 30 6 PAB ( ) PB ( ) nb ( ) 55 Your Turn PE ( È F) PE ( ) + PF ( ) - PE ( Ç F) PE ( Ç F) PE ( Ç F) PE ( Ç F) 0.40 PE ( Ç F) PE ( F) PF ( ) Your Turn 3 Since at least one coin is a tail, the sample space is reduced to {ht, th, tt}. Two of these equally likely outcomes have exactly one head, so P (one head at least one tail). 3 Your Turn 4 Let C represent lives on campus and A represent has a car on campus. Using the given information, PAC ( ) and PC ( ) 4. By the product rule, 4 PA ( Ç C) PAC ( ) PC ( ) Your Turn 5 Using Figure 3, we follow the A branch and then the U branch. Multiplying along the tree we find the probability of the composite branch, which represents A Ç U or the event that A is in charge of the campaign and it produces unsatisfactory results. Your Turn 6 PA ( Ç U) There are two paths that result in one NY plant and one Chicago plant: C first and NY second, and NY first and C second. From the tree, P (C, NY), and P (NY,C). So P(one NY plant and one Chicago plant) P(C, NY) + P(NY, C) Your Turn 7 Successive rolls of a die are independent events, so P(two fives in a row) P(five) P(five) Your Turn 8 Let M represent the event you do your math homework and H represent the event you do your history assignment. PM ( ) 0.8 PH ( ) 0.7 PM ( È H) 0.9 Now solve for PM ( Ç H). PM ( È H) PM ( ) + PH ( ) - PM ( Ç H) PM ( Ç H) PM ( Ç H) PM ( Ç H) 0.6 However, PM ( ) PH ( ) (0.8)(0.7) 0.56, so PM ( Ç H) ¹ PM ( ) PH ( ). Since M and H do not satisfy the product rule for independent events, they are not independent. 7.5 Exercises. Let A be the event the number is and B be the event the number is odd. The problem seeks the conditional probability PAB ( ). Use the definition PA ( Ç B) PAB ( ). PB ( ) Here, PA ( Ç B) 0 and PB ( ). Thus, 0 PAB ( ) 0. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

39 Section P(4 even) P(4 Ç even) P(even) n(4 Ç even) n(even) 3 4 P (heart on nd heart on st) P (second is black first is a spade) 5, since 5 there are 5 black cards left out of 5 cards. Note that the sample space is reduced from 5 cards to 5cards after the first card is drawn. 3. Let A be the event the number is even and B be the event the number is 6. Then PA ( Ç B) PAB ( ) 6. PB ( ) 6 4. Let A be the event the number is odd and B be the event the number is 6. Since 6 is an even number, PAB ( ) P(sum of 8 greater than 7) P(8 Ç greater than 7) P(greater than 7) n(8 Ç greater than 7) n(greater than 7) Let A be the event sum of 6 and B be the event double. 6 of the 36 ordered pairs are doubles, so PB ( ) 6. There is only one outcome, 3-3, 36 6 in A Ç B (that is, a double with a sum of 6), so PA ( Ç B). Thus, PAB ( ) The event of getting a double given that 9 was rolled is impossible; hence, P (double sum of 9) n(double and sum is 8) P(double sum is 8) n(sum is 8) 5 9. Use a reduced sample space. After the first card drawn is a heart, there remain 5 cards, of which are hearts. Thus,. Use a reduced sample space. After the first card drawn is a jack, there remain 5 cards, of which are face cards. Thus, P (face card on nd jack on st). 5. P (second is an ace first is not an ace) 4, since 5 there are 4 aces left out of 5 cards. 3. P(a jack and a 0) P(jack followed by 0) + P(0 followed by jack) The probability of drawing an ace and then drawing a 4 is The probability of drawing a 4 and then drawing an ace is Since these are mutually exclusive, the probability of drawing an ace and a 4 is P(two black cards) P(black on st) P(black on nd black on st) Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

40 44 Chapter 7 SETS AND PROBABILITY 6. The probability that the first card is a heart is The probability that the second is a heart, given that the first is a heart, is 5. Thus, the probability that both are hearts is Examine a table of all possible outcomes of rolling a red die and rolling a green die (such as Figure 8 in Section 7.4). There are 9 outcomes of the 36 total outcomes that correspond to rolling red die comes up even and green die comes up even in other words, corresponding to A Ç B. Therefore, 9 PA ( Ç B) We also know that PA ( ) / and PB ( ) /. Since PA ( Ç B) PA ( ) PB ( ), 4 the events A and B are independent. 0. The knowledge that it rains more than 0 days affects the knowledge that it rains more than 5 days. (For instance, if it hasn t rained more than 0 days, it couldn t possibly have rained more than 5 days.) Since PDC ( ) ¹ PD ( ), the events are dependent.. Notice that PFE ( ) ¹ PF ( ): the knowledge that a person lives in Dallas affects the probability that the person lives in Dallas or Houston. Therefore, the events are dependent.. First not that PG ( ) /7 and PH ( ) /. A tree diagram of all the possible outcomes of the experiment would have 4 branches, exactly one of which would correspond to the outcome today is Tuesday and the coin comes up heads, or G Ç H. Therefore, PG ( Ç H) /4. But notice that PG ( Ç H) PG ( ) PH ( ). 4 7 Thus, the events are independent. 3. (a) The events that correspond to sum is 7 are (, 5), (3, 4), (4, 3), and (5, ), where the first number is the number on the first slip of paper and the second number is the number on the second. Of these, only (3, 4) corresponds to first is 3, so P (first is 3 sum is 7). 4 (b) The events that correspond to sum is 8 are (3, 5) and (5, 3). Of these, only (3, 5) corresponds to first is 3, so P (first is 3 sum is 8). 4. (a) Given that the first number is 5, there are four possible sums: 5 + 6, 5 + 7, , and One of the four possible outcomes corresponds to the sum 8. Therefore, P (sum is 8 first number is 5). 4 (b) Given that the first number is 4, there are four possible sums: 4 + 5, 4 + 6, , and None of these corresponds to the sum 8. Therefore, 0 P (sum is 8 first number is 4) (a) Many answers are possible; for example, let B be the event that the first die is a 5. Then PA ( Ç B) P(sum is 7 and first is 5) PA ( ) PB ( ) P(sum is 7) P(first is 5) so, PA ( Ç B) PA ( ) PB ( ). 36 (b) Many answers are possible; for example, let B be the event that at least one die is a 5. PA ( Ç B) P(sum is 7 and at least one is a 5) 36 PA ( ) PB ( ) P(sum is 7) P(at least one is a 5) so, PA ( Ç B) ¹ PA ( ) PB ( ). 9. Since A and B are independent events, PA ( Ç B) PA ( ) PB ( ) Thus, PA ( È B) PA ( ) + PB ( ) - PA ( Ç B) Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

41 Section (a) If A and B are mutually exclusive, then PB ( ) PA ( È B) - PA ( ) (b) If A and B are independent, then PA ( Ç B) PAB ( ) PA ( ) PB ( ) 0.5. Thus, PA ( Ç B) 0.5 PB ( ). Solving PA ( È B) PA ( ) + PB ( ) - PA ( Ç B) for PA ( Ç B) and substituting into the previous equations we get PA ( ) + PB ( ) - PA ( È B) 0.5 PB ( ) PB ( ) PB ( ) PB ( ) 0.4 PB ( ). PB ( ) At the first booth, there are three possibilities: shaker has heads and shaker has heads; shaker has tails and shaker has heads; shaker has heads and shaker has tails. We restrict ourselves to the condition that at least one head has appeared. These three possibilities are equally likely so the probability of two heads is 3. At the second booth we are given the condition of one head in one shaker. The probability that the second shaker has one head is. Therefore, you stand the best chance at the second booth. 3. (a) Let C be the event the coin comes up heads and D be the event 6 is rolled on the die. We have PC ( ), PD ( ), 6 PC ( Ç D). So the probability of winning of the event C Ç D is /. What is the probability that either the head or the 6 occurred in other words, what is PC ( È D)? Use the union for probability. PC ( È D) PC ( ) + PD ( ) - PC ( Ç D) We want to find the probability of head and 6 occurred given that head or 6 occurred that is, P(( C Ç D) ( C È D)). Use the formula for the conditional probability. P(( C Ç D) ( C È D)) P(( C Ç D) ( C È D)) PC ( È D) PC ( Ç D) PC ( È D) 7 7 The probability of winning the original game with the additional information is now /7. On the other hand, the probability of winning the new game, of the event 6 is rolled, is simply PD ( ) /6. Since /6 > /7, it is better to switch. 33. No, these events are not independent. 34. For a two-child family, the sample space is M - M M - F F - M F - F. P (same sex) 4 P (same sex at most one male) 3 The events are not independent. For a three-child family, the sample space is M - M - M M - M - F M - F - M M - F - F F - M - M F - M - F F - F - M F - F - F. P (same sex) 8 4 P (same sex at most one male) 4 The events are independent. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

42 46 Chapter 7 SETS AND PROBABILITY 35. Assume that each box is equally likely to be drawn from and that within each box each marble is equally likely to be drawn. If Laura does not redistribute the marbles, then the probability of winning the Porsche is, since the event of a pink marble being drawn is equivalent to the event of choosing the first of the two boxes. If however, Laura puts 49 of the pink marbles into the second box with the 50 blue marbles, the probability of a pink marble being drawn increases to 74. The probability of the first box 99 being chosen is, and the probability of drawing a pink marble from this box is. The probability of the second box being chosen is, and the probability of drawing a pink marble from this box is 49. Thus, the probability of drawing a pink 99 marble is Therefore Laura increases her chances of winning by redistributing some marbles. 36. PC ( D) is the probability that a customer cashes a check, given that the customer made a deposit. PC ( Ç D) PCD ( ) PD ( ) nc ( Ç D) nd ( ) The probability that a customer cashing a check will fail to make a deposit is nd ( Ç C) 30 PD ( C). nc ( ) PC ( D ) is the probability that a customer does not cash a check, given that the customer did not make a deposit. PC ( Ç D ) PC ( D ) PD ( ) nc ( Ç D ) nd ( ) The probability that a customer making a deposit will not cash a check is nc ( Ç D) 0 PC ( D). nd ( ) P[( C Ç D ) ] is the probability that a customer does not both cash a check and make a deposit. P[( C Ç D) ] - P( C Ç D) (a) Since the separate flights are independent, the probability of 4 flights in a row is (0.773)(0.773)(0.773)(0.773)» Let M represent the event the main computer fails and B the event the backup computer fails. Since these events are assumed to be independent, PM ( Ç B) PM ( ) PB ( ) 0.003(0.005) This is the probability that both computers fail, which means that the company will not have computer service. The fraction of the time it will have service is Independence is a fairly realistic assumption. Situations such as floods or electric surges might cause both computers to fail at the same time. 43. Let W be the event withdraw cash from ATM and C be the event check account balance at ATM. PC ( È W) PC ( ) + PW ( ) - PC ( ÇW) PC ( ÇW) PC ( ÇW) PC ( Ç W) 0.8 PC ( Ç W) PW ( C) PC ( ) » The probability that she uses an ATM to get cash given that she checked her account balance is Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

43 Section Use the following tree diagram for Exercises 44 through 46. Use the following tree diagram for Exercises 49 through Since 60% of production comes off assembly line A, PA ( ) Also P(pass inspection A ) 95, so P(not pass A ) Therefore, P( A Ç not pass) P( A) P(not pass A) 0.60(0.05) Since 40% of the production comes off B, PB ( ) Also, P(pass B ) 0.85, so P(not pass B ) 0.5. Therefore, P(not pass Ç B) P( B) P(not pass B) 0.40(0.5) P(bike did not pass inspection) 0.60(0.05) (0.5) The sample space is { RW, WR, RR, WW }. The event red is {RW, WR, RR}, and the event mixed is {RW, WR}. n(mixed and red) P(mixed red) n(red) By the product rule for independent events, two events are independent if the product of their probabilities is the probability of their intersection. 0.75(0.4) 0.3 Therefore, the given events are independent. 49. P(all girls first is a girl) P(all girls and first is a girl) P(first is a girl) n(all girls and first is a girl) n(first is a girl) P(3 girls 3rd is a girl) P(3 girls and 3rd is a girl) P(3rd is a girl) P(all girls second is a girl) P(all girls and second is a girl) P(second is a girl) n(all girls and second is a girl) n(second is a girl) 4 5. P(3 girls at least girls) P(3 girls and at least girls) P(at least girls) P(3 girls) P(at least girls) P(3 girls) P( girls) + P(3 girls) Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley. 4

44 48 Chapter 7 SETS AND PROBABILITY (Note that and P(3 girls) P( GGG) 8 P( girls) P( GGB) + P( BGG) + P( GBG) ö. 8 ø 53. P(all girls at least girl) P(all girls and at least girl) P(at least girl) n(all girls and at least girl) n(at least girl) PM ( ) 0.487, the total of the M column. 6. By the definition of independent events, C and M are independent if PCM ( ) PC ( ). From Exercises 55 and 59, and PC ( ) PCM ( ) Since PCM ( ) ¹ PC ( ), events C and M are not independent, so we say that they are dependent. This means that red-green color blindness does not occur equally among men and women. 6. (a) From the table, PC ( Ç D) and PC ( ) PD ( ) (0.000) Since PC ( Ç D) PC ( ) PD ( ), C and D are independent events; color blindness and deafness are independent events. 63. First draw a tree diagram. 55. PC ( ) 0.039, the total of the C row overweight 56. PM ( Ç C) 0.035, the entry in the M column and C row man 0.37 not overweight 57. PM ( È C) PM ( ) + PC ( ) - PM ( Ç C) woman 0.64 overweight PM ( Ç C) PM ( C) PC ( ) » PC ( Ç M) PCM ( ) PM ( ) » not overweight (a) P (overweight man) (0.493)(0.683) (b) P (overweight) (0.507)(0.64) (c) The two events man and overweight are independent if P(overweight man) P(overweight) P(man). P(overweight) P(man) (0.667)(0.493) PM ( Ç C) PM ( C) PC ( ) » Since ¹ 0.36, the events are not independent 9 43 PA ( ) Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

45 Section nc ( Ç F) 7 PC ( F) nf ( ) 9 PAH ( ) PB ( H ) P(( AÈ CÈ D) F) n(( A È C È D) Ç F) nf ( ) No, PA ( ) ¹ PAH ( ). 69. Let H be the event patient has high blood pressure, N be the event patient has normal blood pressure, L be the event patient has low blood pressure, R be the event patient has a regular heartbeat, and I be the event patient has an irregular heartbeat. We wish to determine PR ( Ç L). Statement (i) tells us PH ( ) 0.4 and statement (ii) tells us PL ( ) 0.. Therefore, PH ( ) + PN ( ) + PL ( ) PN ( ) + 0. PN ( ) Statement (iii) tells us PI ( ) 0.5. This and statement (iv) lead to PI ( Ç H) PI ( ) (0.5) Statement (v) tells us PN ( Ç I) PN ( ) (0.64) Make a table and fill in the data just found. H N L Totals R I Totals To determine PR ( Ç L), we need to find only PI ( Ç L). PI () PI ( Ç H) + PI ( Ç N) + PI ( Ç L) PI ( Ç L) PI ( Ç L) PI ( Ç L) 0.0 Now calculate PR ( Ç L). PL ( ) PR ( Ç L) + PI ( Ç L) 0. PR ( Ç L) PI ( Ç L) 0.0 The correct answer choice is e. 70. (a) P (letter is in drawer ) 0. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley. (b) (c) (d) P(letter is in drawer letter is not in drawer ) 0.» P(letter is in drawer 3 letter is not in drawer or ) P(letter is in drawers 8 letter is not in drawers through 7) 0.» (e) The probability is increasing. (f) P (letter is in some drawer) 0.8 (g) P(letter is in some drawer not in drawer ) (h) (i) 0.7» P(letter is in some drawer not in drawer or ) P(letter is in some drawer not in drawer through 7) 0.» (j) The probability is decreasing. 7. (a) The total number of males is (5844) ,030. The total number of infants is (7,798) 35, 596. So among infants who are part of a twin pair, the proportion of males is 8, ,596

46 430 Chapter 7 SETS AND PROBABILITY p p To answer parts (b) through (g) we make the assumption that the event twin comes from an identical pair is independent of the event twin is male. We can then construct the following tree diagram. identical not identical P(B) P(B) P(B) P(B)( P(B)) both male both female both male male and female (f) Now only the not identical branch is involved, and the expression is ( - p) PB ( )(- PB ( )). (g) The equation to solve will now be the following: 634 7, 798 ( - p)( )( ) Again the answer will be the same as in part (c). Note that because of our independence assumption these three estimates for p must agree. 7. Draw the tree diagram. ( P(B)) both female (b) The event that the pair of twins is male happens along two compound branches. Multiplying the probabilities along these branches and adding the products we get pp( B) + (- p)( P( B)). (c) Using the values from (a) and (b) together with the fraction of mixed twins, we solve this equation: 5844 p( ) + ( - p)( ) 7, p p p p p 0.87 So our estimate for p is Note that if you use fewer places in the value for P(B) you may get a slightly different answer. (d) Multiplying along the two branches that result in two female twins and adding the products gives p( - P( B)) + ( - p)( - P( B)). (e) The equation to solve will now be the following: 56 7, 798 p( ) + ( - p)( ) The answer will be the same as in part (c) 73. (a) (a) P(married) P( job and married) + P(no job and married) 0.5(0.64) (0.86) (b) P ( job and single) 0.5(0.36) 0.87 (b) (c) (d) (e) The probability that a person is a current smoker is different from the probability that the a person is a current smoker given that that the person has less than a high school diploma. Thus knowing that a person has less than a high school diploma changes our estimate of the probability that the person is a smoker, so these events are not independent. Alternatively, we could note that the product of the answers to (a) and (b), which is P(smoker) P(less than HS diploma) Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

47 Section is equal to (0.059)(0.489) , while according to (c), P(smoker and less than HS diploma) Since these values are different, the two events current smoker and less than a high school diploma are not independent. 74. (a) P(forecast of rain rain) » 83% 80 P(forecast of no rain no rain) 764» 0.83» 83% 90 (b) P (rain forecast of rain) 66» (c) P(no rain forecast of no rain)» (a) In this exercise, it is easier to work with complementary events. Let E be the event at least one of the faults erupts. Then the complementary event E is none of the faults erupts, and we can use PE ( ) - PE ( ). Consider the event E : none of the faults erupts. This means the first fault does not erupt and the second fault does not erupt and... and the seventh fault does not erupt. Letting F i denote the event the i th fault erupts, we wish to find PE ( ) PF ( Ç F Ç F Ç F Ç F Ç F Ç F ) Since we are assuming the events are independent, we have PE ( ) PF ( Ç F Ç F3 Ç F4 Ç F5 Ç F6 Ç F7 ) PF ( ) PF ( ) PF ( 3 ) PF ( 4 ) PF ( 5 ) PF ( 6 ) PF ( 7 ) Now use PF ( i ) - PF ( i) and perform the calculations. PE ( ) PF ( ) PF ( ) PF ( 3 ) PF ( 4 ) PF ( 5 ) PF ( 6 ) PF ( 7 ) ( - 0.7) ( - 0.)... ( ) (0.73)(0.79)(0.89)(0.90)(0.96)(0.97)(0.97)» 0.4 Therefore, PE ( ) - PE ( )» - 0.4» (a) (b) (c) (d) (e) (f) (g) P(at least one component works) - P(no component works) - P(all n components fail) - (0.03) n If this probability is to be at least , then n - (0.03) ³ n - (0.03) ³ n (0.03) , and the smallest whole number value of n for which this inequality holds true is n 4. Therefore, 4-3 backup components must be used, or 4 total components. 357 P (second class)» P (surviving)» P (surviving first class)» P (surviving child and third class) 79» P (woman first class and survived) 03» P (third class man and survived) 46» P(survived man)» P(survived man and third class) 46» 0.63 No, the events are not independent. 78. The events are probably dependent, and the agent used the product rule for independent events. 79. First draw the tree diagram. 76. P (component fails) 0.03 (a) Let n represent the number of these components to be connected in parallel; that is, the component has n - backup components. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

48 43 Chapter 7 SETS AND PROBABILITY (a) P(person is 65 or over and has a loan) P(65 or over) P( has loan 65 or over) 0.0(0.6) 0.05 (b) P(person has a loan) P(65 or over and has loan) + P(under 65 and has loan) 0.0(0.6) (0.53) (a) drinks diet soft drinks is æ4ö æö ç è5 ø 3èçø ; 30 0 (b) diets, but does not drink diet soft drinks is æö ç. 3çè5 ø First draw the tree diagram. 8. From the tree diagram, we see that the probability that an individual from the community is (a) a woman jogger is 0.5(0.40) 0.060; (b) a man who is not a jogger is (0.85)(0.45) (c) a woman is 0.5(0.40) (0.55) (d) The two events woman and jogger are independent if P(woman jogger) P(woman) P( jogger). We are given that P ( jogger) 0.5 so that P(woman) P( jogger) » but we found P (woman jogger) in part (a). Therefore, the events are not independent. (a) (b) (c) P(fails both st and nd tests) P(fails st) P(fails nd fails st) 0.5(0.0) 0.05 P(fais three times in a row) 0.5(0.0)(0.30) 0.05 P(requires at least tries) P(does not pass on st try) Let F i be the event the ith burner fails. The event all four burners fail is equivalent to the event the first burner fails and the second burner fails and the third burner fails and the fourth burner fails that is, the event F Ç F Ç F 3 Ç F 4. We are told that the burners are independent. Therefore PF ( Ç F Ç F3 Ç F4) PF ( ) PF ( ) PF ( 3) PF ( 4) (0.00)(0.00)(0.00)(0.00) First draw a tree diagram. From the tree diagram, we see that the probability that a person Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

49 Section (a) (b) æö P(no ticket) ç + çè3 ø 3 P(no ticket on three consecutive weekends) P(0) (0.4)(0.4) 0.6 P() (0.4)(0.6) 0.48 P() (0.6)(0.6) (a) 85. P(luxury car) 0.04 and P(luxury car CPA) 0.7 Use the formal definition of independent events. Since these probabilities are not equal, the events are not independent. 86. Let A be the event student studies and B be the event student gets a good grade. We are told that PA ( ) 0.6, PB ( ) 0.7, and PA ( Ç B) 0.5. PA ( ) PB ( ) 0.6(0.7) 0.4 (a) Since PA ( ) PB ( ) is not equal to PA ( Ç B), A and B are not independent. Rather, they are dependent events. (b) Let A be the event a student studies and B be the event the student gets a good grade. Then PB ( Ç A) 0.5 PB ( A)» PA ( ) 0.6 (c) Let A be the event a student gets a good grade and B be the event the student studied. Then (b) From the tree diagram, P(win) kr P(tie) ( - k) r P(lose) k( - r) + ( - k)( - r) k - kr + - r - k + kr - r PB ( Ç A) 0.5 PAB ( )» PB ( ) (a) We will assume that successive free throws are independent. P(0) 0.4 P() (0.6)(0.4) 0.4 P() (0.6)(0.6) 0.36 (b) Let p be her season free throw percentage. Solve: p P(0) - p P() p p - p + p p» 0.68 From the tree diagram, P(win) rk P(tie) r( - k) + ( - r) r r - rk + r - r r - rk - r r( - k - r) P(lose) ( - r)( - r) ( - r) (c) P(win) is the same under both strategies. (d) If r <, ( - r) > ( - r). The probability of losing is smaller for the -point first strategy. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

50 434 Chapter 7 SETS AND PROBABILITY 7.6 Bayes Theorem Your Turn Let E be the event passes math exam and F be the event attended review session. PEF ( ) 0.8 PEF ( ) 0.65 PF ( ) 0.6 so PF ( ) 0.4 Now apply Bayes Theorem. PFPEF ( ) ( ) PFE ( ) PFPEF ( ) ( ) + PF ( ) PEF' ( ) (0.6)(0.8) (0.6)(0.8) + (0.4)(0.65) So the probability that the student attended the review session given that the student passed the exam is Your Turn Let F F F3 E in English I in English II in English III received help from writing center Express the given information in terms of these variables. P( F) 0. P( E F) 0.80 P( F) 0.68 P( E F) 0.40 P( F3) 0.0 P( E F3) 0. PF ( ) PEF ( ) PF ( E) PF ( ) PEF ( ) + PF ( ) PEF ( ) + PF ( 3) PEF ( 3) (0.)(0.80) (0.)(0.80) + (0.68)(0.40) + (0.0)(0.) 0.46 Given that a student received help from the writing center, the probability that the student is in English I is Exercises. Use Bayes theorem with two possibilities M and M. PM ( ) PNM ( ) PM ( N) PM ( ) PNM ( ) + PM' ( ) PNM' ( ) 0.4(0.3) 0.4(0.3) + 0.6(0.4) Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

51 Section By Bayes theorem, PM ( N) - PM ( N) PM ( ) PNM ( ) - PM ( ) PNM ( ) + PM ( ) PNM ( ) 0.4(0.3) - 0.4(0.3) + 0.6(0.4) Using Bayes theorem, PR ( ) PQR ( ) PR ( Q) PR ( ) PQR ( ) + PR ( ) PQR ( ) + PR ( 3) PQR ( 3) 0.5(0.40) (0.5)(0.40) (0.0) (0.70) Using Bayes theorem, PR ( ) PQR ( ) PR ( Q) PR ( ) PQR ( ) + PR ( ) PQR ( ) + PR ( 3) PQR ( 3) 0.55(0.0) (0.5)(0.40) (0.0) (0.70) Using Bayes theorem, PR ( 3) PQR ( 3) PR ( 3 Q) PR ( ) PQR ( ) + PR ( ) PQR ( ) + PR ( 3) PQR ( 3) 0.30(0.70) (0.5)(0.40) (0.0) (0.70) This is the complement of the event in Exercise 3, so PR ( Q) - PR ( Q) We first draw the tree diagram and determine the probabilities as indicated below. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

52 436 Chapter 7 SETS AND PROBABILITY We want to determine the probability that if a white ball is drawn, it came from the second jar. This is PJ ( W ). Use Bayes theorem. PJ ( ) PW ( J) PJ ( 3 3 W) PJ ( ) PW ( J) + PJ ( ) PW ( J) + PJ ( 3) PW ( J3) PJ3 PW J W PJ PW J + PJ PW J + PJ3 PW J ( ) ( ) 3 PJ ( ) ( ) ( ) ( ) ( ) ( ) ( ) 7 9. Let G represent good worker, B represent bad worker, S represent pass the test, and F represent fail the test. The given information if PG ( ) 0.70, PB ( ) PG ( ) 0.30, PSG ( ) 0.85 (and therefore PFG ( ) 0.5), and PSB ( ) 0.35 (and therefore PFB ( ) 0.65). If passing the test is made a requirement for employment, then the percent of the new hires that will turn out to be good workers is PGS ( ) 85% of new hires become good workers. PG ( ) PSG ( ) PG ( ) PSG ( ) + PB ( ) PS ( B) 0.70(0.85) 0.70(0.85) (0.35) Let Q be the event person is qualified and A be the event person was approved. Set up a tree diagram. Using Bayes theorem, PQ ( ) PAQ ( ) PQ ( A) PQ ( ) PAQ ( ) + PQ ( ) PAQ ( ) 0.75(0.85) 0.75(0.85) + 0.5(0.0) » Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

53 Section Let Q represent qualified and A represent approved by the manager. Set up the tree diagram. PQ A ( ) PQ ( ) PAQ ( ) PQ ( ) PAQ ( ) + PQ ( ) PAQ ( ) 0.5(0.0) 0.75(0.85) + 0.5(0.0) » Let A be the event the bag came from supplier A, B be the event the bag came from supplier B, and D be the event the bag is damaged. Set up a tree diagram. PA ( ) PDA ( ) PAD ( ) PA ( ) PDA ( ) + PB ( ) PDB ( ) 0.70(0.0) 0.70(0.0) (0.05) 0.070» Let D represent damaged, A represent from supplier A, and B represent from supplier B. Set up the tree diagram. PB ( ) PDB ( ) PBD ( ) PB ( ) PDB ( ) + PA ( ) PDA ( ) 0.30(0.05) 0.30(0.05) (0.0) » Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

54 438 Chapter 7 SETS AND PROBABILITY 4. Let D be the event a defective appliance and A be the event appliance manufactured by company A. Start with a tree diagram, where the first stage refers to the companies and the second to defective appliances. From Bayes theorem, we have PA ( ) PDA ( ) PAD ( ) PA ( ) PDA ( ) + PB ( ) PDB ( ) + PC ( ) PDC ( ) » Start with the tree diagram, where the first state refers to the companies and the second to a defective appliance. PB ( ) PDB ( ) PBD ( ) PA ( ) PDA ( ) + PB ( ) PDB ( ) + PC ( ) PDC ( ) 0.40(0.05) 0.5(0.0) (0.05) (0.0) » Let H be the event high rating, F be the event sponsors college game, F be the event sponsors baseball game, and F 3 be the event sponsors pro football game. First set up a tree diagram. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

55 Section Now use Bayes theorem. PF ( H) PF ( ) PH ( F) PF ( ) PH ( F) + PF ( ) PH ( F) + PF ( ) PH ( F) (0.70) (0.70) + 0.(0.50) + 0.3(0.60) Let H represent high rating, F l represent sponsors college game, F represent sponsors baseball game, and F 3 represent sponsors pro football game. PF ( H) 3 PF ( 3) PH ( F3) PF ( ) PH ( F) + PF ( ) PH ( F) + PF ( ) PH ( F) (0.60) 0.5(0.70) + 0.(0.50) + 0.3(0.60) Let A represent driver is 6 0, B represent driver is 30, C represent driver is 3 65, D represent driver is 66 99, and E represent driver has an accident. We wish to find PAE ( ). The correct choice is b. PA ( ) PEA ( ) PAE ( ) PA ( ) PEA ( ) + PB ( ) PEB ( ) + PC ( ) PEC ( ) + PD ( ) PED ( ) 0.08(0.06) 0.08(0.06) + 0.5(0.03) (0.0) + 0.8(0.04) » Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

56 440 Chapter 7 SETS AND PROBABILITY 9. Using the given information, construct a table similar to the one in the previous exercise. Category of Policyholder Portion of Policyholders Probability of Dying in the Next Year Standard Preferred Ultra-preferred Let S represent standard policyholder, R represent preferred policyholder, U represent ultra-preferred policyholder, and D represent policyholder dies in the next year. We wish to find PU ( D ). The correct answer choice is d. PU ( ) P( DU ) PU ( D) PS ( ) PDS ( ) + PR ( ) PDR ( ) + PU ( ) PDU ( ) 0.0(0.00) 0.50(0.00) (0.005) + 0.0(0.00) 0.000» Let T represent driver is a teen, Y represent driver is a young adult, M represent driver is in midlife, S represent driver is a senior, and C represent driver has been involved in at least one collision in the past year. We wish to find PY ( C ). The correct choice is d. PY ( ) PCY ( ) PY ( C) PT ( ) PCT ( ) + PY ( ) PCY ( ) + PM ( ) PCM ( ) + PS ( ) PCS ( ) 0.6(0.08) 0.08(0.5) + 0.6(0.08) (0.04) + 0.3(0.05) 0.08» Let L be the event the object was shipped by land, A be the event the object was shipped by air, S be the event the object was shipped by sea, and E be the event an error occurred. The correct response is c. PL ( ) PEL ( ) PLE ( ) PL ( ) PEL ( ) + PA ( ) PEA ( ) + PS ( ) PES ( ) 0.50(0.0) 0.50(0.0) (0.04) + 0.0(0.4) Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

57 Section There are a total of mortgages being studied at the bank. (a) (b) ( 800 ) 60 ( ) ( ) ( ) + ( ) P (5% down default) (0.45) 0.05(0.45) (0.5) + 0.0(0.) + 0.0(0.) » A mortgage being paid to maturity is the complement of a mortgage being defaulted. P(0% down paid to maturity ) P(0% down not default) ( 800 ) 60 ( ) ( ) ( ) + ( ) (0.5) 0.95(0.45) (0.5) (0.) (0.) » 3. Let E represent the event hemoccult test is positive, and let F represent the event has colorectal cancer. We are given PF ( ) 0.003, PEF ( ) 0.5, and PEF ( ) 0.03 and we want to find PFE ( ). Since PF ( ) 0.003, PF ( ) Therefore, PF ( ) PEF ( ) PFE ( )» PF ( ) PEF ( ) + PF ( ) PEF ( ) Let H be the event has hepatitis and T be the event positive test. First set up a tree diagram. Using Bayes theorem, PH ( ) PT ( H) 0.8(0.90) PH ( T)» PH ( ) PT ( H) + PH ( ) PT ( H ) 0.8(0.90) + 0.(0.05) Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

58 44 Chapter 7 SETS AND PROBABILITY PT ( D ) PT ( D ) PD ( ) so PD ( ) We can now fill in the complete tree. D T T D T + T (a) (b) (c) + + PD ( T ) PD ( ) PT ( D ) PD ( ) PT ( D ) + PD ( ) PT ( D ) (0.005)(0.796) (0.005)(0.796) + (0.995)(0.098) PD ( ) PT ( D ) PD ( T ) PD ( ) PT ( D ) + PD ( ) PT ( D ) (0.995)(0.90) (0.995)(0.90) + (0.005)(0.04) PD ( ) PT ( D ) PD ( T ) PD ( ) PT ( D ) + PD ( ) PT ( D ) (0.005)(0.04) (0.005)(0.04) + (0.995)(0.90) Alternatively, since PD ( T ) + PD ( T ), we can subtract the answer to (b) from. (d) The right half of the tree stays the same, but PD ( + ) is now 0.05 and PD ( - ) is PD ( T ) PD ( ) PT ( D ) PD ( ) PT ( D ) + PD ( ) PT ( D ) (0.05)(0.796) (0.05)(0.796) + (0.985)(0.098) 0.0 Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

59 Section PT ( D ) PT ( D ) PD ( ) 0.05 so PD ( ) 0.95 We can now fill in the complete tree. D T T 0.95 D T + T (a) PD ( ) PT ( D ) PD ( T ) PD ( ) PT ( D ) + PD ( ) PT ( D ) (0.05)(0.999) (0.05)(0.999) + (0.95)(0.00) (b) The calculation is the same except with different values for PD ( + ) and PD ( - ). Now PD ( + ) 0.00 and PD ( - ) PD ( T ) PD ( ) PT ( D ) PD ( ) PT ( D ) + PD ( ) PT ( D ) (0.00)(0.999) (0.00)(0.999) + (0.999)(0.00) (a) Let H represent heavy smoker, L be light smoker, N be nonsmoker, and D be person died. Let x P( D N), that is, let x be the probability that a nonsmoker died. Then PDL ( ) x and PDH ( ) 4 x. Create a table. We wish to find PH ( D ). Level of Smoking Probability of Level Probability of Death for Level H 0. 4x L 0.3 x N 0.5 x Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

60 444 Chapter 7 SETS AND PROBABILITY The correct answer choice is d. PH ( ) PDH ( ) PH ( D) PH ( ) PDH ( ) + PL ( ) PDL ( ) + PN ( ) PDN ( ) 0.(4 x) 0.(4 x) + 0.3( x) + 0.5( x) 0.8x.9x» Let C represent patient was critical, S be patient was serious. T be patient was stable, and V be patient survived. We want to find PSV ( ). The problem statement gives probabilities that a patient died in the emergency room, but we are interested in finding the probability that a patient survived. So convert the given probabilities to probabilities that a patient survived by finding the probability of the complementary event, using the result and create a table as before. The correct answer choice is b. P(category of patient patient survived) - P(category of patient patient died) Category of Patient Portion of Category Probability of Survival for Category C S T PS ( ) PV ( S) PSV ( ) PC ( ) PV ( C) + PS ( ) PV ( S) + PT ( ) PV ( T) 0.3(0.90) 0.(0.60) + 0.3(0.90) + 0.6(0.99) » Let H represent person has the disease and R be test indicates presence of the disease. We wish to determine PH ( R ). Construct a table as before. Category of Person Probability of Population Probability of Presence of Disease H H The correct answer choice is b. PHR ( ) PH ( ) PRH ( ) PH ( ) PRH ( ) + PH ( ) PRH ( ) 0.0(0.950) 0.0(0.950) (0.005) » Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

61 Section Let H represent male has a circulation problem and S be male is a smoker. We wish to determine PH ( S ). Let x be the probability that the male is a smoker and construct a table as before. Category of Male Portion of Population Probability of Being a Smoker H 0.5 x H 0.75 x PH ( ) PSH ( ) PHS ( ) PH ( ) PSH ( ) + PH ( ) PSH ( ) 0.5( x) 0.5( x) ( x) 0.50x».5x 5 The correct answer choice is c. 3. We start with a tree diagram based on the given information. Let W stand for woman and A stand for abstains from alcohol A W A (a) PA ( ) PAWPW ( ) ( ) + PAW ( ) PW ( ) (0.047)(0.07) + (0.953)(0.86) W A A (b) - PW ( A) - - PW ( ) P( A W ) - - PW ( ) P( A W ) + PW ( ) P( A W) (0.953)(0.86) (0.953)(0.86) + (0.047)(0.07) Let M be the event wife was murdered and G be the event husband is guilty. Set up a tree diagram. PG ( ) PM ( G) 0.00() PGM ( )» PG ( ) PM ( G) + PG ( ) PM ( G ) 0.00() (0.00) Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

62 446 Chapter 7 SETS AND PROBABILITY 33. P(between 35 and 44 never married) (for a randomly selected man) (0.86)(0.04) (0.86)(0.04) + (0.3)(0.90) + (0.86)(0.488) + (0.348)(0.8) + (0.48)(0.044) To find the proportion of women who have been married in each age category, we subtract the numbers in the second column of the table of data for women from, giving the values 0.75, 0.634, 0.853, 0.908, and P(between 8 and 4 has been married) (for a randomly selected woman) (0.)(0.75) (0.)(0.75) + (0.7)(0.634) + (0.78)(0.853) + (0.345)(0.908) + (0.84)(0.960) P (between 45 and 64 never married) (for a randomly selected woman) (0.345)(0.09) (0.345)(0.09) + (0.)(0.85) + (0.7)(0.366) + (0.78)(0.47) + (0.84)(0.040) Let S + and S - stand for wore seat belt and did not wear seat belt and let E + and E - stand for was ejected and was not ejected. We start by constructing the tree corresponding to the given data. S E E 0.36 S 0.36 E + (a) (b) + + PS ( E ) - - PS ( E ) 0.64 E PS ( ) PE ( S ) PS ( ) PE ( S ) + PS ( ) PE ( S ) (0.638)(0.0) (0.638)(0.0) + (0.36)(0.36) PS ( ) PE ( S ) PS ( ) PE ( S ) + PS ( ) PE ( S ) (0.36)(0.64) (0.36)(0.64) + (0.638)(0.98) Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

63 Section In Exercises 37 and 38, let S stand for smokes and S - for does not smoke. 37. P(8-44 S) P(8-44) P( S 8-44) P(8-44) PS ( 8-44) + P(45-64) PS ( 45-65) + P(65-74) PS ( 65-74) + P( > 75) PS ( > 75) (0.49)(0.3) (0.49)(0.3) + (0.34)(0.) + (0.09)(0.) + (0.08)(0.06) P(45-64 S - ) - P(45-64) P( S 45-64) P(45-64) PS ( 45-64) + P(8-44) PS ( 8-44) + P(65-74) PS ( 65-74) + P( > 75) PS ( > 75) (0.34)( - 0.) (0.34)( - 0.) + (0.49)( - 0.3) + (0.09)( - 0.) + (0.08)( ) Category Has terrorist ties Does not have terrorists ties Proportion of Population Probability of Being Picked Up,000, ,999,000, (0.99),000,000 P(Has terrorist ties Picked up) 999,999 (0.99) + (0.0),000,000,000,000 (0.99),000,000,000, ,999 (0.99) + (0.0),000,000,000,000,000, , » Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

64 448 Chapter 7 SETS AND PROBABILITY Chapter 7 Review Exercises. True. True 3. False: The union of a set with itself has the same number of elements as the set. 4. False: The intersection of a set with itself has the same number of elements as the set. 5. False: If the sets share elements, this procedure gives the wrong answer. 6. True 7. False: This procedure is correct only if the two events are mutually exclusive. 8. False: We can calculate this probability by assuming a sample space in which each card in the 5-card deck is equally likely to be drawn. 9. False: If two events A and B are mutually exclusive, then PA ( Ç B) 0 and this will not be equal to P(A)P(B) if P(A) and P(B) are greater than True. False: In general these two probabilities are different. For example, for a draw from a 5-card deck, P (heart queen) /4 and P (queen heart) /3.. True 3. 9 Î {8, 4, -3,- 9, 6} Since 9 is not an element of the set, this statement is false Î/ {3, 9, 7} Since 4 is not an element of the given set, the statement is true. 5. Î/ {0,,,3, 4} Since is an element of the set, this statement is false Î {0,,,3, 4} Since 0 is an element of the given set, the statement is true. 7. {3, 4,5} Í {,3, 4,5, 6} Every element of {3, 4, 5} is an element of {, 3, 4, 5, 6}, so this statement is true. 8. {,, 5, 8} Í {,, 5,0,} Since 8 is an element of the first set but not of the second set, the first set cannot be a subset of the second. The statement is false. 9. {3,6,9,0} Í {3,9,,3} 0 is an element of {3, 6, 9, 0}, but 0 is not an element of {3, 9,, 3}. Therefore, {3, 6, 9, 0} is not a subset of {3, 9,, 3}. The statement is false Í {} The empty set is a subset of every set, so the statement is true.. {,8} Í/ {, 4, 6,8} Since both and 8 are elements of {, 4, 6, 8}, {, 8} is a subset of {, 4, 6, 8}. This statement is false.. 0 Í 0 The empty set contains no elements and has no subsets except itself. Therefore, the statement is false. In Exercises 3 3 U {a,b,c,d,e,f,g,h}, K {c,d,e,f,h}, and R {a,c,d,g}. 3. K has 5 elements, so it has 4. nr ( ) 4, so R has 5 3 subsets. 4 6 subsets. 5. K (the complement of K ) is the set of all elements of U that do not belong to K. K {a, b, g} Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

65 Section 7.R R {b,e,f, h} since these elements are in U but not in R. 7. K Ç R (the intersection of K and R) is the set of all elements belonging to both set K and set R. K Ç R {c, d} 8. K È R {a,c,d,e,f,g,h} since these elements are in K or R, or both. A A B' 40. A Ç B contains all elements in B and not in A. B U 9. ( K Ç R ) {a,b,e,f,g,h} since these elements are in U but not in K Ç R. (See Exercise 7.) 30. ( K È R ) {b} since this element is in U but not in K È R. (See Exercise 8.) 3. / 0 U 3. U / 0, which is always true. 4. ( A Ç B) È C First find A Ç B. 33. A Ç C is the set of all female employees in the K.O. Brown Company who are in the accounting department. 34. B Ç D is the set of all sales employees who have MBA degrees. 35. A È D is the set of all employees in the K.O. Brown Company who are in the accounting department or have MBA degrees or both. A Ç B Now find the union of this region with C. U 36. A Ç D is the set of all employees with MBA degrees who are not in the accounting department. A B 37. B Ç C is the set of all male employees who are not in the sales department. 38. ( B È C ) is the set of all employees who are not either in the sales department or female, that is, all male employees not in the sales department. C (A B) C 4. ( A È B) Ç C includes those elements in C and not in either A or B. 39. A È B is the set of all elements which belong to A or do not belong to B, or both. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.