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Chapter 6
Exercises 6.1 – 6.12
6.1
A chance experiment is any activity or situation in which there is uncertainty about which of two or
more possible outcomes will result.
Consider tossing a coin two times and observing the outcome of the two tosses. The possible
outcomes are (H, H), (H, T), (T, H), and (T, T), where (H, H) means both tosses resulted in Heads,
(H, T) means the first toss resulted in a Head and the second toss resulted in a Tail, etc. This is an
example of a chance experiment with four possible outcomes.
6.2
The collection of all possible outcomes of a chance experiment is the sample space for the
experiment.
The sample space for the experiment described in Problem 6.1 is {(H, H), (H, T), (T, H), (T, T)}.
6.3
a.
Sample space = {(A, A), (A, M), (M, A), (M, M)}.
b.
A
A
M
A
M
M
6.4
c.
B = {(A, A), (A, M), (M, A)}
C = {(A, M), (M, A)}
D = {(M, M)}
Only D is a simple event.
d.
B and C = {(A, M), (M, A)} = C
B or C = {(A, A), (A, M), (M, A)} = B
a.
A = {Head oversize, Prince oversize, Slazenger oversize, Wimbledon oversize, Wilson
oversize}
b.
B = {Wimbledon midsize, Wimbledon oversize, Wilson midsize, Wilson oversize}
c.
In part b, Not B = {Head midsize, Head oversize, Prince midsize, Prince oversize, Slazenger
midsize, Slazenger oversize}
d.
Wimbledon oversize, Wilson midsize, Wilson oversize}
e.
B and C = { Wilson midsize, Wilson oversize }
125
f.
Midsize
Head
Oversize
Prince
Midsize
Oversize
Slazenger
Midsize
Oversize
Wimbledon
Midsize
Oversize
Wilson
Midsize
Oversize
6.5
a.
# of defective tires
# of defective headlights
0
0
1
2
0
1
1
2
0
2
1
2
0
3
1
2
0
4
1
2
126
b.
Ac = {(0,2), (1,2), (2,2), (3,2), (4,2)} where (i,j) means that the number of defective tires is i
and the number of defective headlights is j.
A ∪ B = {(0,0), (1,0), (2,0), (3,0), (4,0), (0,1), (1,1), (2,1), (3,1), (4,1), (0,2), (1,2)} where (i,j)
means that the number of defective tires is i and the number of defective headlights is j.
A ∩ B = {(0,0), (1,0), (0,1), (1,1)}
c.
6.6
a.
C = {(4,0), (4,1), (4,2)}. A and C are not disjoint because the outcomes (4,0) and (4,1) are in
both A and C. B and C are disjoint because the outcomes in B are of the form (i,j) where i =0
or 1 but the outcomes in C are of the form (i,j) where i=4. Recall that the notation (i,j) means
that the number of defective tires is i and the number of defective headlights is j.
First book selected
Second book selected
2
1
3
4
1
2
3
4
1
3
2
4
1
4
2
3
b.
A = {(1,2), (1,4), (2,1), (2,3), (2,4), (3,2), (3,4), (4,1), (4,2), (4,3), (4,4)} where (i,j) means that
copy i was selected for the 2-hour service and copy j was selected for the overnight loan.
c.
B = {(1,3), (1,4), (2,3), (2,4), (3,1), (3,2), (4,1), (4,2)} where (i,j) means that
copy i was selected for the 2-hour service and copy j was selected for the overnight loan.
127
6.7
a.
First book
selected
Second book
selected
Third book
selected
3
2
4
5
1
3
4
5
3
1
4
5
2
3
4
5
3
4
5
6.8
b.
Exactly one book is examined when the first selected book is copy 3, 4, or 5. So A = {3,4,5}.
c.
C = {5, (1,5), (2,5), (1,2,5), (2,1,5)}. The notation used is as follows. The outcome 5 means
that copy 5 is selected first. The outcome (1,5) means that copy 1 is selected first and copy 5
is selected second, and so on.
a.
N, DN, DDN, DDDN, DDDDN are five possible outcomes. Here DDN means that the first
battery selected is defective, the second battery selected is defective, and the third battery
selected is not defective, etc.
b.
The number of outcomes in the sample space is infinite.
c.
E = {DN, DDDN, DDDDDN, ......... }. This event consists of all outcomes in which the first n
batteries selected are all defective with n being an odd number, followed by a nondefective
battery.
128
6.9
6.10
a.
A = {NN, DNN, NDN}
b.
B = {DDNN, DNDN, NDDN}
c.
The number of possible outcomes is infinite.
a.
The 27 possible outcomes are (1,1,1), (1,1,2), (1,1,3), (1,2,1), (1,2,2), (1,2,3), (1,3,1), (1,3,2),
(1,3,3), (2,1,1), (2,1,2), (2,1,3), (2,2,1), (2,2,2), (2,2,3), (2,3,1), (2,3,2), (2,3,3), (3,1,1), (3,1,2),
(3,1,3), (3,2,1), (3,2,2), (3,2,3), (3,3,1), (3,3,2), (3,3,3).
b.
A = {(1,1,1), (2,2,2), (3,3,3)}
c.
B = {(1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1)}
d.
C = {(1,1,1), (1,1,3), (1,3,1), (1,3,3), (3,1,1), (3,1,3), (3,3,1), (3,3,3)}
e.
Bc = {(1,1,1), (1,1,2), (1,1,3), (1,2,1), (1,2,2), (1,3,1), (1,3,3), (2,1,1), (2,1,2), (2,2,1), (2,2,2),
(2,2,3), (2,3,2), (2,3,3), (3,1,1), (3,1,3), (3,2,2), (3,2,3), (3,3,1), (3,3,2), (3,3,3)}
Cc = {(1,1,2), (1,2,1), (1,2,2), (1,2,3), (1,3,2), (2,1,1), (2,1,2), (2,1,3), (2,2,1), (2,2,2), (2,2,3),
(2,3,1), (2,3,2), (2,3,3), (3,1,2), (3,2,1), (3,2,2), (3,2,3), (3,3,2)}
A
∪ B = {(1,1,1), (2,2,2), (3,3,3), (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1)}
A ∩ B = the empty set. There are no outcomes common to A and B.
A ∩ C = {(1,1,1), (3,3,3)}
6.11
a.
E1
E2
E3
129
b.
The required region is the area common to all three circles.
E1
E2
E3
c.
E1
E2
E3
130
d.
E1
E2
E3
e.
E1
E2
E3
131
f.
E1
E2
E3
6.12
a.
( A ∪ B) c
A
B
Ac ∩ B c
A
We see that
B
( A ∪ B ) c = Ac ∩ B c .
132
b.
( A ∩ B) c
A
B
A
B
Ac ∪ B c
We see that
( A ∩ B ) c = Ac ∪ B c .
Exercises 6.13 – 6.27
6.13
6.14
6.15
6.16
a.
A = {(C,N), (N,C), (N,N)}. So P(A) = 0.09 + 0.09 + 0.01 = 0.19.
b.
B = {(C,C), (N,N)}. So P(B) = 0.81 + 0.01 = 0.82.
a.
P(picking 9, then 1, then another 1) =
b.
Classical or relative frequency.
a.
In the long run, 1% of all people who suffer cardiac arrest in New York City survive.
b.
1% of 2329 is .01(2329) = 23.29. So in this study there must have been only 23 or 24
survivors.
1 1 1
= 0.001
• •
10 10 10
The article sounds as though you will fly safely for 26,000 years and crash on the very last day.
Obviously no-one lives that long. In fact, there is a very small risk with each flight.
133
6.17
6.18
6.19
6.20
6.21
6.22
500,000
= .0119
42,005,100
a.
P(twins) =
b.
P(quadruplets) =
c.
P(more than a single child) =
a.
In the long run, 35% of all customers who purchase a tennis racket at this store will have a
grip size equal to 4 ½ inches.
b.
P(grip size is not 4 ½ inches) = 1 - 0.35 = 0.65.
c.
P(oversize head) = 0.20 + 0.15 + 0.20 = 0.55. In the long run, 55% of all customers who
purchase a racket at this store will buy a racket with an oversize head.
d.
P(grip size is at least 4 ½ inches) = 0.20 + 0.15 + 0.15 + 0.20 = 0.70.
a.
P(owns shares in balanced fund) = 0.07.
b.
P(owns shares in a bond fund) = 0.15 + 0.10 + 0.05 = 0.30.
c.
P(does not own shares in a stock fund) = 1 - P(owns shares in a stock fund)
= 1 - (0.18 + 0.25) = 0.57.
a.
P(request is for one of the three B’s) = 0.05 + 0.26 + 0.09 = 0.40.
b.
P(request is not for one of the two S’s) = 1 - P(request is for one of the two S’s)
= 1 - (0.12+0.07) = 0.81.
c.
P(request is for a composer who wrote at least one symphony)
= 1 - P(request is not for Bach or Wagner) = 1-(0.05 + 0.01) = 0.94.
100
= .00000238
42,005,100
505,100
= .012
42,005,100
No, it is not true. The addition rule is for disjoint events. The events “midsize” and “4 3/8 inch grip” are
not disjoint since P(midsize AND 4 3/8 inch grip) = 0.10 is greater than zero.
1287
= 0.000495198 .
2598960
P(hand will consist entirely of a single suit) = P(all spades) + P(all clubs) + P(all diamonds)
+ P(all hearts) = 0.000495198 + 0.000495198 + 0.000495198 + 0.000495198 =
0.001980792.
a.
P(hand will consist entirely of spades) =
b.
P(hand consists entirely of spades and clubs with both suits represented)
63206
= 0.0243197.
=
2598960
c.
P(hand contains cards from exactly two suits) = P(spades and clubs) + P(spades and
diamonds) + P(spades and hearts) + P(clubs and diamonds) + P(clubs and hearts) +
P(diamonds and hearts) = 6(0.0243197) = 0.1459182.
134
6.23
a.
The 24 outcomes (including the outcome (2, 4, 3, 1) ) are
(1, 2, 3, 4), (1, 2, 4, 3), (1, 3, 2, 4), (1, 3, 4, 2), (1, 4, 2, 3), (1, 4, 3, 2)
(2, 1, 3, 4), (2, 1, 4, 3), (2, 3, 1, 4), (2, 3, 4, 1), (2, 4, 1, 3), (2, 4, 3, 1)
(3, 1, 2, 4), (3, 1, 4, 2), (3, 2, 1, 4), (3, 2, 4, 1), (3, 4, 1, 2), (3, 4, 2, 1)
(4, 1, 2, 3), (4, 1, 3, 2), (4, 2, 1, 3), (4, 2, 3, 1), (4, 3, 1, 2), (4, 3, 2, 1)
b.
The required outcomes are
(1, 2, 4, 3), (1, 4, 3, 2), (1, 3, 2, 4), (4, 2, 3, 1), (3, 2, 1, 4), (2, 1, 3, 4).
P(exactly two of the 4 books will be returned to the correct owner) =
6.24
6.25
6.26
6.27
6
= 0.25 .
24
c.
The outcomes corresponding to the event “exactly one students receives his/her book” are
(1, 3, 4, 2), (1, 4, 2, 3), (2, 3, 1, 4), (2, 4, 3, 1), (3, 1, 2, 4), (3, 2, 1, 4), (4, 1, 3, 2), (4, 2, 1, 3).
8
= 0.33333.
Hence the probability of this event =
24
d.
If three students receive their books, then the fourth student must receive his/her book also.
So, the required probability is 0.
e.
The outcomes corresponding to this event are (1, 2, 3, 4), (1, 2, 4, 3), (1, 3, 2, 4), (1, 4, 3, 2),
7
= 0.291667.
(2, 1, 3, 4), (3, 2, 1, 4), (4, 2, 3, 1). Hence the required probability is
24
a.
The simple events in this experiment are
(C, D, P), (C, P, D), (D, C, P), (D, P, C), (P, C, D), (P, D, C).
All six outcomes may be considered equally likely so the probability assigned to any one
outcome is 1/6.
b.
P(C is ranked first) = 2/6 = 1/3.
c.
P(C is ranked first and D is ranked last) = 1/6.
a.
Suppose the students selected are from Mathematics and Physics. We abbreviate this
outcome by {M, P}. With this notation, the ten possible outcomes are {B, C}, {B, M}, {B, P},
{B, S}, {C, M}, {C, P}, {C, S}, {M, P}, {M, S}, {P, S}.
b.
Probability of each simple event = 1/10 = 0.1.
c.
P(one member is from Statistics) = 4/10 = 0.4.
d.
Assuming that the laboratory sciences are Biology, Chemistry, and Physics, the required
probability is 3/10 = 0.3.
a.
The four simple events are (brand M, 2 heads), (brand M, 4 heads), (brand Q, 2 heads),
(brand Q, 4 heads).
b.
P(brand Q with 2 heads) = 32% = 0.32.
c.
P(brand M) = 25% + 16% = 41% = 0.41.
a.
P(O1) + P(O2) + P(O3) + P(O4) + P(O5) + P(O6) = 1 implies that p + 2p + p + 2p + p + 2p =1.
So. 9p = 1 which yields p = 1/9. The probabilities of the six simple events are as follows:
P(O1) = P(O3) = P(O5) = 1/9
P(O2) = P(O4) = P(O6) = 2/9.
135
b.
P(odd number) = P(1 or 3 or 5 ) = 3(1/9) = 3/9 = 1/3 = 0.3333.
P(at most 3) = P(1 or 2 or 3) = 1/9 + 2/9 + 1/9 = 4/9 = 0.4444.
c.
Now, P(O1) + P(O2) + P(O3) + P(O4) + P(O5) + P(O6) = 1 implies that
c + 2c + 3c + 4c + 5c + 6c = 1, from which we get c = 1/21.
Hence P(O1) = 1/21; P(O2) = 2/21; P(O3) = 3/21; P(O4) = 4/21; P(O5) = 5/21; P(O6) = 6/21.
P(odd number) = 1/21 + 3/21 + 5/21 = 9/21 = 0.428571.
P(at most 3) = 1/21 + 2/21 + 3/21 = 6/21 = 0.285714.
Exercises 6.28 – 6.40
6.28
6.29
a.
P (E | F ) =
P (E ∩ F ) 0.54
=
= 0.90.
0.6
P (F )
b.
P (F | E ) =
P (E ∩ F ) 0.54
=
= 0.7714.
P (E )
0.7
a.
Let E = the event that critic 1 gave a “thumbs up” to the chosen movie and F = the event that
critic 2 gave a “thumbs up” to the selected movie. Then P(E ) = 15/60 = ¼, P(F ) = 20/60 =
1/3, and P( E ∩ F ) = 10 / 60 = 1 / 6.
So, the required probability = P (F | E ) =
b.
P (E ∩ F ) 1/ 6
=
= 0.6667.
P (E )
1/ 4
Using the given information we construct the following table showing the various possible
outcomes and their frequencies of occurrence.
Critic 1 “thumbs up”
Critic 1 “thumbs down”
Total
Critic 2
“thumbs up”
10
10
20
Critic 2
“thumbs down”
5
35
40
P (thumbs down from critic 1|thumbs down from critic 2) =
Calculating the totals gives:
6.30
24
= .24
100
a.
P (tattoo ) =
b.
P (tattoo | age 18 − 29) =
c.
d.
Total adults aged 18-29:
Total adults aged 30-50:
Total adults with tattoo:
Total without tattoo:
Total in survey:
18
= .36
50
6
P (tattoo | age 30 − 50) =
= .12
50
18
P (age 18 − 15 | tattoo ) =
= .75
24
136
Total
15
45
60
35
= 0.875.
40
50
50
24
76
100
6.31
The statement in the article implies the following two conditions:
(1) P (D | Y c ) > P (D | Y ) and
(2) P (Y ) > P (Y c ) .
This claim is consistent with the information given in I but not with any of the others. In II and III,
condition (1) is violated; in IV, condition (2) is violated; in V and VI, both conditions are violated.
6.32
It appears to be true that most basketball players are over 6 feet tall, but many people over 6 feet tall
are not basketball players. So it is reasonable to expect P ( A | B ) > P ( B | A) .
6.33
a.
b.
c.
P(F|FP)=
432
= 0.769
562
390
= 0.890
438
Ultrasound appears to be more reliable in predicting boys
P(M|MP)=
6.34
1. P(M) = 0.021 2. P(M|B) = 0.043 3. P(M|W) = 0.07
6.35
a.
i.
ii.
135
= 0.74176.
182
173 + 206
Required conditional probability =
= 0.8594.
210 + 231
Required conditional probability =
b.
Properly restrained
Iowa City (city)
Not Properly restrained
Properly restrained
Iowa City (interstate)
Not properly restrained
6.36
The column total for “very harmful” is 224 and the row totals for “current smoker” is 96, “former
smoker” is 99 and “never smoked” is 99. There were 294 people who questioned.
99
224
60
78
86
a.
= 0.3367 b.
= 0.7619 c.
= 0.625 d.
= 0.7879 e.
= 0.8687
294
294
96
99
99
f . The probabilities in c,d and e differ in that the probability that a randomly selected individual from
that group perceives smoking as very harmful decreases if he or she is a former smoker and
decreases even more if he or she has never smoked. This is probably not a surprise.
6.37
a.
b.
c.
0.1 + 0.175 = 0.275
0.1
= 0.2
0.5
0.325
= 0.65
0.5
137
e.
0.325
= 0.448
0.725
The probabilities in c and d are not equal because the events “wearing a seatbelt” and “being
female” are not independent. It is clearly more likely an adult will regularly use a seatbelt if
they are female than if they are male.
a.
i.
d.
6.38
b.
6.39
6.40
41
100
41
500
=0.167 ii.
= 0.068 iii.
= 0.241
iv.
= 0.833
600
600
600
172
41
18
v.
= 0.41 vi.
= 0.18
100
100
This clearly shows that the probability of randomly chosen senior citizen using a seatbelt is
far lower than someone in the 18 to 24 year age group.
456
215
241
= 0.717 iii. P(S|B) =
= 0.803
= 0.76 ii. P(S|A) =
300
300
600
iv. Drug B appears to have a higher survival rate
a.
i. P(S) =
b.
i. P(S) =
c.
i. P(S) =
d.
Women seem to respond well to both treatments. Men seem to respond very well to
treatment A but not so well to treatment B. When the data is combined, the small quantity of
data collected for men on treatment B - only 20% of those men who had treatment A,
becomes “lost” in the rest of the data.
140
120
20
= 0.600 iii. P(S|B) =
= 0.5
= 0.583 ii. P(S|A) =
240
200
40
iv. Drug A appears to have the higher survival rate.
316
95
221
= 0.95 iii. P(S|B) =
= 0.85
= 0.878 ii. P(S|A) =
360
100
260
iv. Drug A appears to have the higher survival rate
Suppose the probability that a driver that does not live close to a restaurant will have an accident is
x . So P( A | RC ) = x . If a driver who does live close to a restaurant is 30% more likely to have an
accident, then P ( A | R ) = 1.3 x . The difference between these is 0.3x which divided by x; P( A | R ) ,
will give 0.3. The 4th choice is the correct one.
C
Exercises 6.41 – 6.58
6.41
a.
About 85% of all the past calls were for medical assistance.
b.
P(call is not for medical assistance) = 1 – 0.85 = 0.15.
c.
P(two successive calls are both for medical assistance) = (0.85 )(0.85) = 0.7225.
d.
P(first call is for medical assistance and second call is not for medical assistance)
= (0.85)(0.15) = 0.1275.
e.
P(exactly one of two calls is for medical assistance) = P(first call is for medical assistance
and the second is not) + P(first call is not for medical assistance but the second is) =
(0.85)(0.15) + (0.15)(0.85) = 0.255.
f.
Probably not. There are likely to be several calls related to the same event—several reports
of the same accident or fire that would be received close together in time.
138
6.42
a.
In the long run, 30 out of every 100 accidents reported to the police that result in a death
were a single-vehicle rollover.
b.
In the long run, 54 out of every 100 accidents reported to the police that result in a death
were a frontal collision.
c.
P(R|D) = 0.3, P(R) = 0.06. As P(R|D)≠P(R), the events R and D are not independent.
d.
P(F∩D) is only equal to P(F)P(D) if the events F and D are independent. As P(F|D)≠P(F):
0.54≠0.6 , the events F and D are not independent⇒ P(F∩D) ≠ P(F)P(D).
e.
No, F is the event that the selected accident was a frontal collision. Rc is the event that the
selected accident was not a single-vehicle rollover. These are not the same. There are
accidents that are one of these, or both or neither.
6.43
No, a randomly selected adult is more likely to experience pain daily if they are a woman. The events
are dependent.
6.44
Let A = the event that the selected student (from all kindergartners who were tested for TB) has TB;
B = the event that the selected student (selected from all kindergartners who were tested for TB) is a
recent immigrant. Based on the information given, we have P(A | B) = 0.0075 and P(A) = 0.0006. For
independence, we need P(A | B) = P(A) which is not true here. So the two outcomes under
consideration are dependent.
6.45
Let A = the event that the selected adult is a female and B = the event that the selected adult favors
stricter gun control. Based on the given information we estimate P(B | A) = 0.66 and P(B) = 0.56.
Since P(B | A) and P(B) are not equal, we conclude that the two events A and B are dependent
outcomes.
6.46
Let A = the event that the selected smoker who is trying to quit uses a nicotine aid and B = the event
that the selected smoker who has attempted to quit begins smoking again within two weeks. From
the given information we can deduce the following.
P ( A) = 0.113 ; P( B | A c ) = 0.62 ; P ( B | A) = 0.6 . So P ( A ∩ B ) = P ( A) P ( B | A) = 0.0678.
From this we get,
P( A c ∩ B) = P( A c ) P( B | A c ) = (1 − 0.113)(0.62) = 0.54994.
Therefore
P( B) = P( A ∩ B) + P( A c ∩ B) = 0.61774.
Since P( B | A) is not equal to P (B ) , A and B are dependent outcomes.
6.47
P(jury duty call two years in a row) = (0.15)(0.15) = 0.0225.
P(jury duty call three years in a row) = (0.15)(0.15)(0.15) = 0.003375.
6.48
a.
Assuming that whether Jeanie forgets to do one of her ‘to do’ list items is independent of
whether or not she forgets any other of her ‘to do’ list items, the probability that she forgets all
three errands = (0.1)(0.1)(0.1) = 0.001.
b.
P(remembers at least one of her three errands) = 1 – P(she forgets all three errands)
= 1 – 0.001 = 0.999.
c.
P(remembers the first errand, but not the second or the third) = (0.9)(0.1)(0.1) = 0.009.
139
6.49
a.
P(none of the 10 calls lead to a reservation) = (0.7)10 = 0.0282475249.
b.
It was assumed that the outcomes of the different calls are independent, i.e., whether one
call
resulted in a reservation or not has no influence on the results of other calls.
6.50
6.51
6.52
6.53
c.
P(at least one call leads to a reservation) = 1 – 0.0282475249 = 0.9717524751.
a.
P(student has a Visa card) = 7000/10000 = 0.7
b.
P(student has both cards) = 5000/10000 = 0.5
c.
P(student has both cards | student has Visa card) = 5000/7000 = 0.714286
d.
P(student has a Visa card & a Master Card) = 0.5. For the two events to be independent we
must have P(student has a Visa card & a Master Card ) = P(has Visa card) P(has a Master
Card) = (7000/10000) (6000/10000) = 0.42 which is not equal to 0.5. So the two events are
dependent.
e.
If only 4200 has both cards, then the condition P(student has a Visa card & a Master Card ) =
P(has Visa card) P(has a Master Card) holds, so the two events are independent.
a.
P( 1-2 subsystem works) = P(1 works) P(2 works) = (0.9)(0.9) = 0.81.
b.
P( 1-2 subsystem does not work ) = 1 – P( 1-2 subsystem works) = 1 – 0.81 = 0.19.
P( 3-4 subsystem does not work) = 1 – P( 3-4 subsystem works) = 1 – (0.9)(0.9) = 0.19.
c.
Using the results of part b, P( system won’t work) = P( 1-2 subsystem doesn’t work & 3-4
subsystem doesn’t work ) = P( 1-2 subsystem doesn’t work ) P( 3-4 subsystem doesn’t work)
= (0.19)(0.19) = 0.0361.
So P( system will work ) = 1 – P( system won’t work) = 1 – 0.0361 = 0.9639.
d.
If a 5-6 subsystem were added, then P( system will work ) = 1 – P( system won’t work )
= 1 – (0.19)(0.19)(0.19) = 0.993141.
e.
If there were 3 components in series in each subsystem, then P( system will work )
= 1 – P( system won’t work) = 1 – P( first subsystem doesn’t work) P( second subsystem
doesn’t work). The probability that the first subsystem doesn’t work = 1- (0.9)(0.9)(0.9) = 1 –
0.729 = 0.271. This is also the probability that the second subsystem won’t work. So
P(system works) = 1 – (0.271)(0.271) = 0.926559.
a.
i.
ii.
iii.
iv.
v.
b.
P(F|C) is not equal to P(F) so F and C are dependent.
c.
P(F|O) is not equal to P(F) so F and O are dependent.
a.
The expert assumed that a valve stem could be in any one of the “clock positions” with equal
probability, so that there is a 1 in 12 chance for each valve to be in any given clock hour
position. In addition, the expert assumed that the positions of the two valve stems are
independent. Hence, the expert computed the probability of a “match” on both valves to be
the product (1/12)(1/12) = 1/144.
P(F) = 0.51
P(F | C) = 0.56
P(F | Cc ) = 0.45
P(F | O) = 0.36
P(F | Y) = 0.72
140
6.54
6.55
6.56
b.
The assumption of “independence” of the two valve stems is highly questionable because the
valve stems turn in a “synchronized” manner. This would make the probability of the valve
stems being in their original positions greater than the 1/144 value computed by the expert.
a.
P(A wins both matches and B beats C) = P(A beats B) P(A beats C) P(B beats C) =
(0.7)(0.8)(0.6) = 0.336.
b.
P(A wins both her matches) = P(A beats B) P(A beats C) = (0.7)(0.8) = 0.56.
c.
P(A loses both her matches) = P(A loses to B) P(A loses to C) = (1 – 0.7)(1 – 0.8) = 0.06.
d.
There are two ways this can happen – A beats B, B beats C, C beats A OR A beats C, B
beats A, and C beats B. So the required probability = (0.7) (0.6) (1 – 0.8) + (0.8) ( 1 – 0.7) (1
– 0.6) = 0.084 + 0.096 = 0.18.
a.
If the first board is defective, then the probability that the second board is defective =
39/4999. If the first board is not defective, then the probability that the second board is
defective = 40/4999. Since the probability that the second board is defective changes
depending on whether or not the first board is defective, the events E1 and E2 are dependent.
b.
P(not E1 ) = 1 – P(E1) = 1 - (40/5000) = 0.992.
c.
P(E2 | E1 ) = 39/4999 and P(E2 | not E1 ) = 40/4999. They are nearly equal.
d.
Yes.
P(small size of brand B1 ) = P(small) P(brand B1 ) = (0.30)(0.40) = 0.12. The other probabilities are
obtained in a similar manner. The results are shown in the following table.
B1
S
0.12
Size
M
0.20
L
0.08
0.40
B2
0.18
0.30
0.12
0.60
0.30
0.50
0.20
1.00
Brand
6.57
6.58
a.
P(CC) = (50/800)(50/800) = 0.00391.
b.
P(CC) = P(first answer is correct) P(second answer is correct | first answer is correct)
= (50/800)(49/799) = 0.00383. This is very close to the probability computed in part a.
For sampling with replacement, P(both answers are correct) = (50/100)(50/100) = 0.25.
For sampling without replacement P(both answers are correct) = (50/100) (49/99) = 0.24747.
Again, the two probabilities are quite close to each other, but not as close as the two probabilities in
parts b and c of Problem 6.57..
Exercises 6.59 – 6.74
6.59
a.
P(E) = 6/10 = 0.6
b.
P(F | E) = 5/9 = 0.5556
c.
P(E and F) = P(E) P(F|E) = (6/10)(5/9) = 30/90 = 1/3
141
6.60
6.61
a.
P(E1 and E2) = (0.4)(0.3) = 0.12
b.
P(not E1 and not E2) = [1 – P(E1)][1 - P(E2)] = (0.6)(0.7) = 0.42
c.
P(successful in at least one of the two bids) = 1 – (both bids unsuccessful)
= 1 – 0.42 = 0.58. (using the result of part b)
a.
P(individual has to stop at at least one light) = P(E ∪ F) = P(E) + P(F) – P(E ∩ F)
= 0.4 + 0.3 – 0.15 = 0.55.
b.
P(individual doesn’t have to stop at either traffic light) = 1 – P(must stop at at least one light)
= 1 – 0.55 = 0.45 (using the result of part a).
c.
P(stop at exactly one of the two lights) = P(stop at least one of the two lights) – P(stop at both
lights) = 0.55 – 0.15 = 0.40.
d.
P(E only) = P(E and not F) = P(E) – P(E ∩ F) = 0.4 – 0.15 = 0.25.
6.62
The event E ∩ F means that a randomly selected registered voter in a certain city has signed a
petition to recall the mayor and also actually votes in the recall election. P(E ∩ F) = P(F) P(E | F) =
(0.10)(0.80) = 0.08.
6.63
The likelihood of using a cell phone is different for of each type of vehicle; for instance if you drive a
van or SUV, you are more likely to use a cell phone than if you drive a pick-up truck.
6.64
a.
6.65
b.
4.5
4.1
= 0.523, P(S) =
= 0.477, P(R|F) = 0.018, P(R|S) = 0.03
8.6
8.6
P (S )P (R | S )
0.01431
=
= 0.6032
P(S|R) =
P (S )P (R | S ) + P (F )P (F | S ) 0.01431 + 0.0094
This is the probability that a randomly selected gun purchase background check resulted in a
blocked sale by a state or local agency.
a.
i. 0.5
b.
P(TD) = P(TD∩D) + P(TD∩C) = P(TD|D)P(D) + P(TD|C)P(C) =
= (.99)(.01) + (.99)(.01) = 2(.99)(.01) = .0198
P (C )P (TD | P (C )
(.99)(.01)
=
= 0.5
P(C|TD) =
P (TD | P (C ) + P (TD ) | P (D ) 2(.99)(.01)
Yes, it is consistent with the argument given in the quote.
c.
6.66
P(F) =
ii. 0.5
iii. 0.99
iv. .01
Letting P( E ∩ C ∩ L) = x , P ( E ∪ C ∪ L) = .49 so using the other segments of a Venn Diagram and
solving for x :
(.40-.23-.22+ x ) + (.23- x ) +(.30-.23-.21+ x ) +(.22- x ) + x +(.21- x )+(.25-.22-.21+ x ) =.49
x =20.
a P ( E ) = .4
b P (C ) = .3
c P( L) = .25 d P ( E and C) =.23
P( E c and C c and Lc ) = .51 f P ( E or C or L ) = .49 g P ( E | L) = .88
h P ( L | C ) = .70
i P ( E and C and L) = .20
j P ( E and L ) = .22
k P (C and L ) = .21 l P (C | E and L ) = .91
e
142
6.67
6.68
a.
In 4 consecutive years, there are 3 years of 365 days (1095 non leap days) and 1 year of 366
days (365 non leap days and one leap day) – a total of 1461 days, of which one, Feb 29th is a
leap day. Hence a leap day occurs once in 1461 days.
b.
If babies are induced or born by C-section, they are less likely to be born at weekends or on
holidays so they are not equally likely to be born on the 1461 days.
c.
1 in 2.1 million is the probability of picking a mother and baby randomly and finding both to be
a leap-day mom-baby (1/14612). The probability of a leap year baby becoming a leap year
mom means that the mom’s birthday is already fixed, so the hospital spokesperson’s
probability is too small.
a.
Let E = the event that the selected employee uses drugs and F = the event that the selected
employee tests positive. Then P(F|Ec ) = P(false positive) = 0.05 and P(Fc |E) = 0.10. So
P(F|E) = 1 – P(Fc |E) = 1 – 0.10 = 0.90. Also, P(E) = 0.10. We need P(E and F). We have P(E
and F) = P(E)P(F | E) = (0.10)(0.90) = 0.09.
b.
P(Ec ∩ F) = P(Ec)P(F | Ec ) = (1 – 0.10)(0.05) = (0.9)(0.05) = 0.045.
c.
P(F) = P(E ∩ F) + P(Ec ∩ F). Now, P(E ∩ F) = 0.09 from part a and P(Ec ∩ F) = 0.045
from part b. Therefore P(F) = 0.09 + 0.045 = 0.135.
d.
P(uses drugs | tests positive) = P(E|F) =
P(E ∩ F)
0.09
=
= 0.67 , using the results from
0.135
P(F)
parts a and c.
6.69
a.
Assume that the results of successive tests on the same individual are independent of one
another. Let F1 = the event that the selected employee tests positive on the first test and F2 =
the event that he/she tests positive on the second test. Then, P(employee uses drugs and
test positive twice) = P(E)P(F1|E)P(F2|E) = (0.1)(0.90)(0.90) = 0.081.
b.
P(employee tests positive twice)
= P(employee uses drugs and tests positive twice)
+ P(employee doesn’t use drugs and tests positive twice)
= P(E) P(F1|E)P(F2|E) + P(Ec) P(F1|Ec)P(F2|Ec)
= (0.1)(0.90)(0.90) + (0.9)(0.05)(0.05) = 0.081 + 0.00225 = 0.08325.
c.
P(uses drugs | tested positive twice) =
P(uses drugsandtestspositive twice)
0.081
=
= 0.97297.
P(testspositive twice)
0.08325
d.
P(tests negative on the first test OR tests positive on the first test but tests negative on the
second test | does use drugs) = P(F1c | E ) + P(F1 and F2c |E )
= [1 – P(F1 |E )] + P(F1 |E ) [1 – P(F2 |E )]
= [1 – 0.9] + (0.9)(1 – 0.9) = 0.1 + 0.09 = 0.19
(using P(F1 |E ) = 0.90, from part a of Problem 6.68).
e.
The benefit of using a retest is that it reduces the rate of false positives from 0.05 to 0.02703
(obtained as 1 – 0.97297, using the result from part c), but the disadvantage is that the rate
of false negatives has increased from P(Fc |E) = 0.10 to 0.19 (see part d above). Retests also
involve additional expenses.
143
6.70
a.
Let a = the number of adults who are full-time workers and drug users, b = number of adults
who work full-time and do not use drugs, c = number of adults who do not work full-time but
use drugs, and d = number of adults who do not work full time and do not use drugs. The
given statements imply that a/(a+b) = 0.08 and a/(a+c) = 0.7. This is possible if
b = (0.92/0.08)a = (11.5)a and a = (7/3)c. So, this is indeed possible – for instance, take
c = 30, a = 70, b = 805, d = 7095. See the following table for an example. (There are other
possible values of a, b, c, d, for which the given statements hold also.)
Use drugs
Does not use drugs
Total
6.71
Works full-time
70
805
875
Does not work full-time
30
7095
7125
Total
100
7900
8000
b.
Based on the given information, P(D|E) is estimated to be 0.08 and P(E|D) is estimated to be
0.70.
c.
It is not possible to determine P(D) from the information given because the given information
translates into the two equations -- a/(a+b) = 0.08 and a/(a+c) = 0.7, which do not have a
unique solution. See part a. If, for instance, any one of the values a, b, c, d is given, then we
can determine the rest of the values and thereby obtain P(D) = (a+b)/(a+b+c+d) =
(a+b)/8000.
Based on the given information we can construct the following table.
Basic
Deluxe
Model
Model
Buy extended warranty
12%
30%
Do not buy extended warranty
28%
30%
Total
40%
60%
Total
42%
58%
100%
So P(Basic model | extended warranty) = 12/42 = 0.285714.
6.72
a.
P(satisfied) = P(student took the class during fall quarter and was satisfied with book) +
P(student took the class during winter quarter and was satisfied with the book) + P(student
took the class during spring quarter and was satisfied with the book) = (200/1000) +
(150/1000) + (160/1000) = 0.51.
b.
P(took the class during fall quarter | satisfied with book)
=
P(took the class during fall & satisfied)
(200 /1000) 0.2
=
=
= 0.3922 .
0.51
0.51
P(satisfied)
P(took the class during winter quarter | satisfied with book)
=
P(took the class during winter & satisfied)
(150 /1000) 0.15
=
=
= 0.2941 .
0.51
0.51
P(satisfied)
P(took the class during spring quarter | satisfied with book)
=
P(took the class during spring & satisfied)
(160 /1000) 0.16
=
=
= 0.3137 .
0.51
0.51
P(satisfied)
Hence, given that the selected student was satisfied with the textbook, the most likely book
used was by Professor Mean. This has the highest conditional probability (0.3922) among the
three possibilities. The least likely would be the book by Professor Median.
144
6.73
Let A = the event that he takes the small car; B = the event that he takes the big car; C = the event
that he is on time. Then, based on the given information we have P(A) = 0.75, P(B) = 0.25, P(C|A) =
0.9, and P(C|B) = 0.6. We need P(A|C). Using Bayes’ Rule we get
P(C|A)P(A)
(0.9)(0.75)
P(A|C) =
=
= 0.8182.
P(C|A)P(A) + P(C|B)P(B) (0.9)(0.75) + (0.6)(0.25)
6.74
a.
Let A = the event that the selected individual has the disease and B = the event that the test
result is positive. Then it is given that P(A) = 0.001, P(B|A) = 0.95, and P(Bc|Ac) = 0.90.
So P(B|Ac)= 1 – 0.90 = 0.10.
Positive Test
0.95
Has disease
0.001
0.05
Negative Test
Positive Test
0.10
0.999
Doesn’t have disease
0.90
Negative Test
b.
P( has disease and positive test) = (0.001)(0.95) = 0.00095.
c.
P(positive test) = P(has disease and positive test) + P(doesn’t have disease and positive test)
=(0.001)(0.95) + (0.999)(0.10) = 0.00095 + 0.0999 = 0.10085.
d.
Using the definition of conditional probability, we get
P(has disease & positive test) 0.00095
=
= 0.00942.
P(has disease | positive test) =
P(positive test)
0.10085
Yes, the result is surprising because, the test has a fairly high probability of coming up
positive when the individual has the disease and a fairly small probability of coming up
positive when the individual doesn’t have the disease. So, given that the test result is
positive, one would have thought there is a fairly high chance that the subject has the
disease. However, this probability is only 0.00942. The reason this probability is so small is
that the incidence rate of the disease is very small and the positive test result is getting
“downweighted” because of this prior knowledge that the disease is “rare”.
Exercises 6.75 – 6.83
6.75
a.
P(on time delivery in Los Angeles) = 425/500 = 0.85.
b.
P(late delivery in Washington, D.C) = (500 – 405)/500 = 95/500 = 0.19.
c.
Assuming that whether or not one letter is delivered late is “independent” of the on-time
delivery status of any other letter, we calculate P(both letters delivered on-time in New York)
= (415/500)(415/500) = 0.6889. (The independence assumption may not be a valid
assumption).
d.
P(on-time delivery nationwide) = 5220/6000 = 0.87.
145
6.76
a.
Probatio
n
Yes
No
Total
6.77
6.78
2.5 -<
3.0
0.10
0.09
0.19
High School GPA
3.0 -< 3.5
3.5 and Above
0.11
0.27
0.38
0.06
0.37
0.43
Total
0.27
0.73
1.00
b.
P(academic probation) = 0.10 + 0.11 + 0.06 = 0.27.
c.
P(high GPA of 3.5 or above) = 0.43.
d.
P(GPA of 3.5 or above AND academic probation) = 0.06.
P(GPA of 3.5 or above)P(academic probation) = (0.43)(0.27) = 0.1161.
Since P(GPA of 3.5 or above AND academic probation) is not equal to P(GPA of 3.5 or
above)P(academic probation), we conclude that the events are dependent.
e.
P(GPA between 2.5 and 3.0 | academic probation) = 0.10/0.19 = 0.5263
f.
P(GPA of 3.5 or above AND academic probation) = 0.06/0.43 = 0.1395.
a.
P(male) = (200+800+1500+1500+900+1500+300)/17000 = 6700/17000 = 0.3941.
b.
P(agriculture) = 3000/17000 = 0.1765.
c.
P(male AND from agriculture) = 900/17000 = 0.0529.
d.
P(male AND not from agriculture) = 5800/17000 = 0.3412.
Total Monterey County = 369,000
Total San Luis Obispo County = 236,000
Total Santa Barbara County = 392,000
Total Ventura County = 736,000
Total Caucasian = 1,003,000
Total Hispanic = 528,000
Total Black = 61,000
Total Asian = 122,000
Total American Indian = 19,000
Total Count = 1,733,000
a.
736,000
= 0.4247
1,733,000
b.
231,000
= 0.3139
736,000
c.
d.
e.
231,000
= 0.4375
528,000
9000
= 0.0052
1,733,000
P (Asian or from San Luis Obispo) =
122000
236000
9000
+
−
= 0.2014
1,733,000 1,733,000 1,733,000
146
f.
P (Asian or from San Luis Obispo, but not both) =
122000
236000
18000
+
−
= 0.1962
1,733,000 1,733,000 1,733,000
g.
P (both Caucasian) = (
h.
P (neither Caucasian) = (
i.
P (exactly one Caucasian) = P (first is Caucasian, second is not) + P (first is not
Caucasian, second is Caucasian) =
1003,000 730000
730000 1003000
(
)(
)+(
)(
) = .2438 + .2438 = 0.4876
1733000 1732999
1733000 1732999
j.
P (both from same county) = P (both from Monterey) + P (both from San Luis Obispo) +
P ( both from Santa Barbara) + P (both from Ventura) =
369,000 368,999
236,000 235,999
392000 391999
736000 735999
(
)(
)+(
)(
)+(
)(
)+(
)(
)
1,733,000 1,732,999
1,733,000 1732999
1733000 1732999
1733000 1732999
= 0.0453 + 0.0185 + 0.0512 + 0.1804 = 0.2954
k.
P (both from different racial/ethnic group) = 1 – P (both from the same racial/ethnic
group = 1 – P (both Caucasian) – P (both Hispanic) – P (both Black) – P (both Asian) – P
(both American Indian)
528,000 527999
P (both Caucasian) = 0.335
P (both Hispanic) = (
)(
) = 0.0928
1733000 1732999
61000
60999
P (both Black) = (
)(
) = 0.0012
1733000 1732999
122000 121999
P (both Asian) = (
)(
) = 0.0050
1733000 1732999
19000
18999
P (both American Indian) = (
)(
) = 0.0001
1733000 1732999
1,003,000 1,002,999
)(
) = 0.335
1,733,000 1,732,999
730000 729999
)(
) = 0.1774
1733000 1732999
P (both from different racial/ethnic groups) =
1- 0.335 – 0.0928 – 0.0012 – 0.0050 – 0.0001 = 0.5659
6.79
6.80
The results will vary from one simulation to another. The approximate probabilities given were
obtained from a simulation done by computer with 100,000 trials.
a.
0.71293
a.
Assume that the requests are numbered from 1 to 20 as follows:
1−5
requests for a single license
6−14 requests for 2 licenses
15−20 requests for 3 licenses
b.
0.02201
Select a two digit random number from the table. If it is over 20 ignore it. If it is a number
from 1 to 20 retain it. If it is a number from 1 to 5 then 1 license will be granted on this
selection. If it is a number from 6−14 then two licenses will be granted for this selected
number. If it is a number from 15−20 then three licenses will be granted on this selected
number.
Select a second random number between 1 and 20 and grant the license request.
Select a third random number between 1 and 20 and grant the license request.
147
Select a fourth random number between 1 and 20 and grant the request if there are enough
licenses remaining to satisfy the request.
Repeat the previous step until all 10 licenses have been granted.
The simulation results will vary from one simulation to another. The approximate probability
obtained from a simulation with 20,000 trials done by computer was .3467.
b.
An alternative procedure for distributing the licenses might be as follows. The twenty
individuals requested a total of 41 licenses. Create a box containing 41 slips of paper. If an
individual requested 1 license, his name is placed on only 1 piece of paper. If an individual
requested two licenses, his name is placed on two slips. If he requested three licenses, his
name is placed on three slips. Choose 10 slips of paper at random from the collection.
Licenses are granted to those individuals whose names are selected. A person is granted as
many licenses as the number of times their name is selected.
6.81
The simulation results will vary from one simulation to another. The approximate probability should
be around .8504.
6.82
The simulation results will vary from one simulation to another. The approximate probability should
be around .8468.
6.83
a.
The simulation results will vary from one simulation to another. The approximate probability
should be around 0.6504.
b.
The decrease in the probability of on time completion for Jacob made the biggest change in
the probability that the project is completed on time.
Exercises 6.84 – 6.98
6.84
R
a.
0.05
Line A1
0.95
N
0.5
0.08
R
0.3
Line A2
0.92
N
0.2
0.10
R
Line A3
0.90
N
148
b.
P(came from line 1 and needs rework) = (0.5)(0.05) = 0.025.
c.
P(needs rework) = P(came from line 1 and needs rework) + P(came from line 2 and needs
rework) + P(came from line 3 and needs rework) = (0.5)(0.05) + (0.3)(0.08) + (0.2)(0.1) =
0.069.
6.85
P(parcel went via A1 and was late) = P(went via A1)P(late | went via A1) = (0.4)(0.02) = 0.008.
6.86
a.
P(arrived late) = P(sent via A1 and arrived late) + P(sent via A2 and arrived late) + P(sent via
A3 and arrived late) = (0.4)(0.02) + (0.5)(0.01) + (0.10)(0.05) = 0.018.
b.
We use the definition of conditional probability to compute each of the following.
6.87
6.88
P(A1 |L) =
P(sent via A1 and arrived late) (0.4)(0.02) 4
=
= = 0.44444 .
P(arrived late)
0.018
9
P(A2 |L) =
P(sent via A 2 and arrived late) (0.5)(0.01) 5
=
=
= 0.27778 .
P(arrived late)
0.018
18
P(A3 |L) =
P(sent via A 3 and arrived late) (0.1)(0.5) 5
=
=
= 0.27778 .
P(arrived late)
0.018
18
a.
(.3 )5 = .00243
b.
(.3 )5 + (.2)5 = .00243 + .00032 = .00275
c.
Select a random digit. If it is a 0, 1, or 2, then award A one point. If it is a 3 or 4, then award
B one point. If it is 5, 6, 7, 8, or 9, then award A and B one-half point each. Repeat the
selection process until a winner is determined or the championship ends in a draw (5 points
each). Replicate this entire procedure a large number of times. The estimate of the
probability that A
wins the championship would be the ratio of the number of times A wins to the total number
of replications.
d.
It would take longer if no points for a draw are awarded. It is possible that a number of
games would end in a draw and so more games would have to be played in order for a player
to earn 5 points.
a.
Choose a random digit. If it is a 1, 2, 3, 4, 5, 6, 7 or 8, then seed 1 defeats seed 4. If it is a 9
or 0, then seed 4 defeats seed 1.
b.
Choose a random digit. If it is a 1, 2, 3, 4, 5 or 6, then seed 2 defeats seed 3. If it is a 7, 8, 9,
or 0, then seed 3 defeats seed 2.
c.
If seeds 1 and 2 have won in the first round, then they are competing in game 3. Select a
random digit. If it is a 1, 2, 3, 4, 5 or 6, then seed 1 defeats seed 2 in game 3. If it is a 7, 8,
9, or 0, then seed 2 defeats seed 1 in game 3.
If seeds 1 and 3 have won in the first round, then they are competing in game 3. Select a
random digit. If it is a 1, 2, 3, 4, 5, 6, or 7, then seed 1 defeats seed 3 in game 3. If it is an 8,
9, or 0, then seed 3 defeats seed 1 in game 3.
149
If seeds 2 and 4 have won in the first round, then they are competing in game 3. Select a
random digit. If it is a 1, 2, 3, 4, 5, 6, or 7, then seed 2 defeats seed 4 in game 3. If it is an 8,
9, or 0, then seed 4 defeats seed 2 in game 3.
If seeds 3 and 4 have won in the first round, then they are competing in game 3. Select a
random digit. If it is a 1, 2, 3, 4, 5, or 6, then seed 3 defeats seed 4 in game 3. If it is a 7, 8,
9, or 0, then seed 4 defeats seed 3 in game 3.
d.
For game 1, selected digit is 7. Seed 1 wins game 1.
For game 2, selected digit is 9. Seed 3 wins game 2.
So seeds 1 and 3 compete in game 3.
For game 3, selected digit is 6. Seed 1 wins game 3 and the tournament.
e.
Summarize the result of part d as follows: digits selected (7,9,6) → seed 1 wins tournament.
Replication
s
2
3
4
5
6
7
8
9
10
Digits
Selected
(5,2,9)
(0,6,0)
(7,5,8)
(0,4,1)
(6,6,9)
(3,1,2)
(2,6,2)
(7,5,0)
(5,6,4)
Winner of
Tournament
seed 2
seed 4
seed 2
seed 2
seed 2
seed 1
seed 1
seed 2
seed 1
Based on this simulation of 10 trials, the estimate of the probability that the first seed wins the
tournament is 4/10 = .4.
f.
Answers will vary.
g.
The answers for parts e and f differ because they are based on different random selections
and different number of trials. Generally, the larger the number of trials, the better the
estimate is. So one would believe that the estimate from part f is better than the estimate of
part e.
6.89
P(need to examine at least 2 disks) = 1 – P(get a blank disk in the first selection) = 1 – 15/25 = 0.4.
6.90
a.
P(Ec) = 1 – 0.60 = 0.40
b.
Events E and F are disjoint, so P(E ∪ F) = P(E) + P(F) = 0.60 + 0.15 = 0.75.
c.
P(Ec ∩ Fc) = P( (E ∪ F)c ) = 1 – P(E ∪ F) = 1 – 0.75 = 0.25.
6.91
Let the five faculty members be A, B, C, D, E with teaching experience of 3, 6, 7, 10, and 14 years
respectively. The two selected individuals will have a total of at least 15 years of teaching experience
if the selected individuals are A and E, or B and D, or B and E, or C and D, or C and E, or D and E.
Since there are a total of 10 equally likely choices and 6 out of these meet the specified requirement,
P(two selected members will have a total of at least 15 years experience) = 6/10 = 0.6.
6.92
P(reads at least one of the three magazines) = P( A or B or C)
= 0.14 + 0.23 + 0.37 – 0.08 – 0.09 – 0.13 + 0.05 = 0.49.
150
6.93
Total number of viewers = 2517.
179 + 87
a.
P(saw a PG movie) =
= 0.1057.
2517
420 + 323 + 179 + 114 + 87
b.
P(saw a PG or a PG-13 movie) =
= 0.4462.
2517
c.
P(did not see an R movie) =
2517 − 600 − 196 − 205 − 139
= 0.5471.
2517
6.94
R1 and R2 are not independent events because P(R2|R1) = 1139/2516 and this is not equal to
P(R2|R1c) = 1140/2516. However, these two probabilities are very nearly equal, so from a practical
point of view these two events may be regarded as independent (approximately).
6.95
a.
P(neither is selected for testing | batch has 2 defectives) = (8/10)(7/9) = 56/90 = 0.6222.
b.
P(batch has 2 defectives and neither selected for testing) = P(batch has 2 defectives)
P(neither is selected for testing | batch has 2 defectives) = (0.20)(0.6222) = 0.1244.
P(neither component selected is defective) = P(batch has 0 defectives & neither selected
component is defective) + P(batch has 1 defective & neither selected component is defective)
+ P(batch has 2 defectives & neither selected component is defective) = (0.5)(1) +
(0.3)(9/10)(8/9) + (0.2)(8/10)(7/9) = 0.5 + 0.24 + 0.1244 = 0.8644.
c.
6.96
a.
P(E) = 20/25 = 0.8.
b.
P(F|E) = 19/24 = 0.79167.
c.
P(G|E ∩ F) = 18/23 = 0.78261.
d.
P(E ∩ F ∩ G) = P(E) P(F | E) P(G | E ∩ F) = (20/25)(19/24)(18/23) = 0.49565.
6.97
P(E ∩ F ∩ G ∩ H) = P(E)P(F|E)P(G| E ∩ F)P(H| E ∩ F ∩ G) = (20/25)(19/24)(18/23)(17/22) = 0.3830.
P(at least one bad bulb ) = 1 – P(all 4 bulbs are good ) = 1 – 0.383 = 0.617.
6.98
a.
P(1 is sent by all relays | 1 sent by transmitter) = (0.8)(0.8)(0.8) = 0.512.
b.
P(1 is received by the receiver | 1 is sent by transmitter) = P(relay 1 sends 1, relay 2 sends 1,
relay 3 sends 1) + P(relay 1 sends 1, relay 2 sends 0, relay 3 sends 1) + P(relay 1 sends 0,
relay 2 sends 1, relay 3 sends 1) + P( relay 1sends 0, relay 2 sends 0, relay 3 sends 1) =
(0.8)(0.8)(0.8) + (0.8)(0.2)(0.2) + (0.2) (0.2) (0.8) + (0.2)(0.8)(0.2) = 0.608.
151