CHAPTER 2
PROBABILITY
2-1.
Objective and subjective.
2-2.
An event is a set of basic outcomes of an experiment. The union of two events is the set
containing all basic outcomes that are either in one event or in the other, or in both. The
intersection of two events is the set of basic outcomes that are members of both events.
2-3.
The sample space is the universal set pertinent to a given experiment. It is the set of all possible
outcomes of an experiment.
2-4.
The probability of an event is a measure of the likelihood of the occurrence of the event. When
sample points are equally likely, the probability of the event is the relative size of the set
comprising the event within the sample space.
2-5.
The union G  F is the event that the baby is either a girl, or is over 5 pounds (of either sex).
The intersection G  F is the event that the baby is a girl over 5 pounds.
2-6.
The union is the event that the player scores in the game with A, or in the game with B, or in
both. The intersection is the event that the player scores in both games.
2-7.
Sample
Space
first toss
second toss > first
First Toss Second Toss > First
1
2
3
4
5
6
2
3
4
5
6
3
4
5
6
4
5
6
5
6
6
none
There are 36 possible outcomes tossing two dice.
There are 15 possible outcomes where the second
toss is greater than the first.
P(Second Toss > First) = 15/36 = 0.417
33
2-8.
R  T is the event that a randomly chosen person is exposed to the ad on the radio or the ad on
television, or both. R  T is the event that a randomly chosen person is exposed to the ad on the
radio and the ad on television.
2-9.
S  B : purchase stock or bonds, or both. S  B : purchase stock and bonds.
2-10.
Since there are only 37 possible outcomes, the probability of winning is 1/37 = 0.027, better odds
(for the player) than the American version. Because of this, the house admission fee makes sense.
2-11.
The third sample space is correct. It lists all possible outcomes in tossing a coin twice. The first
two sample spaces do not contain all the possible outcomes of the experiment.
2-12.
a)
b)
c)
d)
P(exploring the products) = 1790 / 2385 = 0.7505
P(purchase) = 387 / 2385 = 0.1623
P(purchase | explored the products) = 387 / 1790 = 0.2162
(a) approximately 75 percent of the visitors go beyond the homepage.
2-13.
a) P(Ace of Spades) = 1 / 52
b) P(Friend’s card is Ace of Spades| that you have drawn a card first) = 1 / 52
c) 1 / 51
2-14.
Based on murder statistics for a given time period: total murders per 100,000 population.
2-15.
0.85, for example, is a typical "very likely" probability.
2-16.
More likely to occur than not to occur.
2-17.
The team is very likely to win.
2-18.
P(first shopper detected) + P(second detected) - P(both detected) = 0.98 + 0.94 - 0.93 = 0.99
2-19.
a. The two events are mutually exclusive.
b. Let 0, D be the events: machine is out-of-control, down (respectively). Then we
need P(O  D)  P(O)  P(D)  P(O  D) = 0.02 + 0.015 - 0 = 0.035
c.
2-20.
P(D)  I  P(D)  I  0.015  0.985 . This event and D are mutually exclusive.
P(F) + P(>50) – P(F & >50) = 12/20 + 2/20 – 2/20 = 0.6
P(< 30) = 2/20 = 0.1
34
2-21.
P(T  R)  P(T )  P( R)  P(T  R) = 0.25 + 0.34 – 0.10 = 0.49
2-22.
P( S  B)  P( S )  P( B)  P( S  B) = 0.85 + 0.33 - 0.28 = 0.90
2-23.
P(VT CE) = 380/550 + 412/550 - 357/550 = 0.7909
2-24.
a.
b.
c.
d.
e.
f.
g.
2-25.
P(M  H)  P(M)  P(H)  P(M  H) = (11 + 8 – 5) / 28 = 1 / 2 = 0.50
2-26.
25%: Reading the market shares as 2 %, 2%, and 4%, the answer is 2(2+2+4) = 25%
2-27.
0.34
Given that P(R | B) = 0.85 and P(B) = 0.40,
P(R  B) = P(R | B) P(B) = (.85)(.40) = 0.34
2-28.
P(M | R) = 0.32 Given that P(M  R) = 0.80 and P(R) = 0.4,
1268 / 2074, found by: 597 / 2074 + 962 / 2074 – 291 / 2074 = 1268 / 2074
230 / 2074
419 / 2074, found by: 230 / 2074 + 189 / 2074 = 419 / 2074
306 / 1112
189 / 427
773 / 2074, found by: 310 / 2074 + 291 / 2074 + 172 / 2074
773 / 1647, found by: (310 + 291 + 172) / (648 + 597 + 402) = 773 / 1647
P(M | R) = P(M / R) P(R) = (0.80)(0.40) = 0.32
2-29.
P(D | L) = 0.60 Given that P(L D) = 0.12 and P(L) = 0.20,
P(D | L) = P(L D) / P(L) = .12 / .20 = 0.60
2-30.
2.5% of the packages are late. Given that P(N | D) = 0.25 and P(D) = 0.10,
P(N D) = P(N | D) P(D) = (.25)(.10) = 0.025
2-31.
a. P(M) = 198 / 1976 = 0.1002
b. P(E) = 408/1976 = 0.2065
c. Mutually exclusive events: P(M S) = P(M) + P(S) = (198 + 968) / 1976 = 0.59
d. P(H) = 284/1976 = 0.144
e. P(S | H) = P(S H) / P(H) = (128 / 1976) / (284 / 1976) = 0.451
f. P(P | E) = P(P E) / P(E) = (233 / 1976) / (408 / 1976) = 0.571
g. P(W | O) = P(W O) / P(O) = (99 / 1976) / (590 / 1976) = 0.168
h. P(E  O) = P(E) + P(O) – P(E  O) = (408 + 590 – 100) / 1976 = 0.454
i. P(H S) = P(H) + P(S) – P( H  S ) = (284 + 968 – 128) / 1976 = 0.569
35
2-32.
61.1% successfully completed. Given that P(A | H) = 0.94 and P(H) = 0.65,
P(A H) = P(A | H) P(H) = (.94)(.65) = 0.611
2-33.
a. P(I) = 119 / 246 = 0.484
b. P(D) = 112/ 246 = 0.455
c. P(I  D) = 34 / 246 = 0.138
d. P(I  D) = 49/ 246 = 0.199
e. P(D | I) = P(D I) / P(I) = 0.138 / 0.484 = 0.285
f. P(I | D) = P(I D)/P(D) = (85 / 246) / (134 / 246) = 0.634
g. P(D I) = P(D) + P(I) – P( D  I ) = 0.455 + 0.484 – 0.138 = 0.801
2-34.
Let E,S denote the events: top Executive made over $1M, Shareholders made money,
respectively. Then:
a. P(E) = 3 / 10 = 0.30
b. P(S ) = 3 / 10 = 0.30
c. P(E | S ) = P(I S )/P(S ) = (2 / 10) / (3 / 10) = 2/3 = 0.667
d. P(S | E) = P(S E)/P(E) = (1/10)/ (3/10) = 1/3 = 0.333
2-35.
a. P(Male | 2003) = P(Male ∩ 2003) / P(2003) = 0.59 / (0.59 + 0.65) = 0.4758
b. P(Male | 2002) = P(Male ∩ 2002) / P(2002) = 0.47 / (0.47 + 0.55) = 0.4608
2-36.
(Use template: Probability of at least 1.xls)
For any single person, the probability that the person is obese is 65% and the probability that the
person is not obese is 35%. The probability that none of the five people chosen is obese would
be (0.35)5 = 0.00525. The probability that at least one of the people chosen is overweight and
obese is:
P(≥ 1 overweight & obese) = 1 – P(none are overweight and obese) = 1 – 0.00525 = 0.99475.
Probability of at least one success from many independent trials.
Success Probs
1
2
3
4
5
0.65
0.65
0.65
0.65
0.65
Prob. of at least one success
36
0.9947
2-37.
 2  2  2  5  5  19  19 
  (by independence)
 3  3  3  6  6  20  20 
P(at least one job) = 1 – P(no jobs) = 1 –      
= 0.8143
2-38.
Assume independence:
P(at least one arrives on time) = 1 – P(all three fail to arrive)
= 1 – (1 – .90)(1 – .88)(1 – .91) = 0.99892
2-39.
P(increase sales all three countries) = P(increase sales in US) x P(increase sales Australia) x
P(increase sales in Japan)
P(increase sales all three countries) = (0.95)(0.90)(0.85) = 0.72675
2-40.
P(device works satisfactorily) = 1 – P(both components fail)
= 1 – (0.02)(0.1) = 1 – 0.002 = 0.998
2-41.
P(at least one benefit) = 1 – P(no benefits) = 1 – (1 – 0.93)(1 – 0.55)(1 – 0.70) = 1 - 0.00945 =
0.99055
2-42.
(Use template: Probability of at least 1.xls)
For any single cellphone user, the probability that he/she is a Verizon customer is (36 / 148.4) =
24.26%. The probability that the user is not a Verizon customer is 75.74%. The probability that
none of the six cellphone users are Verizon customers would be (0.7574)6 = 0.1888. Therefore,
the probability that at least one of the cellphone users is a Verizon customer is:
P(≥ 1 Verizon customer) = 1 – P(none are Verizon customers) = 1- 0.1888 = 0.8112
Probability of at least one success from many independent trials.
Success Probs
1
2
3
4
5
6
0.2426
0.2426
0.2426
0.2426
0.2426
0.2426
Prob. of at least one success
37
0.8112
2-43.
Problem 2-43:
E1: The chosen box is A, P(A) = 0.50
E2: The chosen coin is a dime, P(dime)=0.167
E3: The outcome of the coin toss is Heads, P(H)=0.50
1. Are events E1 and E2 independent? No
P(E1)P(E2) = (0.50)(0.167) = 0.833
while: P(E1∩E2) = P(E1|E2)P(E2) = (1.0)(0.167) = 0.167
They are not equal.
2. Are events E1 and E3 independent? Yes
P(E1)P(E3) = (0.50)(0.5) = 0.25
while: P(E1∩E3) = P(E1|E3)P(E3) = (0.5)(0.5) = 0.25
They are equal.
3. Are events E2 and E3 independent? Yes
P(E2)P(E3) = (0.167)(0.5) = 0.0833
while: P(E2∩E3) = P(E2|E3)P(E3) = (0.167)(0.5) = 0.0833
They are equal.
2-44.
P(H)P(M) = (284/1976)(198/1976) = 0.01440
P(H M) = 29/1976 = 0.01468  0.01440
The two events are not independent (but they are close)
2-45.
P(D)P(I) = (112/246)(119/246) = 0.2202
P(D I) = 34/246 = 0.1382  0.2202
The two events are not independent.
2-46.
P(E)P(S ) = (3/10)(3/10) = 0.09
P( E  S ) = 2/10 = 0.20  0.09
The two events are not independent. There may be some sort of relationship between them as a
general rule, if seen in all firms.
2-47.
P(getting at least one disease) = 1 – P(getting none of the three)
= 1 – P(not getting Mal.)P(not getting schist.)P(not getting s.s.)
= 1 – (1 – P(getting mal.))(1 – P(getting schist.))(1 – P(getting s.s.))
= 1 – (1 – (110/2100))(1 – (200/600))(1 – (.025/50))
= 1 – (0.9476)(0.6667)(0.9995) = 0.3686
2-48.
The device works if at least one out of three works.
P(device works) = 1 – P(all components fail)
= 1 – (1 – 0.96)(1 – 0.91)(1 – 0.80) = 0.99928
38
2-49.
(Use template: Probability of at least 1.xls)
The probability of dying in a car crash in France in 2003 is 5732 / 59625919 = 0.000096. The
probability on not dying in a car crash is 0.999904. The probability of not dying in a car crash
for the next five years, assuming conditions stay the same as in 2003, would be (0.999904)5 =
0.99952. Therefore, the probability of dying in a car crash over the next five years is:
1 – P(not dying) = 1 – 0.99952 = 0.00048.
Probability of at least one success from many independent trials.
Success Probs
1
2
3
4
5
2-50.
0.000096
0.000096
0.000096
0.000096
0.000096
Prob. of at least one success
0.0005
P(at least one drives home safely) = 1 – P(none drive home safely) = 1 – (0.50)(0.75)(0.80)
= 1 – 0.30 = 0.70
Success
Probs
1
2
3
2-51.
0.5
0.25
0.2
Prob. of at least one success
0.7000
Since 1/4 of the items are in any particular quartile, and assuming independent random samples
with replacement so that all four choices have the same probability of being in the top
quartile, P(all four in top quartile) = (1/4)4 = 1/256 = 0.0039
P(at least one from bottom quartile) = 1 – P(all four from top three quartiles) =
1 – (3/4)4 = 175/256 = 0.684
2-52.
(55)(30)(21)(13) = 450,450 sets of representatives
2-53.
9! = (9)(8)(7)(6)(5)(4)(3)(2)(1) = 362,880 different orders.
2-54.
nPr = n! / (n–r)! = 15! / (15-8)! = (15)(14)(13)(12)(11)(10)(9)(8) = 259,459,200
Permutation
n
15
2-55.
r
8
nPr
259459200
6P3 = 6! / (6–3)! = (6)(5)(4) = 120 ordered choices
39
2-56.
7! / [(7–2)!2!] = 21 pairs.
Combination
n
r
7
2
nCr
21
2-57.
Only one possible combination of 3 elements chosen from the 14 parts consists of the 3 faulty
ones. So since any 3-element combination is equally likely to be picked, the probability
= 1 / (14! / 3!11!) = 1 / 364 = 0.00275
2-58.
Only one of the combinations wins, so the probability of guessing it is 1 / (36! / 6!30!) = 1 /
1,947,792 = 0.000000513
2-59.
How many ways of guessing a set of 6 of the numbers from 1 to 36 will have 5 correct and 1
wrong? If W = {WI, W2, - - -, W6} is the winning combination, then there are 6 choices of which
wi is not in the guessed combination, and 30 possible wrong guesses in place of wi (since the one
wrong guess can be any of the numbers from 1 to 36 that are not in W). So (6) (30) = 180
possible combinations match exactly 5 of the winning numbers. Thus the probability of making
such a guess is = 180/1,947,792 = 0.0000924
2-60.
Let T,R be the events: successful takeover, resignation of a board member.
P(T | R) = 0.65 P(T | R) = 0.30 P(R) = 0.70
P(T) = P(T | R)P(R) + P(T | R)P(R) = (.65)(.7) + (.30)(.30) = 0.545
2-61.
Let A,S be the events: the drug is approved, the drug has side effects.
P(A | S) = 0.5 P(A | S ) = 0.95 P(S) = 0.20 P(S ) = 0.8
P(A) = P(A | S)P(S) + P(A | S )P(S ) = (.5)(.2) + (.95)(.80) = 0.86
2-62.
Let D,B be the events: deal is concluded, competitor makes a bid.
P(D | B) = 0.25
P( D | B ) = 0.45 P(B) = 0.40 P(B) = 0.6
P(D) = P(D | B)P(B) + P(D | B)P(B) = (.25)(.4) + (.45)(.6) = 0.37
2-63.
Let S,E be the events: property is sold, economy improves.
P(S | E) = 0.9
P(S | E) = 0.50
P(E) = 0.70
P(S) = P(S | E)P(E) + P(S | E)P(E) = (.9)(.7) + (.50)(.30) = 0.78
2-64.
Let F,A be the events: ships sail full this summer, dollar appreciates against European
currencies.
P(F | A) = 0.75
P( F | A = 0.92 P(A) = 0.23
P(F) = P(F | A)P(A) + P(F | A)P(A) = (.75)(.23) + (.92)(.77) = 0.8809
40
2-65.
Let O,G be the events: door should open, green light appears.
P(O) = 0.9 P(G | O) = 0.98
P(G | O) = 0.05
P(G | O) P(O)
(.98)(.9)
P(O | G) =
=
P(G | O) P(O) + P(G | O)P(O)
(.98)(.9) + (.05)(.1)
= 0.9944
2-66.
Let S,E be the events: the alarm sounds, there is an emergency situation.
P(E) = 0.004
P(S | E) = 0.95 P(S | E) = 0.02
P(S | E) P(E)
(.95)(.004)
P(E | S) =
=
P(S | E) P(E) + P(S | E)P(E)
(.95)(.004) + (.02)(.996)
= 0.1602
2-67.
Let I,H,M,L be the events: indicator rises, economic situation is high, medium, low.
P(I | H) = 0.6 P(I | M) = 0.3 P(I | L) = 0.1 P(H) = 0.15 P(M) = 0.7 P(L) = 0.15
P(I | H) P(H)
(.6)(.15)
P(H | I) =
=
P(I | H) P(H) + P(I |M) P(M) + P(I | L) P(L)
(.6)(.15)+ (.3)(.7)+ (.1)(.15)
= 0.2857
2-68.
Let I,O be the events: test indicates oil, oil really is present
P(O) = 0.4
P(I | O) = 0.85
P(I | O) = 0.10
P(I | O) P(O)
(.85)(.4)
P(O | I) =
=
P(I | O) P(O) + P(I | O)P(O)
= 0.85
41
(.85)(.4) + (.10)(.6)
2-69.
Let I,S be the events: test indicates success, product really is successful.
P(S) = 0.6
P(I | S) = 0.75
P(I | S ) = 0.15
P(I | S) P(S)
(.75)(.6)
P(S | I) =
=
P(I | S) P(S) + P(I | S )P(S )
(.75)(.6) + (.15)(.4)
= 0.8824
2-70.
Let MM, FF, MF denote the events of reaching the two men, the two women, the married couple
(respectively). Let W be the event that a woman answers the door.
Then:
P(MF) = 1/3
P(W | MF) = ½
P(MF W) = 1/6
P(MM) = 1/3
P(W | MM) = 0
P(MM W) = 0
P(FF) = 1/3
P(W | FF) = 1
P(FF W) = 1/3
P(W) = ½
P(MF | W) = P(MF W)/P(W) = [ (1/6) / (1/2) ] = 1/3
2-71.
P(  1 out of 5 misdirected) = 1 – P (none misdirected)
= 1 – [P(a single call not misdirected)]5 = 1 – (199/200)5 = 0.0248
2-72.
(0.02)(1/200) = 1/10,000 = 0.0001
2-73.
P(A B) = 0.4 + 0.3 – 0.1 = 0.6
2-74.
a. P(probability all work in retailing) = 0.12 x 0.12 x 0.09 x 0.09 = 0.000117
b. P(at least one works in retail) = 1 – (.88)(.88)(.91)(.91) = 0.359
2-75.
P(C S) = 0.15 + 0.10 – 0.05 = 0.20
2-76.
Let C, J be the events: pass CPA exam, get job offer.
P(C) = 0.6 P(C J) = 0.4
P(J | C) = P(C J) / P(C) = 0.4/0.6 = 0.667
2-77.
P(A) = 0.20
P(A B) = 0.12
P(B | A) = P(A B) / P(A) = 0.12/0.20 = 0.60
42
2-78.
Let P, I be the events: production increases, interest rates decline more than half a point.
P(P | I) = 0.72 P(I) = 0.25
P(P I) = P(P | I ) P(I) = (.72)(.25) = 0.18
2-79.
Let EQ, SD be the events: engineering quality, sporty design rated among most important
features.
P( EQ  SD) = 0.25
P(EQ) = 0.35
P(SD) = 0.50
(.35)(.50) = .175  .25, therefore the two events are not independent
2-80.
The assumption of independence is justified by random sampling.
P(all 3 people consider EQ) = (.35)3 = 0.0429
P(at least one person considers EQ) = 1 – P(none of the three consider EQ)
= 1 – (.65)3 = 0.7254
2-81.
P(exposed to at least one mode of advertising)
= 1 – P (not exposed to any of the 3)
= 1 – (.90)(.85)(.80) = 0.388
Success
Probs
1
2
3
2-82.
0.1
0.15
0.2
Prob. of at least one success
0.3880
Let S, R be the events: see the ad in the Wall Street Journal, remember it.
P(S) = 0.6 P(R | S) = 0.85
P( S  R) = P(R | S) P(S) = (.85)(.6) = 0.51
51% of the people see and remember the advertisement
2-83.
P(defective) = 0.1 5 chips are chosen at random.
P(none of the 5 are defective) = (0.90)5 = 0.59049
P(at least one is defective) = 1 – P(none are defective) = 0.40951
2-84.
P(at least one color) = 1 – P(none of the colors) By independence:
= 1 – (1– 0.3)(1 – 0.2)(1 – 0.15) = 0.524
Success
Probs
1
2
3
0.3
0.2
0.15
Prob. of at least one success
43
0.5240
2-85.
(2/3)5 = 0.132
2-86.
Let S, F, N, U be the events: subsidiary will be successful, political situation is favorable,
neutral, unfavorable.
P(S | F) = 0.55 P(S / N) = 0.3 P(S / U) = 0.1
P(F) = 0.6
P(N) = 0.2 P(U) = 0.2
P(S) = P(S | F) P(F) + P(S | N) P(N) + P(S | U)P(U)
= (0.55)(0.6) + (0.3)(0.2) + (0.1)(0.2) = 0.41
2-87.
Let L, A be the events: legislation is passed, authorization is granted.
P(L A) = 0.5 P(L) = 0.75
The choice is probably not random.
P(A | L) = P(L A) / P(L) = 0.5/0.75 = 0.6667
2-88.
Let D, H be the events: customer defaults, economy is high.
P(D | H) = 0.04
P(D | H) = 0.13
P(H) = 0.65
P(D) = P(D | H) P(H) + P(D | H)P(H)
= (.04)(.65) + (.13)(.35) = 0.0715
2-89.
a) P(assembly line worker) = P(skilled worker on assembly line) + P(unskilled worker on
assembly line) = (0.30)(0.15) + (0.70)(0.30) = 0.045 + 0.21 = 0.255
b) P(unskilled worker | assembly line worker) = 0.21 / 0.255 = 0.8235
44
2-90.
P(at least one booking) = 1 – P(none) = 1 – (0.92)20 = 0.8113
Success
Probs
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
0.08
0.08
0.08
0.08
0.08
0.08
0.08
0.08
0.08
0.08
0.08
0.08
0.08
0.08
0.08
0.08
0.08
0.08
0.08
0.08
Prob. of at least one success
0.8113
2-91.
P(none are in error) = (0.95)10 = 0.5987
2-92.
Let A, F, So, J, Se be the events: student got an A, student is a freshman, sophomore,
junior, senior
P(Se | A)
P(A | Se) P(Se)
=
P(A | Se) P(Se) + P(A |F) P(F) + P(A | So) P(So) + P(A | J)P(J)
(.40)(.15)
=
= 0.2034
(.40)(.15) + (.20)(.30) + (.30)(.35) + (.35)(.20)
2-93.
Let S, C be the events: product is successful, competitor produces similar product.
P(S | C) = 0.67 P(S | C) = 0.42 P(C) = 0.35
P(S) = (.67)(.65) + (.42)(.35) = 0.5825
45
2-94.
The probability of being killed by a dog is 12 / 280,000,000. The probability of not being killed
by a dog is 1- [12/280,000,000]. The probability of being killed by a dog over the next 20 years
is: 1- P(not being killed) = 1 – [1-(12/280,000,000)]20 = 0.00000086.
2-95.
Conceptually, in the hole-card game there is a card missing from the space of cards available for
play by the players-the hole card itself, which is removed from play at the beginning. So, for
example, if a player wants to make a draw and needs anything lower than a 4 in order not to go
over 21, she can't be sure of the probability that the next card is in fact smaller than a 4 (given the
cards already drawn and known to everyone), since the hole card may or may not be one of the
remaining cards smaller than 4. In other words, the probability space represented by the deck (in
a randomized order) used for the main play isn't exactly specified in the hole-card game (from
any player's point of view), while this space is always based on the full 52-card deck in the
regular game. Practically speaking, however, the probabilities involved vary only slightly, and
their role in the outcome of the game should be more or less unnoticeable when averaged over a
span of games.
2-96.
The probability of dying in a car crash in the U.S. is 40676 / 280000000 = 0.000145. The
probability of dying in a car crash in France is 0.000096 (from problem 2-49). The rate in the
U.S. is 1.5 times larger than the rate in France. The probability of not dying in a car crash in the
U.S. is 0.999855. The probability of dying in a car crash in the U.S. over the next 20 years is:
1- P(not dying) = 1 – (0.999855)20 = 0.002896.
2-97.
Assume a large population so that the sampling can be considered as being done with
replacement (i.e., removing one item does not appreciably alter the remaining number of data
points on either side of the median). Then the first item drawn is on a particular side of the
median, after which the second item has a 1/2 probability of coming from the other side, since
half of the points in the population are on one side and half are on the other by the definition of
median. Thus the probability that the median will lie between the two points drawn is 1/2.
2-98.
Making the same assumptions as in Problem 2-97, first count the number of ways that all n
elements drawn could lie on the same side of the median: since the choices are independent (this
being a random sample) and each of the n – 1 choices after the first has probability 1/2 of being
on the same side of the median as the first choice, the number of such ways is (1/2)n-1. The
median will lie somewhere between the smallest and largest values drawn exactly when the
above situation does not occur, so the desired probability is 1 – (1/2).
2-99.
P (at least one paper accepted) = 1 – P (no papers accepted)
= 1 – P(first paper rejected) P(second paper rejected) ...
= 1 – [1 – .14][1 – (.14)(.9)][1 – (.14)(.9)2] [1 – (.14)(.9)3] ...
= 1 – (.86)(.874)(.8866)(.89794)(.908146) ...
Using 800 iterations on a computer: = 0.766660928
46
2-100. We need a method that is fair even if the caller knows which way the coin is biased (since
otherwise a single random call would still have probability 1/2 of matching the actual outcome).
Here it is: the caller pre-selects one of H-T or T-H as the guessed sequence of outcomes of two
consecutive flips. The coin is then flipped twice: if it results in two different outcomes, the caller
attends the meeting iff his or her choice of the order of outcomes was correct. But if the two flips
have the same outcome, then the coin is flipped twice again, repeated as needed until a pair of
different outcomes is obtained. Note that even if the coin is much more likely to come up heads
(and even if the caller knows this), the sequence H-T is equally likely as T-H: they both have
probability P(H) P(T) , since the two tosses are independent.
2-101.
open
close
total
Kwik Save Somerfield
893
0
107
424
1000
424
total
893
531
1424
a. 107 / 531 = .202
b. 531 / 1424 + 1000 / 1424 – 107 / 1424 = 1.00
c. 0
2-102. 0.089 = 1700 /19100
2-103. Since there are 38 members, a simple majority would be 20 votes. P(yes) = 0.25
P(pass) = 0.0002
P(Northwest compete successfully) = 0.00014
2-104. S = Species Survives G = Project goes ahead
Given Probabilities: P(–S | G) = 0.6 P(G) = 0.7 P(S | –G) = 1.0
P(S) = P(S | G) P(G) + P(S | –G) P(–G)
= (0.4)(0.7) + (1.0)(0.3) = 0.58
47
2-105. (Use template: Bayesian Revision.xls)
Bayesian Revision based on Empirical Conditional
Probabilities
Machine A Machine B Machine C
s1
s2
s3
s4 s5 s6
Prior Probability
0.26
0.38
0.36
s7
s8
Total
1
Conditional Probabilities
defective
good
P(I1 | .)
P(I2 | .)
P(I3 | .)
P(I4 | .)
P(I5 | .)
Total
s1
0.08
0.92
s2
0.05
0.95
s3
s4 s5 s6
0.04
0.96
s7
s8
1
1
1
0 0 0
0
0
s2
0.0190
0.3610
s3
0.0144
0.3456
s4 s5 s6
s7
s8
s2
0.3506
0.3817
s3
0.2657
0.3654
s4 s5 s6
s7
s8
Joint Probabilities
s1
I1 0.0208
I2 0.2392
I3
I4
I5
Posterior Probabilities
s1
P(. | I1) 0.3838
P(. | I2) 0.2529
P(. | I3)
P(. | I4)
P(. | I5)
1) Given the part is defective, the probability it is from machine A is 0.3838.
2) Given the part is good, the probability that it is from machine B is 0.3817.
48
Marginal
0.0542
0.9458