Chapter 4: Probability and Probability Distributions
4.1
a
b
c
d
4.2
a
This experiment involves tossing a single die and observing the outcome. The sample space for this
experiment consists of the following simple events:
E1: Observe a 1
E4: Observe a 4
E2: Observe a 2
E5: Observe a 5
E3: Observe a 3
E6: Observe a 6
Events A through F are composed of the simple events in the following manner:
A: (E2)
D: (E 2)
B: (E 2, E 4, E 6)
E: (E 2, E 4, E6)
C: (E 3, E 4, E 5, E 6)
F: contains no simple events
Since the simple events Ei, i = 1, 2, 3, …, 6 are equally likely, P( Ei )  1 6 .
To find the probability of an event, we sum the probabilities assigned to the simple events in that
event. For example,
1
P( A)  P( E2 ) 
6
4 2
3 1
Similarly, P( D)  1 6; P( B)  P( E )  P( E2 )  P( E4 )  P( E6 )   ; and P(C )   . Since
6 3
6 2
event F contains no simple events, P( F )  0.
It is given that P( E1 )  P( E2 )  0.15 and P( E3 )  0.40. Since
 P( E )  1 , we know that
i
S
(i)
P( E4 )  P( E5 )  1  0.15  0.15  0.40  0.30
Also, it is given that
(ii)
P( E4 )  2P( E5 )
We have two equations in two unknowns which can be solved simultaneously for P(E4) and P(E5).
Substituting equation (ii) into equation (i), we have
2 P( E5 )  P( E5 )  0.3
3P( E5 )  0.3 so that P( E5 )  0.1
Then from (ii),
P( E4 )  2 P( E5 )  0.2.
To find the necessary probabilities, sum the probabilities of the simple events:
P( A)  P( E1 )  P( E3 )  P( E4 )  0.15  0.4  0.2  0.75
P( B)  P( E2 )  P( E3 )  0.15  0.4  0.55
c–d The following events are in either A or B or both: { E1, E2, E3, E4}. Only event E3 is in both A and B.
b
4.3
It is given that P( E1 )  0.45 and that 3P( E2 )  0.45, so that P( E2 )  0.15. Since
 P( E )  1 , the
i
S
remaining 8 simple events must have probabilities whose sum is
P( E3 )  P( E4 )  ...  P( E10 )  1  0.45  0.15  0.4
0.4
 0.05 for i  3, 4, ..., 10
Since it is given that they are equiprobable, P( Ei ) 
8
4.4
a
It is required that
 P( E )  1 . Hence,
i
P( E2 )  1  0.49  0.21  0.09  0.21
S
b
NEL
The player will hit on at least one of the two free throws if he hits on the first, the second, or both. The
associated simple events are E1, E2, and E3 and
P(hits on at least one)  P( E1 )  P( E2 )  P( E3 )  0.91
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4-1
Instructor’s Solutions Manual to Accompany Introduction to Probability and Statistics, 3CE
4.9
The four possible outcomes of the experiment, or simple events, are represented as the cells of a 2  2 table,
and have probabilities as given in the table.
a
P(adult judged to need glasses) = 0.44 + 0.14 = 0.58
b
P(adult needs glasses but does not use them) = 0.14
c
P(adult uses glasses) = 0.44 + 0.02 = 0.46
4.14
a
Visualize a tree diagram with four stages—selecting the runner who places first, second, third, and
fourth, respectively. There are four choices at the first stage, three choices (branches) at the second stage, two
choices (branches) at the third stage, and only one choice (branch) for the last runner. The total number of simple
events is 4(3)(2)(1)  24. The simple events are listed below:
JBED
BJED
EBJD
DBEJ
JEBD
BEJD
EJBD
DEBJ
JEDB
BEDJ
EJDB
DEJB
JBDE
BJDE
EBDJ
DBJE
JDEB
BDEJ
EDJB
DJEB
JDBE
BDJE
EDBJ
DJBE
b
If all runners are equally qualified, the simple events will be equally likely with probability
P( Ei )  1 24.
c–e Sum the probabilities of the appropriate simple events:
6 1
2
1
6 1
P(Dave wins) 
 , P(Dave is first, John is second) 
 , and P(Ed is last) 

24 4
24 12
24 4
4.15
Similar to Exercise 4.9. The four possible outcomes of the experiment, or simple events, are represented as
the cells of a 2  2 table, and have probabilities (when divided by 300) as given in the table.
a
P(normal eyes and normal wing size)  140 300  0.467
b
P(vermillion eyes)  (3  151) 300  154 300  0.513
c
P(either vermillion eyes or miniature wings or both)  (3  151  6) 300  160 300  0.533
4.40
Each simple event is equally likely, with probability 1 5.
4.41
4.42
4-2
a
AC  E2 , E4 , E5 
P  AC   3 5
b
A  B  E1
P  A  B  1 5
c
B  C  E4 
PB C  1 5
d
A  B  S  E1 , E2 , E3 , E4 , E5 
P  A  B  1
e
B | C  E4 
P( B | C )  1 2
f
A | B  E1
P( A | B)  1 4
g
A B C  S
P A B C 1
h
 A  B
P  A  B  4 5
a
P  AC   1  P  A  1 
b
P  A  B  1 P  A  B  1
a
P  A | B 
C
 E2 , E3 , E4 , E5 
C
2 3

5 5
C
1 4

5 5
P( A  B) 1 5 1


P( B)
45 4
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Instructor’s Solutions Manual to Accompany Introduction to Probability and Statistics, 3CE
P( B  C ) 1 5 1


P(C )
25 2
b
PB | C 
a
P  A  B   P( A)  P( B)  P( A  B)  2 5  4 5  1 5  5 5  1
b
P( A  B)  P( A | B) P( B)  1 4  4 5  1 5
c
P( B  C )  P( B | C ) P(C )  1 2  2 5  1 5
4.44
a
From Exercise 4.40, P  A | B   1 4 while P( A)  2 5. Therefore, A and B are not independent.
4.47
b
a
b
From Exercise 4.40, P( A  B)  1 5. Since P( A  B)  0, A and B are not mutually exclusive.
Since A and B are independent, P( A  B)  P( A) P( B)  0.4(0.2)  0.08.
P( A  B)  P( A)  P( B)  P( A  B)  0.4  0.2  (0.4)(0.2)  0.52
4.48
a
b
Since A and B are mutually exclusive, P( A  B)  0.
P( A  B)  P( A)  P( B)  P( A  B)  0.3  0.5  0  0.8
4.49
a
Use the definition of conditional probability to find
P( A  B) 0.12
P  B | A 

 0.3
P( A)
0.4
Since P( A  B)  0, A and B are not mutually exclusive.
If P( B)  0.3, then P( B)  P( B | A), which means that A and B are independent.
4.43
b
c
4.50
a
The event A will occur whether event B occurs or not (B C). Hence,
P( A)  P( A  B)  P( A  BC )  0.34  0.15  0.49
b
c
d
e
Similar to part a. P( B)  P( A  B)  P( AC  B)  0.34  0.46  0.80
The contents of the cell in the first row and first column is P( A  B)  0.34.
P( A  B)  P( A)  P( B)  P( A  B)  0.49  0.80  0.34  0.95
Use the definition of conditional probability:
P( A  B) 0.34
P  A | B 

 0.425
P( B)
0.80
Similar to part e.
P( A  B) 0.34
P  B | A 

 0.694
P( A)
0.49
f
4.51
a
From Exercise 4.50, since P( A  B)  0.34, the two events are not mutually exclusive.
From Exercise 4.50, P  A | B   0.425 and P( A)  0.49. The two events are not independent.
Define the following events:
P: test is positive for drugs
N: test is negative for drugs
D: employee is a drug user
It is given that, on a given test, P( P | D)  0.98, P( N | DC )  0.98.
b
4.52
a
For a given test, P( P | DC )  1  0.98  0.02. Since the tests are independent
P(fail both test | DC )  (0.02)(0.02)  0.0004
NEL
b
P(detection | D)  P( N  P | D)  P( P  N | D)  P( P  P | D)
c
 (0.98)(0.02)  (0.02)(0.98)  (0.98)(0.98)  0.9996
P(pass both | D)  P( N  N | D)  (0.02)(0.02)  0.0004
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Instructor’s Solutions Manual to Accompany Introduction to Probability and Statistics, 3CE
4.53
Define the following events:
A: project is approved for funding
D: project is disapproved for funding
For the first group, P( A1 )  0.2 and P( D1 )  0.8. For the second group,
P(same decision as first group)  0.7 and P(reversal)  0.3. That is, P( A2 | A1 )  P( D2 | D1 )  0.7 and
P( A2 | D1 )  P( D2 | A1 )  0.3.
a
P( A1  A2 )  P( A1 ) P( A2 | A1 )  0.2(0.7)  0.14
b
P( D1  D2 )  P( D1 ) P( D2 | D1 )  0.8(0.7)  0.56
c
P( D1  A2 )  P( A1  D2 )  P( D1 ) P( A2 | D1 )  P( A1 )P(D2 | A1 )  0.8(0.3)  0.2(0.3)  0.30
4.58
Define the following events:
4.61
Define the events:
S: student chooses Starbucks
T: student chooses Tim Hortons
D: student orders a decaffeinated coffee
Then P(S )  0.3; P(T )  0.7; P( D | S )  P( D | T )  0.60
a
Using the given probabilities, P(T  D)  P(T ) P( D | T )  0.7(0.6)  0.42
b
Since P( D)  0.6 regardless of whether the student visits Starbucks or Tim Hortons, the two events
are independent.
P ( S  D) P ( S ) P ( D | S )
c
P ( S | D) 

 P( S )  0.3
P ( D)
P ( D)
d
P(T  D)  P(T )  P( D)  P(T  D)  0.7  0.6  (0.7)(0.6)  0.88
D: person dies
S: person smokes
It is given that P(S )  0.2, P( D)  0.006, and P( D | S )  10P( D | S C ). The probability of interest is
P( D | S ) . The event D, whose probability is given, can be written as the union of two mutually exclusive
intersections. That is, D  ( D  S )  ( D  S C ) .
Then, using the Addition and Multiplication Rules,
P  D   P( D  S )  P( D  S C )  P( D | S ) P(S )  P( D | S C ) P(S C )  P( D | S )(0.2)  1 10  P( D | S )  (0.8)
Since P( D)  0.006, the above equation can be solved for P( D | S )
P( D | S )(0.2  0.08)  0.006
P( D | S )  0.006 0.28  0.0214
4.63
Define
A: smoke is detected by device A
B: smoke is detected by device B
If it is given that P( A)  0.95, P( B)  0.98, and P( A  B)  0.94
a
P( A  B)  P( A)  P( B)  P( A  B)  0.95  0.98  0.94  0.99
b
4.64
4-4
P( AC  BC )  1  P( A  B)  1  0.99  0.01
Using the probability table given in the question, we find
a
P( A)  0.42
b
P( A  G)  0.22
0.22
P( A | G ) 
 0.4231
c
0.52
d
P(G)  0.52
e
P( A  G)  P( A)  P(G)  P( A  G)  0.42  0.52  0.22  0.72
0
P ( D | O) 
0
f
0.06
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Instructor’s Solutions Manual to Accompany Introduction to Probability and Statistics, 3CE
g
h
i.
j.
4.67
If events A and G are independent then must P( A | G)  P( A) be true. From parts c and a, we can see
these probabilities are almost equal. Thus, A and G are independent.
D and O are mutually exclusive since P( D O)  0.
D and O are not independent. If these two events are independent then P( D | O)  P( D) must be true.
P( D)  0.09 whereas P( D | O)  0. Also, mutually exclusive events are always dependent.
If A and G are mutually exclusive, then P( A  G) must be zero. This probability is 0.22; thus, these
two events are not mutually exclusive.
Similar to Exercise 4.54.
54
a
P( A) 
1029
517
b
P( F ) 
1029
37
c
P A F  
1029
P( F  A) 37 1029 37
P  F | A 


d
P( A)
54 1029 54
e
P  F | B 
P( F  B) 64 1029 64


P( B)
99 1029 99
f
PF | C 
P( F  C ) 138 1029 138


P(C )
241 1029 241
g
P C | M  
P(C  M ) 103 1029 103


P( M )
512 1029 512
h
P( B C )  1  P( B)  1 
99
930 310


1029 1029 343
4.74
Use the Law of Total Probability: P( A)  P(S1 ) P( A | S1 )  P(S2 ) P( A | S2 )  0.6(0.3)  0.4(0.5)  0.38
4.75
Define the following events:
V: crime is violent
R: crime is reported
C
It is given that P(V )  0.2, P(V )  0.8, P( R | V )  0.9, P( R | V C )  0.7.
a
The overall reporting rate for crimes is
P( R)  P(V ) P( R | V )  P(V C ) P( R | V C )  0.2(0.9)  0.8(0.7)  0.74
P(V ) P( R | V ) 0.2(0.9)
b
Use Bayes’ Rule:
P(V | R) 

 0.24
P( R)
0.74
P(V C ) P( R | V C ) 0.8(0.7)

 0.76
P( R)
0.74
Notice that the proportion of non-violent crimes (0.8) is much larger than the proportion of violent
crimes (0.2). Therefore, when a crime is reported, it is more likely to be a non-violent crime.
and
c
4.76
NEL
P(V C | R) 
Define
A: machine produces a defective item
B: worker follows instructions
Then P( A | B)  0.01, P( B)  0.90, P( A | BC )  0.03, P( BC )  0.10
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Instructor’s Solutions Manual to Accompany Introduction to Probability and Statistics, 3CE
The probability of interest is
P( A)  P( A  B)  P( A  B C )
 P( A | B) P( B)  P( A | B C ) P( B C )
4.77
 0.01(0.90)  0.03(0.10)  0.012
Define the following events:
A: passenger uses airport A
B: passenger uses airport B
C: passenger uses airport C
D: a weapon is detected
Suppose that a passenger is carrying a weapon. It is given that
P( D | A)  0.9
P( A)  0.5
P( D | B)  0.5
P( B)  0.3
P( D | C )  0.4
The probability of interest is
P(C )  0.2
P( A) P( D | A)
0.5(0.9)

 0.6618
P( A) P( D | A)  P( B) P( D | B)  P(C ) P( D | C ) 0.5(0.9)  0.3(0.5)  0.2(0.4)
0.2(0.4)
0.08
Similarly, P(C | D) 

 0.1176
0.5(0.9)  0.3(0.5)  0.2(0.4) 0.68
P ( A | D) 
4.80
The probability of interest is P( A | H ), which can be calculated using Bayes’ Rule and the probabilities
given in the exercise.
P( A) P( H | A)
P( A | H ) 
P( A) P( H | A)  P( B) P( H | B)  P(C ) P( H | C )
0.01(0.90)
0.009

 0.3130
0.01(0.90)  0.005(0.95)  0.02(0.75) 0.02875
Using the probability table,
P( D)  0.08  0.02  0.10
P( DC )  1  P( D)  1  0.10  0.90

4.82
a
P( N  DC ) 0.85
P( N  D) 0.02

 0.94
P ( N | D) 

 0.20
P( DC )
0.90
P ( D)
0.10
P ( D) P ( N | D)
0.10(0.20)
Using Bayes’ Rule, P( D | N ) 

 0.023
P( D) P( N | D)  P( DC ) P( N | DC ) 0.10(0.20)  0.90(0.94)
Using the definition of conditional probability,
P( N  D) 0.02
P( D | N ) 

 0.023
P( N )
0.87
P( N | D C ) 
b
c
d
e
f
4-6
P( P  DC ) 0.05

 0.056
P( DC )
0.90
P( N  D) 0.02
P  false negative   P( N | D) 

 0.20
P ( D)
0.10
The probability of a false negative is quite high, and would cause concern about the reliability of the
screening method.
P  false positive   P( P | DC ) 
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