Download Yes

Exercises 6.1 – 6.13
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
b
Sample space = {(A, A), (A, M), (M, A), (M, M)}.
A
A
M
A
M
M
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
143
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
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
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.
144
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
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.
6.11 a
E2
E1
145
b
The required region is the area common to all three circles.
E1
E2
E3
c
E1
E2
E3
146
d
E1
E2
E3
e
E1
E2
E3
f
147
E1
E2
E3
148
Exercises 6.14 – 6.28
6.14 a
P(picking 9, then 1, then another 1) =
1 1 1
=
 
10 10 10
0.001
b
Relative frequency because we are looking at the long term
behavior.
6.15 a
In the long run, 1% of all people who suffer cardiac arrest in New
York City survive.
b
6.17 a
b
6.18 a
1% of 2329 is .01(2329) = 23.29. So in this study there must have been only 23 or 24 survivors.
A = {(C,N), (N,C), (N,N)}. So P(A) = 0.09 + 0.09 + 0.01 = 0.19.
B = {(C,C), (N,N)}. So P(B) = 0.81 + 0.01 = 0.82.
P(twins) =
500,000
42,005,100
= .0119
100
42,005,100
b
P(quadruplets) =
c
P(more than a single child) =
6.19 a
= .00000238
505,100
42,005,100
= .012
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
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.
149
6.22 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.
6.23 a
P(hand will consist entirely of spades) =
1287
 0.000 495198
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.
b
P(hand consists entirely of spades and clubs with both suits
represented)
= 63206 = 0.0243197.
2598960
c
6.25 a
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.
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.
6.27 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.
150
c
P(brand M) = 25% + 16% = 41% = 0.41.
6.28 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.
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.
151
Exercises 6.29 – 6.39
6.29 a
P( E | F ) 
P( E  F ) 0.54

 0.90 .
P( F )
0.6
b
P( F | E ) 
P( E  F ) 0.54

 0.7714 .
P( E )
0.7
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.
P(F|FP)=
432
562
= 0.769
390
438
b.
P(M|MP)=
c.
Ultrasound appears to be more reliable in predicting boys
6.35 a
= 0.890
i
Required conditional probability =
ii
Required conditional probability =
b
135
 0.74176 .
182
173  206
 0.8594 .
210  231
Properly restrained
Iowa City (city)
Not Properly restrained
Properly restrained
Iowa City (interstate)
152
Not properly restrained
6.37 a
b
c
d
e
6.39 a
0.1 + 0.175 = 0.275
0.1
 0.2
0. 5
0.325
 0.65
0.5
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.
i. P(S) =
456
 0.76
600
ii. P(S|A) =
215
300
= 0.717
iii. P(S|B) =
241
=
300
0.803
iv Drug B appears to have a higher survival rate
b
i P(S) =
140
 0.583
240
ii P(S|A) =
120
200
= 0.600
iii P(S|B) =
20
=
40
0.5
iv Drug A appears to have the higher survival rate.
c.
i P(S) =
316
 0.878
360
ii P(S|A) =
95
100
= 0.95
iii P(S|B) =
221
=
260
0.85
iv Drug A appears to have the higher survival rate
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.
153
Exercises 6.40 – 6.58
6.41 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.
6.42 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 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.
6.45 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 | Ac )  0.62 ; P( B | A)  0.6 . So P( A  B)  P( A) P( B | A)  0.0678 .
From this we get,
P( Ac  B)  P( Ac ) P( B | Ac )  (1  0.113)(0.62)  0.54994.
154
Therefore
P( B)  P( A  B)  P( Ac  B)  0.61774.
Since
6.47 a
P( B | A) is
not equal to
P(B) ,
A and B are dependent outcomes.
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.
6.49 a
P(student has a Visa card) = 7000/10000 = 0.7
b
P(student has both cards) = 5000/10000 = 0.5
c
0.714286
P(student has both cards | student has Visa card) = 5000/7000 =
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.
6.51 a
i
P(F) = 0.51
155
ii
iii
iv
v
P(F | C) = 0.56
P(F | Cc ) = 0.45
P(F | O) = 0.36
P(F | Y) = 0.72
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.
6.53 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
= 0.56.
P(A wins both her matches) = P(A beats B) P(A beats C) = (0.7)(0.8)
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.
6.55 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
156
6.56 P(E1 | E2 ) = 0 because, given that the fire is a two-alarm fire, it cannot be
a single-alarm blaze. But P(E1) = 0.4 which is not equal to P(E1 | E2 ), so
E1 and E2 are dependent.
More generally, if E1 and E2 are disjoint events, then P(E1 | E2 ) = 0
because given that E2 has occurred, there is no way E1 could also have
occurred. The only way P(E1 | E2 ) could equal P(E1 ) in this situation is
if P(E1 ) = 0. Likewise, P(E2 | E1) = P(E2) is if P(E2 ) = 0. So, disjoint events
with positive probabilities cannot be independent.
6.57 a
b
CC, CN, NC, NN.
P(CC) = (50/800)(50/800) = 0.00391.
c
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 b.
6.58 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
157
6.60 a
b
P(E1 and E2) = (0.4)(0.3) = 0.12
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)
6.61 a
P(individual has to stop at at least one light) = P(E
P(F) – P(E  F)
= 0.4 + 0.3 – 0.15 = 0.55.
 F)
= P(E) +
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
b
P(F) =
4 .5
8 .6
P(S|R) =
= 0.523, P(S) =
4 .1
8 .6
= 0.477, P(R|F) = 0.018, P(R|S) = 0.03
P(S)P( R|S)
0.01431

 0.6032
P(S)P( R|S)  P( F )P( F |S) 0.01431 0.0094
158
This is the probability that a randomly selected gun purchase
background check resulted in a blocked sale by a state or local
agency.
6.65 a
i 0.5
ii 0.5
iii 0.99
iv .01
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
c
P(C|TD) =
P(C )P(TD | P(C )
(.99)(. 01)

 0.5
P(TD | P(C )  P(TD)| P( D) 2(.99)(. 01)
Yes, it is consistent with the argument given in the quote.
6.66 a
The events need to be mutually exclusive.
b
Complement rule: the higher the probability
c
Conditions 1 and 2 demonstrate the addition rule: the probability
of a randomly selected race being either one race or another race is
the sum the two probabilities. Multiplication rule: The probability
of two randomly selected people being the same race and ethnicity
is the product of the two probabilities. The complement rule: the
formula after “1 –“ is built by calculating the probability of finding
the probability that two randomly selected individuals are racially
or ethnically the same. Subtracting this probability from 1 (or
finding the complement) will give the probability that two
randomly selected individuals are racially or ethnically different.
6.67
a
b
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.
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.
159
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.
6.68 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
c
P(Ec

P(Ec

F) = P(Ec)P(F | Ec ) = (1 – 0.10)(0.05) = (0.9)(0.05) = 0.045.
P(F) = P(E  F) + P(Ec  F). Now, P(E  F) = 0.09 from part a and
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 ,
0.135
P(F)
using the
results from parts a and c.
6.69 a
b
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.
P(employee tests positive twice)
= P(employee uses drugs and tests positive twice)
+ P(employee doesn’t use drugs and tests positive twice)
160
= 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 drugs and tests positive twice)
0.081

 0.97297 .
P(tests positive 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.
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.)
Works
Does not Total
full-time
work fulltime
Use drugs
70
30
100
161
Does not use
drugs
Total
805
7095
7900
875
7125
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.
6.71 Based on the given information we can construct the following table.
Basic
Deluxe
Total
Model
Model
Buy extended warranty 12%
30%
42%
Do not buy extended
28%
30%
58%
warranty
Total
40%
60%
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(satisfie d)
162
.
P(took the class during winter quarter | satisfied with book)
=
(150 / 1000 ) 0.15
P(took the class during winter & satisfied)

 0.2941

0.51
0.51
P(satisfie d)
.
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(satisfie d)
.
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.
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(A | C) 
6.74 a
P(C | A)P(A)
(0.9)( 0.75)

 0.8182 .
P(C | A)P(A)  P(C | B)P(B) (0.9)( 0.75)  (0.6)( 0.25)
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
163
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) =
P(has disease & positive test) 0.00095

 0.00942 .
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 =
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) =
0.19.
164
(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.
6.76 a
Probatio
n
Yes
No
Total
High School GPA
2.5 -<
3.0 -< 3.5
3.5 and
3.0
Above
0.10
0.11
0.06
0.09
0.27
0.37
0.19
0.38
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
0.5263
P(GPA between 2.5 and 3.0 | academic probation) = 0.10/0.19 =
f
0.1395.
P(GPA of 3.5 or above AND academic probation) = 0.06/0.43 =
165
6.77 a
= 0.3941.
P(male) = (200+800+1500+1500+900+1500+300)/17000 = 6700/17000
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.
6.78 Total Monterey County = 369,000
Total Caucasian = 1,003,000
Total San Luis Obispo County = 236,000
Total Hispanic = 528,000
Total Santa Barbara County = 392,000
Total Black = 61,000
Total Ventura County = 736,000
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
231,000
 0.4375
528 ,000
9000
 0.0052
1,733 ,000
d
122000
236000
9000


 0.2014
1,733 ,000 1,733 ,000 1,733 ,000
e
P (Asian or from San Luis Obispo) =
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
Caucasian) = (1,003,000 )(1,002 ,999 )  0.335
1,733 ,000 1,732 ,999
g
P (both
h
P (neither Caucasian) = (
730000
729999
)(
)  0.1774
1733000 1732999
166
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)
P (both Caucasian) = 0.335 P (both Hispanic) = ( 528 ,000 )( 527999 ) =
1733000
0.0928
P (both Black) =
P (both
P (both
61000
60999
)(
) = 0.0012
1733000 1732999
Asian) = ( 122000 )( 121999 ) = 0.0050
1733000 1732999
American Indian) = ( 19000 )( 18999 ) =
1733000 1732999
1732999
(
0.0001
P (both from different racial/ethnic groups) =
1- 0.335 – 0.0928 – 0.0012 – 0.0050 – 0.0001 = 0.5659
6.79 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
b
0.02201
167
6.80 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
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.
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
168
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
b
The simulation results will vary from one simulation to another.
The approximate probability should be around 0.6504.
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 a
R
0.05
Line A1
0.95
N
0.5
R
0.08
0.3
Line A2
0.92
0.2
N
R
169
0.10
Line A3
0.90
N
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.
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
170
.
.
P(A3 |L) =
6.87 a
P(sent via A 3 and arrived late) (0.1)( 0.5) 5


 0.27778
P(arrived late)
0.018
18
.
.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.
6.88 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.
171
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.
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.
Replicati
ons
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
172
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
c
P(Ec
 F)
= P(E) + P(F) = 0.60 + 0.15 =
0.75.
 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.
6.93
Total number of viewers = 2517.
173
179  87
2517
a
P(saw a PG movie) =
= 0.1057.
b
P(saw a PG or a PG-13 movie) =
c
P(did not see an R movie) =
420  323  179  114  87
2517
2517  600  196  205  139
2517
= 0.4462.
= 0.5471.
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
c
6.96 a
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.
P(E) = 20/25 = 0.8.
b
P(F|E) = 19/24 = 0.79167.
c
P(G|E  F) = 18/23 = 0.78261.
d
0.49565.
P(E  F  G) = P(E) P(F | E) P(G | E  F) = (20/25)(19/24)(18/23) =
174
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
0.512.
b
P(1 is sent by all relays | 1 sent by transmitter) = (0.8)(0.8)(0.8) =
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.
175