CHAPTER a. S = { 1324, 1342, 1423, 1432, 2314, 2341, 2413, 2431, 3124, 3142, 4123, 4132, 3214, 3241, 4213, 4231 }

Transcription

1 CHTER Section.. a. S { 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4 } b. Event contains the outcomes where is first in the list: { 4, 4, 4, 4 } c. Event B contains the outcomes where is first or second: B { 4, 4, 4, 4, 4, 4, 4, 4 } d. The compound event B contains the outcomes in or B or both: B {4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4 }. a. Event { 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 CD C: CD { RRL, RRS, RLR, RSR, LRR, SRR } 47

2 . a. Event { SSF, SFS, FSS } b. Event B { SSS, SSF, SFS, FSS } c. For Event C, the system must have component working S in the first position, then at least one of the other two components must work at least one S in the nd and rd positions: Event C { SSS, SSF, SFS } d. Event C { SFF, FSS, FSF, FFS, FFF } Event C { SSS, SSF, SFS, FSS } Event C { SSF, SFS } Event B C { SSS, SSF, SFS, FSS } Event BC { SSS SSF, SFS } 4. a. Home Mortgage Number Outcome 4 F F F F F F F V F F V F 4 F F V V 5 F V F F F V F V 7 F V V F 8 F V V V 9 V F F F 0 V F F V V F V F V F V V V V F F 4 V V F V 5 V V V F V V V V b. Outcome numbers,, 5,9 c. Outcome numbers, d. Outcome numbers,,, 5, 9 e. In words, the UNION described is the event that either all of the mortgages are variable, or that at most all of them are variable: outcomes,,,5,9,. The INTERSECTION described is the event that all of the mortgages are fixed: outcome. f. The UNION described is the event that either exactly three are fixed, or that all four are the same: outcomes,,, 5, 9,. The INTERSECTION in words is the event that exactly three are fixed ND that all four are the same. This cannot happen. There are no outcomes in common : b c. 48

3 5. a. Outcome Number Outcome b. Outcome Numbers, 4, 7 c. Outcome Numbers, 8,,, 0, d. Outcome Numbers,, 7, 9, 9,, 5, 7 49

4 . a. Outcome Number Outcome b. Outcomes, 4, 5 c. Outcomes,, 9,, 5 d. Outcomes 0,,,, 4, 5 7. a. S {BBB, BBB, BBB, BBB, BBB, BBB, BBB, BBB, BBB, BBB, BBB, BBB, BBB, BBB, BBB, BBB, BBB, BBB, BBB, BBB, BBB, BBB, BBB, BBB, BBB, BBB, BBB, BBB, BBB, BBB, BBB, BBB, BBB, BBB, BBB} b. {BBB, BBB, BBB, BBB, BBB} 50

5 8. a. b. c. 5

6 d. e. 5

7 9. a. In the diagram on the left, the shaded area is B. On the right, the shaded area is, the striped area is B, and the intersection B occurs where there is BOTH shading and stripes. These two diagrams display the same area. b. In the diagram below, the shaded area represents B. Using the diagram on the right above, the union of and B is represented by the areas that have either shading or stripes or both. Both of the diagrams display the same area. 0. a. {Chev, ont, Buick}, B {Ford, Merc}, C {lym, Chrys} are three mutually exclusive events. b. No, let E {Chev, ont}, F {ont, Buick}, G {Buick, Ford}. These events are not mutually exclusive e.g. E and F have an outcome in common, yet there is no outcome common to all three events. 5

8 Section.. a..07 b c. Let event selected customer owns stocks. Then the probability that a selected customer does not own a stock can be represented by This could also have been done easily by adding the probabilities of the funds that are not stocks.. a. B b. B c. B ; B B a. awarded either # or # or both: b. awarded neither # or #: [ ] c. awarded at least one of #, #, #: d. awarded none of the three projects: awarded at least one e. awarded # but neither # nor #:

9 f. either neither # nor # or #: [ ] shaded region awarded none lternatively, answers to a f can be obtained from probabilities on the accompanying Venn diagram 55

10 4. a. B + B - B, so B + B - B b. shaded region B - B Shaded region event of interest B B 5. a. 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 E b. Let event be the event that all five purchase gas. Let event B be the event that all five purchase electric. ll other possible outcomes are those in which at least one of each type is purchased. Thus, the desired probability B a. There are six simple events, corresponding to the outcomes CD, CD, DC, DC, CD, and DC. The probability assigned to each is. b. C ranked first {CD, CD} +. c. C ranked first and D last {CD} 5

11 7. a. The probabilities do not add to because there are other software packages besides SSS and SS for which requests could be made. b c. B + B since and B are mutually exclusive events d. B [ B ] De Morgan s law - B This situation requires the complement concept. The only way for the desired event NOT to happen is if a 75 W bulb is selected first. Let event be that a 75 W bulb is selected first, and 5. Then the desired event is event. 9 So Let event be that the selected joint was found defective by inspector. 0, 000. Let event B be analogous for inspector B. B 75 0, 000. Compound event B is the event that the selected joint was found defective by at least one of the two inspectors. B 59 0, 000. a. The desired event is B, so we use the complement rule: 59 B - B - 0, , b. The desired event is B. B B - B. B + B - B, So B B - B Let S, S and S represent the swing and night shifts, respectively. Let C and C represent the unsafe conditions and unrelated to conditions, respectively. a. The simple events are {S,C}, {S,C}, {S,C}, {S,C},{S,C}, {S,C}. b. {C} {S,C},{S,C},{S,C} c. {S} - {S,C}, {S,C}

12 . a. {M,H}.0 b. low auto [{L,N}, L,L, L,M, L,H}] Following a similar pattern, low homeowner s c. same deductible for both [{ LL, MM, HH }] d. deductibles are different same deductibles e. at least one low deductible [{ LN, LL, LM, LH, ML, HL }] f. neither low at least one low a b c. exactly one ssume that the computers are numbered as described. lso assume that computers and are the laptops. ossible outcomes are,,,4,5,,,4,5,,4,5, 4,5 4, and 5,. a. both are laptops [{,}] 5.07 b. both are desktops [{,4,5, 4,5 4, 5,}] 5.40 c. at least one desktop no desktops both are laptops.07.9 d. at least one of each type both are the same both laptops both desktops

13 4. Since is contained in B, then B can be written as the union of and B, two mutually exclusive events. See diagram. From xiom, [ B ] + B. Substituting B, B + B or B - B. From xiom, B 0, so B or B. For general events and B, B, and B. 5. B + B - B.5 C.55, B C.0 B C B C B C + B + C + B C a. B C.98, as given. b. none selected - B C c. only automatic transmission selected.0 from the Venn Diagram d. exactly one of the three

14 . a b c d. at most two errors all three types Outcomes:,B,C,C,F B, B,C B,C B,F C, C,B C,C C,F C, C,B C,C C,F F, F,B F,C F,C a. [,B or B,]. 4 7 b. at least one C c. at least 5 years at most 4 years [, or, or,7 or 7, or,0 or 0, or,7 or 7,] There are 7 equally likely outcomes. a. all the same [,, or,, or,,] 7 9 b. at most are assigned to the same station all are the same c. all different [{,,,,,,,,,,,,}] 7 9 0

15 Section. 9. a choices for president, 4 remain for vice president b !!! c. 0 No ordering is implied in the choice 0. a. Because order is important, we ll use 8, 87. b. Order doesn t matter here, so we use C 0, 59, c. From each group we choose : 8, 0 8,0 59,775 d. The numerator comes from part c and the denominator from part b:. 4 e. 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 , a. n n 97 4 b. n n n , so such a policy could be carried out for 45 successive nights, or approximately 0 years, without repeating exactly the same program.

16 . a b. 4 c d. # with at least on Sony total # - # with no Sony e. at least one Sony exactly one Sony only Sony is receiver + only Sony is CD player + only Sony is deck ! a. 5, 0 5 5!0! 8 7 b , 0 c. exactly 4 have cracks. 0 d. at least 4 exactly 4 + exactly

17 4. a.. 8,70 0 all from day shift ,45,00 8, b. all from same shift c. at least two shifts represented all from same shift d. Let day shift unrepresented, swing shift unrepresented, and graveyard shift unrepresented. Then we wish. day unrepresented all from swing and graveyard 45 5, 45 0, 45 5, all from graveyard , 45 0, 0, So

18 5. There are 0 possible outcomes -- 5 ways to select the positions for B s votes: BB, BB, BB, BB, BB, BB, BB, BB, BB, and BB. Only the last two have ahead of B throughout the vote count. Since the outcomes are equally likely, the desired probability is a. n, n 4, n 5, so n n n 0 runs b. n, just one temperature, n, n 5 implies that there are 0 such runs. 7. There are 5 0 ways to select the 5 runs. Each catalyst is used in different runs, so the number of ways of selecting one run from each of these 5 groups is 5. Thus the desired probability is a. selecting - 75 watt bulbs b. all three are the same c

19 d. To examine exactly one, a 75 watt bulb must be chosen first. ways to accomplish this. To examine exactly two, we must choose another wattage first, then a 75 watt. 9 ways. Following the pattern, for exactly three, 9 8 ways; for four, ; for five, examine at least bulbs examine 5 or less examine exactly or or or 4 or 5 [one + two + + five] [ ] a. We want to choose all of the 5 cordless, and 5 of the 0 others, to be among the first serviced, so the desired probability is. 089 b. Isolating one group, say the cordless phones, we want the other two groups represented in the last 5 serviced. So we choose 5 of the 0 others, except that we don t want to include the outcomes where the last five are all the same. 0 5 So we have But we have three groups of phones, so the desired probability is c. We want to choose of the 5 cordless, of the 5 cellular, and of the corded phones:

20 40. a. If the s are distinguishable from one another, and similarly for the B s, C s and D s, then there are! ossible chain molecules. Six of these are: B C C D C D D B B, B C C D C D D B B B C C D C D D B B, B C C D C D D B B B C C D C D D B B, B C C D C D D B B These! differ only with respect to ordering of the s. In general, groups of chain molecules can be created such that within each group only the ordering of the s is different. When the subscripts are suppressed, each group of collapses into a single molecule B s, C s and D s are still distinguishable. t this point there are! molecules. Now suppressing subscripts on the B s, C s and D s in turn gives!! ultimately 9, 00 chain molecules. 4! b. Think of the group of s as a single entity, and similarly for the B s, C s, and D s. Then there are 4! Ways to order these entities, and thus 4! Molecules in which the s are contiguous, the B s, C s, and D s are also. Thus, all together! a. at least one F among st no F s among st n alternative method to calculate no F s among st would be to choose none of the females and of the 4 males, as follows: , obviously producing the same result b. all F s among st c. orderings are different orderings are the same for both semesters # orderings such that the orders are the same each semester/total # of possible orderings for semesters

21 4. Seats: 4 J& in & J& next to each other J& in & + + J& in 5& at least one H next to his W no H next to his W We count the # of ways of no H next to his W as follows: # if orderings without a H-W pair in seats # and and no H next to his W * 4 * # 48 * pair, # can t put the mate of seat # here or else a H-W pair would be in #5 and. # of orderings without a H-W pair in seats # and, and no H next to his W 4 # 9 # can t be mate of person in seat # or #. So, # of seating arrangements with no H next to W nd no H next to his W, so at least one H next to his W # of 0 high straights s, 4 9 s, etc ,598, high straight straight Multiply by 0 because there are 0 different card values that could be high: ce, King, etc. There are only 40 straight flushes 0 in each suit, so straight flush

22 n n! n! n 44. k k! n k! n k! k! n k The number of subsets of size k the number of subsets of size n-k, because to each subset of size k there corresponds exactly one subset of size n-k the n-k objects not in the subset of size k. Section a , C C.00 C C b. C If we know that the individual came from ethnic group, the probability that he has type blood is.40. C C from ethnic group is.447. If a person has type blood, the probability that he is c. Define event D {ethnic group selected}. We are asked for D B D B B.500 [ ].909. DB , B B 4. Let event be that the individual is more than feet tall. Let event B be that the individual is a professional basketball player. Then B the probability of the individual being more than feet tall, knowing that the individual is a professional basketball player, and B the probability of the individual being a professional basketball player, knowing that the individual is more than feet tall. B will be larger. Most professional BB players are tall, so the probability of an individual in that reduced sample space being more than feet tall is very large. The number of individuals that are pro BB players is small in relation to the # of males more than feet tall. 8

23 47. B.5.50 a. B. 50 B.5.50 b. B. 50 B B.5.40 c. B. 5 B B.5.40 d. B. 875 [ B] B.50.5 e. B a b c. We want [exactly one at least one]. at least one lso notice that the intersection of the two events is just the st event, since exactly one is totally contained in at least one So [exactly one at least one] d. The pieces of this equation can be found in your answers to exercise section.:

24 49. The first desired probability is both bulbs are 75 watt at least one is 75 watt. at least one is 75 watt none are 75 watt Notice that [both are 75 wattat least one is 75 watt] both are 75 watt So both bulbs are 75 watt at least one is 75 watt. 74 Second, we want same rating at least one NOT 75 watt. at least one NOT 75 watt both are 75 watt Now, [same ratingat least one not 75 watt] both 40 watt or both 0 watt. both 40 watt or both 0 watt Now, the desired conditional probability is a. M LS R.05, directly from the table of probabilities b. M r M,r,LS + M,r,SS c. SS sum of 9 probabilities in SS table 5, LS.5.44 d. M r

25 M SS l SS l e. M SS l. 5 SS M l M l f. SS M l. 444 LS M l - SS M l a. R from st R from nd R from nd R from st R from st b. same numbers both selected balls are the same color both red + both green Let be the event that # fails and be the event that # fails. We assume that q and that r. Then one approach is as follows: rq rq + -rq.07 These two equations give q , from which q.04 and r.5. lternatively, with t, t t.07, implying t.0 and thus q.04 without reference to conditional probability. 5. B B B since B is contained in, B B 7

26 7 54..,.5,.8,.,.05,.07,.0 a b c. ] [ ] [ d This is the probability of being awarded all three projects given that at least one project was awarded. 55. a. B B b. two other H s next to their wives J and M together in the middle..... ] [ in the middle J M M or J H W orw H and J M M or J and H W orw H numerator! denominator! so the desired probability 48.

27 7 c. all H s next to W s J & M together all H s next to W s including J&M/J&M together ! 4 5! 4 5. If B > B, then B < B. roof by contradiction. ssume B B. Then B B. - B B. B B. This contradicts the initial condition, therefore B < B B B B B B B B B B B B 58. ] [ [ C C B C C C B C B C C B C B C + C + B C B C

28 B B. 5.. B B a. B. b. B B + B + B.455 c. B. B. 4 B.455. B. 4, B a. not. disc has. loc.0 not disc has loc. 07 has. loc b. disc no. loc.8 disc no loc. 509 no. loc.55 74

29 . 0 def in sample 0 def in batch 0 def in sample def in batch def in sample def in batch 0 def in sample def in batch def in sample def in batch def in sample def in batch a. 0 def in batch 0 def in sample def in batch 0 def in sample def in batch 0 def in sample 44 75

30 b. 0 def in batch def in sample 0.0 def in batch def in sample def in batch def in sample 54. Using a tree diagram, B basic, D deluxe, W warranty purchase, W no warranty. 4.. B W B W..5.0 D W B W W..5.0 D W We want B W

31 . a. b. B C c. B C B C + B C d. C B C+ B C + B C + B C B C B C.54.8 e. B C

32 4. a b. has d c. doesn t have d satis.5. mean satis. 9.5 median satis.94 mode satis.7 So Mean and not Mode! is the most likely author, while Median is least. 78

33 . Define events,, and as flying with airline,, and, respectively. Events 0,, and are 0,, and flights are late, respectively. Event DC the event that the flight to DC is late, and event L the event that the flight to L is late. Creating a tree diagram as described in the hint, the probabilities of the second generation branches are calculated as follows: For the branch, 0 [DC L ] [DC ] [L ].7.9.; [DC L DCL ] ; [DCL] [DC] [L]...0 Follow a similar pattern for and. From the law of total probability, we know that + + from tree diagram below We wish to find,, and /. 4 ; / 88 ; /. 47 ; 79

34 7. a. U F Cr.0 b. r NF Cr.05 c. r Cr d. F Cr e. Cr.55 r Cr Cr.5.55 f. R Cr. 80

35 Section.5 8. Using the definition, two events and B are independent if B ; B.5;.50;.5.50, so and B are dependent. Using the multiplication rule, the events are independent if B B; B.5; B , so and B are dependent. 9. a. Since the events are independent, then and B are independent, too. see paragraph below equation.7. B. B -.7. b. B + B B B B B. c. B B. 4 B B ,.055. and are not independent..05,.0. and are not independent..07,.07. and are independent. 7. B B B B - B [ ] B B. B B B lternatively, B B B B B. B 7. Using subscripts to differentiate between the selected individuals, O O O O two individuals match +B B + B B + O O Let event E be the event that an error was signaled incorrectly. We want at least one signaled incorrectly E E E 0 - E E E 0. E For 0 independent points, E E E 0 E E E 0 so E E E 0 - [.95] Similarly, for 5 points, the desired probability is - [E ]

36 74. no error on any particular question.9, so no error on any of the 0 questions Then at least one error For p replacing., the two probabilities are -p n and -p n. 75. Let q denote the probability that a rivet is defective. a. seam need rework.0 seam doesn t need rework no rivets are defective st isn t def 5 th isn t def q 5, so.80 q 5, q.80 /5, and thus q b. The desired condition is.0 q 5, i.e. q 5.90, from which q at least one opens none open at least one fails to open all open Let older pump fails, newer pump fails, and x. Then.0 + x,.05 + x, and x.0 + x.05 + x. The resulting quadratic equation, x -.85x , has roots x.0059 and x.844. Hopefully the smaller root is the actual probability of system failure. 78. system works works 4 works works + 4 works - works 4 works works works + works 4 works

37 79. Using the hints, let i p, and x p, then system lifetime exceeds t 0 p + p p 4 p p 4 x x. Now, set this equal to.99, or x x.99 x x Use the ± ±. quadratic formula to solve for x: ±..99 or.0 Since the value we want is a probability, and has to be, we use the value of Event : {,,,,4,5, }, ; Event B: {,4,4,44,45,4,4 }, B ; Event C: {,,5,44,5,, }, C ; Event B: {,4 }; B ; Event C: {,4 }; C ; Event BC: {,4 }; C ; Event BC: {,4 }; BC ; B B C C B C BC The events are pairwise independent. B C BC The events are not mutually independent 8

38 8. both detect the defect at least one doesn t -..8 a. st detects nd doesn t st detects st does nd does Similarly, st doesn t nd does., so exactly one does.+.. b. neither detects a defect [both do + exactly does] [.8+.] 0 so all escape pass.70 a b. all pass c. exactly one passes d. # pass 0 pass + exactly one passes e. pass or more pass. pass. pass. pass. pass. pass a. Let D detection on st fixation, D detection on nd fixation. detection in at most fixations D + D D D + D D D p + p p p p. b. Define D, D,, D n as in a. Then at most n fixations D + D D + D D D + + D D D n- D n p + p p + p p + + p p n- p [ + p + p + + p n- ] n p p p p lternatively, at most n fixations at least n+ are req d no detection in st n fixations c. no detection in fixations p D D D n p n n 84

39 d. passes inspection {not flawed} {flawed and passes} not flawed + flawed and passes e. flawed passed.9 + passes flawed flawed.9+ p. flawed passed. p passed.9 +. p For p.5, flawed passed , B B + B 0,000 a B + B B.09984; since B B, the events are not independent. b. B.04. Very little difference. Yes. c. B., B.04, but B B. 0, so the 9 0 two numbers are quite different. In a, the sample size is small relative to the population size, while here it is not. 85. system works works 4 5 works 7 works works 4 5 works 7 works With the subsystem in figure.4 connected in parallel to this subsystem, system works

40 8. a. For route #, late stopped at or or 4 crossings stopped at 0 or [ ].05 For route #, late stopped at or crossings stopped at none thus route # should be taken. b. 4 crossing route late 4cros sin g late late π π π π π π π π ππ π at most is lost both lost π exactly lost π - π exactly at most π exactly at. most π π π 8

41 Supplementary Exercises 88. a b c. # having at least of the 0 best 40 - # of crews having none of 0 best d. best will not work ; a. line. Crack b. Blemish line.5 c. Surface Defect line and Surface Defect So line Surface Defect a. The only way he will have one type of forms left is if they are all course substitution forms. He must choose all of the withdrawal forms to pass to a subordinate. The desired probability is b. He can start with the wd forms: W-C-W-C or with the cs forms: C-W-C-W: # of ways: ; The total # ways to arrange the four forms: The desired probability is 70/

42 9. B + B B. + B -.44 So + B.770 and B.44. Let x and y B, then using the first equation, y.77 x, and substituting this into the second equation, we get x.77 x.44 or x -.77x Use the quadratic formula to solve:.77 ± ±.. So.45 and B. or a b sent received received c. sent received

43 9. a. There are 54 0 possible orderings, so BCDEF b. # orderings in which F is rd 4* 4, * because F must be here, so F rd c. F last F hasn t heard after 0 times not on # not on # not on # When three experiments are performed, there are different ways in which detection can occur on exactly of the experiments: i # and # and not # ii # and not # and #; iii not# and # and #. If the impurity is present, the probability of exactly detections in three independent experiments is If the impurity is absent, the analogous probability is Thus present detected in exactly out of det ected. in. exactly. present det ected. in. exactly exactly selects category # all are different exactly.. selects# all. are. different all. are. different 5 4 Denominator Numerator contestant # selects category # and the other two select two different categories The desired probability is then

44 97. a. pass inspection pass initially passes after recrimping pass initially + fails initially goes to recrimping is corrected after recrimping following path bad-good-good on tree diagram.974 b. needed no recrimping passed inspection passed. initially passed. inspection a. both + carrier both + + not a carrier both + both + carrier x carrier + both + not a carrier x not a carrier both tests agree carrier both. positive both. positive b. carrier both + ve Let st functions, B nd functions, so B.9, B.9, B.75. Thus, B + B - B , implying.8. B.75.8 This gives B E late late E E

45 0. a. The law of total probability gives late E late i E i i b. E on time E on time E on. time on. time Let B denote the event that a component needs rework. Then B i B i i Thus B B B a. all different at least two the same b. at least two the same.47 for k, and.507 for k c. at least two have the same SS number all different Thus at least one coincidence BD coincidence SS coincidence

46 04..5 a. G R < R < R b. G R < R < R G R < R < R , B R < R < R., classify as granite. <.05, so classify as basalt., so classify as basalt. c. erroneous classif B classif as G + G classif as B classif as G BB + classif as B GG R < R < R B.75 + R < R < R or R < R < R G

47 d. For what values of p will G R <R <R >.5, G R < R < R >.5, G R < R < R >.5?.p.p.p +. p.+.5p G R < R < R >. 5 iff.5p.5p +. p.5p >..5p +.7 p G R < R < R >. 5 iff 4 p > 9 4 > 7 p > 7 G R < R < R 5 iff p most restrictive If 4 > 7 p always classify as granite. 05. detection by the end of the nth glimpse not detected in st n G G G n - G G G n p p p n - π p n i i 0. a. walks on 4 th pitch st 4 pitches are balls b. walks on th of the st 5 are strikes, # is a ball of the st 5 are strikes# is a ball [0.5 5 ].5.55 c. Batter walks walks on 4 th + walks on 5 th + walks on th d. first batter scores while no one is out first 4 batters walk a. all in correct room b. The 9 outcomes which yield incorrect assignments are: 4, 4, 4, 4, 4, 9 4 4, 4, 4, and 4, so all incorrect. 75 9

48 08. a. all full B C at least one isn t full all full b. only NY is full B C B C.8 Similarly, only tlanta is full. and only L is full.08 So exactly one full Note: s 0 means that the very first candidate interviewed is hired. Each entry below is the candidate hired for the given policy and outcome. Outcome s0 s s s Outcome s0 s s s s 0 hire# So s is best. 0. at least one occurs none occur p p p p 4 p p p p p p p p 4 + p p p p p p p p 4 + p p p p 4. draw slip or 4 ½; draw slip or 4 ½; draw slip or 4 ½; draw slip 4 ¼; draw slip 4 ¼; draw slip 4 ¼ Hence ¼, ¼, ¼, thus there exists pairwise independence draw slip 4 ¼ /8 p, so the events are not mutually independent. 94