d. Event D = { RRL, RRS, RLR, RSR, LRR, SRR, LLR, LLS, LRL, LSL, RLL, SLL, SSR, SSL, SRS, SLS, RSS, LSS }

Transcription

1 Section 1 a Event A { RRR, LLL, SSS } b Event B { RLS, RSL, LRS, LSR, SRL, SLR } c Event C { RRL, RRS, RLR, RSR, LRR, SRR } d Event D { RRL, RRS, RLR, RSR, LRR, SRR, LLR, LLS, LRL, LSL, RLL, SLL, SSR, SSL, SRS, SLS, RSS, LSS } e Event D contains outcomes where all cars go the same direction, or they all go different directions: D { RRR, LLL, SSS, RLS, RSL, LRS, LSR, SRL, SLR } Because Event D totally encloses Event C, the compound event C D D: C D { RRL, RRS, RLR, RSR, LRR, SRR, LLR, LLS, LRL, LSL, RLL, SLL, SSR, SSL, SRS, SLS, RSS, LSS } Using similar reasoning, we see that the compound event C D C: C D { RRL, RRS, RLR, RSR, LRR, SRR } 9 In the diagram on the left, the shaded area is (A B) On the right, the shaded area is A, the striped area is B, and the intersection A B occurs where there is BOTH shading and stripes These two diagrams display the same area In the diagram below, the shaded area represents (A B) Using the diagram on the right above, the union of A and B is represented by the areas that have either shading or stripes or both Both of the diagrams display the same area

2 Section 11 f 7 g h Let event A selected customer owns stocks Then the probability that a selected customer does not own a stock can be represented by A ) 1 - A) 1 (18 + ) 1-7 This could also have been done easily by adding the probabilities of the funds that are not stocks 1 i A B) + - j (A B) ) 1 - A B) 1 - k A B ; A B ) A) A B) - 1 l awarded either 1 or (or both): A 1 A ) A 1 ) + A ) - A 1 A ) m awarded neither 1 or : A 1 A ) P[(A 1 A ) ] 1 - A 1 A ) 1 - n awarded at least one of 1,, : A 1 A A ) A 1 ) + A ) + A ) - A 1 A ) - A 1 A ) A A ) + A 1 A A ) o awarded none of the three projects: A 1 A A ) 1 awarded at least one) 1-7 p awarded but neither 1 nor : A 1 A A ) A ) - A 1 A ) A A ) + A 1 A A ) q either (neither 1 nor ) or : P[( A 1 A ) A ] shaded region) awarded none) + A )

3 Alternatively, answers to a f can be obtained from probabilities on the accompanying Venn diagram 1 r Let event E be the event that at most one purchases an electric dryer Then E is the event that at least two purchase electric dryers E ) 1 E) s Let event A be the event that all five purchase gas Let event B be the event that all five purchase electric All other possible outcomes are those in which at least one of each type is purchased Thus, the desired probability 1 A) B) t The probabilities do not add to 1 because there are other software packages besides SPSS and SAS for which requests could be made u A ) 1 A) 1-7 v A B) A) + B) + 8 (since A and B are mutually exclusive events) w A B ) P[(A B) ] (De Morgan s law) 1 - A B) x {M,H}) 1 y low auto) P[{(L,N}, (L,L), (L,M), (L,H)}] Following a similar pattern, low homeowner s) z same deductible for both) P[{ LL, MM, HH }]

4 aa deductibles are different) 1 same deductibles) bb at least one low deductible) P[{ LN, LL, LM, LH, ML, HL }] cc neither low) 1 at least one low) (A B) A) + B) - A B) A C), B C) A B C) A B C) A) B) C) + A B) + A C) + B C) dd A B C) 98, as given ee none selected) 1 - A B C) 1-98 ff only automatic transmission selected) from the Venn Diagram gg exactly one of the three) Section 9 a ()() 7; ()() 19 b () 17,7; (), c (),97; () 1,79,1 d 1 97,78/() 9 e Because order is important, we ll use P 8, 8(7)() f Order doesn t matter here, so we use C, 9,77 g From each group we choose : 8, 1

5 h The numerator comes from part c and the denominator from part b: 1 9,77 8,1 i We use the same denominator as in part d We can have all zinfandel, all merlot, or all cabernet, so all same) all z) + all m) + all c) 9, j (n 1 )(n ) (9)(7) k (n 1 )(n )(n ) (9)(7)(1), so such a policy could be carried out for successive nights, or approximately 1 years, without repeating exactly the same program l 9,88 m 9 Starting line-ups and 9 batting orders (9)(9) 11,81,89, n 1 (1)(1),1 9 o We want to choose all of the cordless, and of the 1 others, to be among the first 1 serviced, so the desired probability is p Isolating one group, say the cordless phones, we want the other two groups represented in the last serviced So we choose of the 1 others, except that we don t want to include the outcomes where the last five are all the same So we have ' 1 1 But we have three groups of phones, so the desired probability is 98 () 1 1 ' ' ( ) * * +, - q We want to choose of the cordless, of the cellular, and of the corded phones: r at least one F among 1 st ) 1 no F s among 1 st ) An alternative method to calculate no F s among 1 st ) would be to choose none of the females and of the males, as follows: 71 8, obviously producing the same result

6 1 s all F s among 1 st ) 71 8 t orderings are different) 1 orderings are the same for both semesters) There are 8 possible orderings for the second semester, only one of which would match the first semester (whatever ordering that was) So, orderings are different) 1 1/ Section u A) , C) A C) A C) v A C) If we know that the individual came from ethnic group, the probability that he has C) A C) type A blood is C A) 7 If a person has type A blood, the probability that he is from A) 7 ethnic group is 7 D B ) 19 w Define event D {ethnic group 1 selected} We are asked for D B ) 11 D B )8 + B ) , B ) 1 B) 1 [ ] 99 7 x A B) B A) A) y A B ) B A) A) A B) z A B) 1 B) A B) 1 aa A B) 87 B) P[ A ( A B)] bb A A B) 79 A B)

7 9 The first desired probability is both bulbs are 7 watt at least one is 7 watt) at least one is 7 watt) 1 none are 7 watt) 1 ' 9 1 ' 1 Notice that P[(both are 7 watt) (at least one is 7 watt)] both are 7 watt) 1 watt at least one is 7 watt) Second, we want same rating at least one NOT 7 watt) at least one NOT 7 watt) 1 both are 7 watt) Now, P[(same rating) (at least one not 7 watt)] both watt or both watt) both watt or both watt) Now, the desired conditional probability is So both bulbs are 7 cc M LS PR), directly from the table of probabilities dd M Pr) M,Pr,LS) + M,Pr,SS) +7 1 ee SS) sum of 9 probabilities in SS table, LS) 1 - ff M) Pr) M SS Pl) 8 gg M SS Pl) SS Pl) SS M Pl) 8 hh SS M Pl) M Pl) LS M Pl) 1 - SS M Pl) 1 - B A) A B) A) B) (since B is contained in A, A B B) A)

8 8 Let A {carries Lyme disease} and B {carries HGE} We are told A) 1, B) 1, and A B A B) 1 From this last statement and the fact that A B is contained in A B, A B) 1 A B) 1A B) 1[A) + B) A B)] 1[1 + 1 A B)] 11A B) A B) A B) Finally, the desired probability is A B) A B) B) P ( A1 B) A ) B A) 1 P ( A B) 1 P ( A B) ii A B) 1 jj B) A 1 B) + A B) + A B) A 1 kk A 1 B) 1 B) 1, A B), A B) 1 7 B) Section 71 ll Since the events are independent, then A and B are independent, too (See paragraph below equation 7) B A ) B ) 1-7 mm A B)A) + B) A) B) + 7 ()(7) 8 AB ( A B)) A B) AB ) A B) 1 8 nn AB A B) 1

9 8 system works) 1 works works) 1 works) + works) - 1 works works) 1 works works) + works works) 1 ) ) ( ) + (9)(9) (9+9-81)(9)(9) both detect the defect) 1 at least one doesn t) 1-8 oo 1 st detects nd doesn t) 1 st detects) 1 st does nd does) Similarly, 1 st doesn t nd does) 1, so exactly one does) 1+1 pp neither detects a defect) 1 [both do) + exactly 1 does)] 1 [8+] so all escape) ()()() 18 qq walks on th pitch) 1 st pitches are balls) () rr walks on th ) of the 1 st are strikes, is a ball) of the 1 st are strikes) is a ball) [1() ]() 1 ss Batter walks) walks on th ) + walks on th ) + walks on th ) + () tt first batter scores while no one is out) first batters walk) (7) 1

10 Section 1 1 S: FFF SFF FSF FFS FSS SFS SSF SSS X: a Possible values are, 1,,, 1; discrete b With N on the list, values are, 1,,, N; discrete c Possible values are 1,,,, ; discrete d { x: < x < } if we assume that a rattlesnake can be arbitrarily short or long; not discrete e With c amount earned per book sold, possible values are, c, c, c,, 1,c; discrete f { y: < y < 1} since is the smallest possible ph and 1 is the largest possible ph; not discrete g With m and M denoting the minimum and maximum possible tension, respectively, possible values are { x: m < x < M }; not discrete h Possible values are,, 9, 1, 1, -- ie (1), (), (), (), giving a first element, etc,; discrete

11 Section 11 i x 8 x) 1 j Relative Frequency 1 x 7 8 k x ) + 1 x > ) 1 1 l X ) p() + p(1) + p() + p() m X < ) X ) p() + p(1) + p() n X) p() + p() + p() + p() o X ) p() + p() + p() + p() 71 p The number of lines not in use is X, so X is equivalent to X, X to X, and X to X Thus we desire X ) p() + p() + p() q X if X, ie X, or X, and X ) 1+1+

12 17 r ) Y ) 1 st batteries are acceptable) AA) (9)(9) 81 s p() Y ) UAA or AUA) (1)(9) + (1)(9) [(1)(9) ] 1 t The fifth battery must be an A, and one of the first four must also be an A Thus, p() AUUUA or UAUUA or UUAUA or UUUAA) [(1) (9) ] u Y y) p(y) the y th is an A and so is exactly one of the first y 1) (y 1)(1) y- (9), y,,,, Section 9 x p( x) v E (X) x ()(8) + (1)(1) + ()() + ()(7) + ()() w V(X) x x σ x ( x ) p( x) ( ) (8) + + ( ) () y V(X) ' ) x x ( p( x) () E (Y) ; E (Y ) 11 V(Y) E(Y ) [E(Y)] 11 () 7 σ y 7 8 E (Y) ± σ y ± 8 (-, 1) or (, 1) Y ) + Y 1) 8 z k 1 1 k 11 1

13 ' µ, * ) x ( p( x) µ 7, 1 x x aa x p( x) Thus µ - σ -, and µ + σ 7, so x-µ σ) X is at least sd s from µ) X is either - or 7) X ) Chebyshev s bound of is much too conservative For k,,, and 1, x-µ kσ), here again pointing to the very conservative nature of the bound 1 k 1 bb µ and, so x-µ σ) X 1) X -1 or +1) , identical to the upper bound cc Let p(-1) 1 1, p ( + 1) ), p( Section dd b(;8,) 8 () () 79 ee b(;8,) 8 () () 79 ff X ) b(;7,) + b(;7,) + b(;7,) 7 gg 1 X) 1 X ) 1-9 (1) (9) 9 1 (9) hh B(;1,) 1 ii b(;1,) B(;1,) B(;1,) 19 jj b(;1,7) B(;1,7) B(;1,7) 1 kk X ) B(;1,) B(1;1,) 8 ll X) 1 - X 1) 1 B(1;1,) 9 mm X 1) B(1;1,7) nn < X < ) < X ) B(;1,) B(;1,) 9 9 X ~ Bin(, 1) oo X 1) n x n x ' 1 p p x ( ) (1 ) (1) (9) 1 pp X ) 1 [X ) + X 1)]

14 From a, we know X 1), and X ) (1) (9) 1 Hence X ) 1 [ + 1] 11 qq Either or goblets must be selected i) Select goblets with zero defects: X ) (1) (9) 1 ii) Select goblets, one of which has a defect, and the th is good: '-* 1 + (1) (9) 9 1 (, ) So the desired probability is X ~ Bin(,) rr X ) B(;,) 1 ss X ) b(;,) 18 tt X ) 1 X ) 1 B(;,) uu X > ) 1 X ) Let S represent a telephone that is submitted for service while under warranty and must be replaced Then p S) replaced submitted) submitted) ()() 8 Thus X, the number among the company s 1 phones that must be 1 8 replaced, has a binomial distribution with n 1, p 8, so p() X) (8) (9) X the number of flashlights that work Let event B {battery has acceptable voltage} Then flashlight works) both batteries work) B)B) (9)(9) 81 We must assume that the batteries voltage levels are independent X Bin (1, 81) X 9) X9) + X1)

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