Part IV: Complete Solutions, Chapter 4
296
Chapter 4: Elementary Probability Theory
Section 4.1
1.
Equally likely outcomes, relative frequency, intuition
2.
The complement is “not rain today.” This probability is 100% – 30% = 70%.
3.
(a) The probability of a certain event is 1.
(b) The probability of an impossible event is 0.
4.
The law of large numbers states that in the long run, as the sample size or number of trials increases, the
relative frequency of outcomes approaches the theoretical probability of the outcome. Five hundred trials
are better because the law of large numbers works better for larger samples.
5.
No. The probability of throwing tails on the second toss is 0.50 regardless of the outcome on the first toss.
6.
(a)
(b)
(c)
(d)
7.
The resulting relative frequency can be used as an estimate of the true probability of all Americans who can
wiggle their ears.
8.
The resulting relative frequency can be used as an estimate of the true probability of all Americans who can
raise one eyebrow.
15
71
124
131
34
, P(1) 
, P(2) 
, P(3) 
, P(4) 
(a) P(no similar preferences)  P(0) 
375
375
375
375
375
15  71  124  131  34 375

 1, yes
(b)
375
375
Personality types were classified into four main preferences; all possible numbers of shared preferences
were considered. The sample space is 0, 1, 2, 3, and 4 shared preferences.
9.
Probabilities must be between 0 and 1 inclusive. –0.41 < 0
Probabilities must be between 0 and 1 inclusive. 1.21 > 1
120% = 1.20, and probabilities must be between 0 and 1 inclusive. 1.20 > 1
Yes, 0 ≤ 0.56 ≤ 1.
10. (a) The sample space would be 1, 2, 3, 4, 5, and 6 dots. If the die is fair, all outcomes will be equally
likely.
1
(b) P(1)  P(2)  P (3)  P (4)  P (5)  P (6)  because the die faces are equally likely, and there are six
6
1 1 1 1 1 1 6

outcomes. The probabilities should and do add to 1         1 because all
6 6 6 6 6 6 6

possible outcomes have been considered.
1 1 1 1
4 2
 
(c) P(number of dots < 5) = P(1 or 2 or 3 or 4 dots) = P(1) + P (2) + P(3) + P(4)    
6 6 6 6
6 3
1 2
or P(dots < 5) = 1 – P(5 or 6 dots)  1   (The applicable probability rule used here will be
3 3
discussed in the next section of the text; rely on your common sense for now.)
(d) Complementary event rule: P(A) = 1 – P(not A)
P(5 or 6 dots) = 1 – P(1 or 2 or 3 or 4 dots)  1 
2 1
1 1 2 1
 , or P(5 or 6) = P(5) + P(6)    
3 3
6 6 6 3
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Part IV: Complete Solutions, Chapter 4
297
11. (a) Note: “Includes the left limit but not the right limit” means 6 A.M. ≤ time t < noon, noon ≤ t < 6 P.M.,
6 P.M. ≤ t < midnight, midnight ≤ t < 6 A.M.
290
 0.30
966
135
 0.14
P(best idea 12 noon–6 P.M.) 
966
319
 0.33
P(best idea 6 P.M.–12 midnight)
966
222
 0.23
P(best idea from 12 midnight to 6 A.M.) 
966
(b) The probabilities add up to 1. They should add up to 1 provided that the intervals do not overlap and
each inventor chose only one interval. The sample space is the set of four time intervals.
P(best idea 6 A.M.–12 noon) 
12. (a) P(germinate) =
number germinated 2, 430

 0.81
number planted
3, 000
3,000  2, 430
570

 0.19
3,000
3,000
(c) The sample space is two outcomes, germinate and not germinate.
(b) P(not germinate) =
P(germinate) + P(not germinate) = 0.81 + 0.19 = 1
The probabilities of all the outcomes in the sample space should and do sum to 1.
(d) No because P(germinate) = 0.81 ≠ P(not germinate) = 0.19
1
If they were equally likely, each would have probability  0.5.
2
n

13. (a) Given: Odds in favor of A are n:m  i.e.,  .
m


n
Show: P( A) 
mn
P( A)
Proof: Odds in favor of A are
by definition
P(not A)
P (not A)  1  P ( A)
complementary events
n
P ( A)
P ( A)


substitution
m P (not A) 1  P ( A)
n[1  P ( A)]  m[ P ( A)]
cross multiply
n  n[ P ( A)]  m[ P ( A)]
n  n[ P ( A)]  m[ P ( A)]
n  (n  m)[ P ( A)]
So
n
 P ( A), as was to be shown.
nm
(b) Odds of a successful call are 2 to 15. Now 2 to 15 can be written as 2:15 or
From part (a): if the odds are 2:15 (let n = 2, m = 15), then P(sale) 
Copyright © Houghton Mifflin Company. All rights reserved.
2
.
15
n
2
2


 0.118.
n  m 2  15
17
Part IV: Complete Solutions, Chapter 4
298
(c) Odds of free throw are 3 to 5, i.e., 3:5.
Let n = 3 and m = 5 here; then, from part (a):
P(free throw) 
n
3
3

  0.375
n  m 35 8
 a
14. (a) Given: Odds against W are a:b  or  .
 b
Show: P(not W) 
a
.
ab
P (not W )
by definition.
P (W )
P(W )  1  P(not W )
complementary events
P(not W ) a

substitution
P(W )
b
P(not W )
a

substitution
1  P(not W ) b
b[ P(not W )]  a[1  P(not W )]
cross-multiply
b[ P(not W )]  a  a[ P(not W )]
b[ P(not W )]  a[ P(not W )]  a
(a  b)[ P(not W )]  a
a
P(not W ) 
ab
a
P (not W ) 
, as was to be shown.
ab
Proof: Odds against W are
(b) Point Given’s betting odds are 9:5. Betting odds are based on the probability that the horse does not
P(not PG wins)
.
win, so odds against Point Given (PG) winning are
P(PG wins)
Let a = 9 and b = 5 in part (a) formula. From part (a), P(not PG wins) 
“not PG wins” is the same as “PG loses,” so P(PG loses) 
 1
a
9
9

 , but event
a  b 9  5 14
9
 0.64, and P(PG wins)
14
9
5

 0.36.
14 14
(c) Betting odds for Monarchos are 6:1. Betting odds are based on the probability that the horse does not
win; i.e., the horse loses.
Let W be the event that Monarchos wins. From part (a), if the events against W are given as a:b, the
a
. Let a = 6 and b = 1 in the part (a) formula, so
P(not W) 
ab
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Part IV: Complete Solutions, Chapter 4
P(not W ) 
299
6
6

6 1 7
6
 0.86
7
P(Monarchos wins)  P(W )  1  P(not W )
6 1
 1    0.14
7 7
P(not W )  P(Monarchos loses)=
(d) Invisible Ink was given betting odds of 30 to 1; i.e., odds against Invisible Ink winning were
30
.
1
Let W denote the event that Invisible Ink wins. Let a = 30, b = 1 in formula from part (a).
Then, from part (a), P(not W) 
a
30
30
, P (not Invisible Ink wins) 

; i.e.,
ab
30  1 31
30
 0.97
31
P(Invisible Ink wins)  1  P(Invisible Ink loses)
30 1
 1

 0.03
31 31
P(Invisible Ink loses) 
15. One approach is to make a table showing the information about the 127 people who walked by the store.
Came into the store
Did not come in
Column total
Buy
25
0
25
Did Not Buy
58 – 25 = 33
69
102
Row Total
58
127 – 58 = 69
127
If 58 came in, 69 didn’t; 25 of the 58 bought something, so 33 came in but didn’t buy anything. Those who
did not come in couldn’t buy anything. The row entries must sum to the row totals, the column entries must
sum to the column totals, and the row totals, as well as the column totals, must sum to the overall total, i.e.,
the 127 people who walked by the store. Also, the four inner cells must sum to the overall total: 25 + 33 + 0
+ 69 = 127.
number outcomes favorable to A
.
This kind of problem relies on formula (2), P(event A) 
total number of outcomes
(a) P ( A) 
58
 0.46
127
Here, we divide by 127 people.
(b) P ( A) 
25
 0.43
58
Here, we divide by 58 people (only those who entered).
(c) P( A)  P(Enter and buy) 
58 25 25


 0.20
127 58 127
Or similarly, read from the table that 25 people both entered and bought something. Divide this by the
total number of people, namely, 127.
33
 0.57
(d) P( A)  P(Buy nothing) 
Here, we divide by 58 people.
58
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Part IV: Complete Solutions, Chapter 4
300
Section 4.2
1.
No. Mutually exclusive events cannot occur at the same time.
2.
If A and B are independent, then P(A) = P(A | B). Therefore, P(A | B) = 0.3.
3.
(a) Event A cannot occur if event B has occurred. Therefore, P(A | B) = 0.
(b) Since we are told that P(A) ≠ 0, and we have determined that P(A | B) = 0, we can deduce that
P(A) ≠ P(A | B). Therefore, events A and B are not independent.
4.
(a) P(A and B) = P(A) × P(B) if events A and B are independent. This product can equal zero only if either
P(A) = 0 or P(B) = 0 (or both). We are told that P(A) ≠ 0 and that P(B) ≠ 0.
Therefore, P(A and B) ≠ 0.
(b) By the preceding line, the definition of mutually exclusive events is violated. Thus A and B are not
mutually exclusive.
5.
6.
7.
(a) P(A and B)
(b) P(B | A)
(d) P(A or B)
(e) P(A or Bc)
(a) P(Ac or B)
(b) P(B | A)
(d) P(A and Bc)
(e) P(A and B)
(c) P(Ac | B)
(c) P(A | B)
(a) Green and blue are mutually exclusive because each M&M candy is only one color.
P(green or blue) = P(green) + P(blue) = 10% + 10% = 20% = 0.20.
(b) Yellow and red are mutually exclusive once again because each candy is only one color.
P(yellow or red) = P(yellow) + P(red) = 20% + 20% = 40% = 0.40.
(c) Use the complementary event.
P(not purple) = 1 – P(purple) = 1 – 0.20 = 0.80 = 80%
8.
The total number of arches tabled is 288. Arch heights are mutually exclusive.
111
(a) P(3 to 9 feet) 
288
30
33
18
81



(b) P(30 feet or taller) = P(30 to 49) + P(50 to 74) + P(75 and higher) 
288 288 288 288
111 96
30 237



(c) P(3 to 49 feet) = P(3 to 9) + P(10 to 29) + P(30 to 49) 
288 288 288 288
96
30
33 159



(d) P(10 to 74 feet) = P(10 to 29) + P(30 to 49) + P(50 to 74) 
288 288 288 288
18
(e) P(75 feet or taller) 
288
Hint: For Problems 9–12, refer to Figure 4-2 if necessary. Think of the outcomes as an (x, y) ordered pair.
9.
(a) Yes, the outcome of the red die does not influence the outcome of the green die.
 1  1  1
(b) P(5 on green and 3 on red) = P(5 on green) · P(3 on red)     
 0.028 .
 6  6  36
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Part IV: Complete Solutions, Chapter 4
301
 1  1  1
(c) P(3 on green and 5 on red) = P(3 on green) · P(5 on red)     
 0.028
 6  6  36
(d) P[(5 on green and 3 on red) or (3 on green and 5 on red)]
= P(5 on green and 3 on red) + P(3 on green and 5 on red)
1
1
2
1



 0.056 (because they are mutually exclusive outcomes).
=
36 36 36 18
10. (a) Yes.
 1  1  1
(b) P(1 on green and 2 on red) = P(1 on green) · P(2 on red)     
 6  6  36
 1  1  1
(c) P(2 on green and 1 on red) = P(2 on green) · P(1 on red)     
 6  6  36
(d) P[(1 on green and 2 on red) or (2 on green and 1 on red)]
= P(1 on green and 2 on red) + P(2 on green and 1 on red)
=
1
1
2
1



(because they are mutually exclusive outcomes).
36 36 36 18
11. (a) We can obtain a sum of 6 as follows:
1+5=6
2+4=6
3+3=6
4+2=6
5+1=6
P (sum  6)  P[(1, 5) or (2, 4) or (3 on red, 3 on green) or (4, 2) or (5, 1)]
 P (1, 5)  P (2, 4)  P(3, 3)  P(4, 2)  P(5, 1)
because the (red, green) outcomes are mutually exclusive
 1  1   1  1   1  1   1  1   1  1 
                   
 6  6   6  6   6  6   6  6   6  6 
because the red die outcome is independent of the green die outcome
1
1
1
1
1
5

   

36 36 36 36 36 36
(b) We can obtain a sum of 4 as follows:
1+3=4
2+2=4
3+1=4
P (sum is 4)  P[(1, 3) or (2, 2) or (3, 1)]
 P(1, 3)  P(2, 2)  P(3, 1)
because the (red, green) outcomes are mutually exclusive
 1  1   1  1   1  1 
           
 6  6   6  6   6  6 
because the red die outcome is independent of the green die outcome
1
1
1
3
1

 


36 36 36 36 12
(c) You cannot roll a sum of 6 and a sum of 4 at the same time. These are mutually exclusive events.
5
3
8 2



P(sum of 6 or 4) = P(sum of 6) + P(sum of 4) =
36 36 36 9
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Part IV: Complete Solutions, Chapter 4
302
12. (a) We can obtain a sum of 7 as follows:
1+6=7
2+5=7
3+4=7
4+3=7
5+2=7
6+1=7
P (sum is 7)  P[(1, 6) or (2, 5) or (3, 4) or (4, 3) or (5, 2) or (6, 1)]
 P(1, 6)  P (2, 5)  P (3, 4)  P (4, 3)  P(5, 2)  P(6, 1)
because the (red, green) outcomes are mutually exclusive
 1  1   1  1   1  1   1  1   1  1   1  1 
                       
 6  6   6  6   6  6   6  6   6  6   6  6 
because the red die outcome is independent of the green die outcome
1
1
1
1
1
1
6 1

    


36 36 36 36 36 36 36 6
(b) We can obtain a sum of 11 as follows: 5 + 6 = 11 or 6 + 5 = 11
P (sum is 11)  P[(5, 6) or (6, 5)]
 P(5, 6)  P(6, 5)
because the (red, green) outcomes are mutually exclusive
 1  1   1  1 
       
 6  6   6  6 
because the red die outcome is independent of the green die outcome
1
1
2
1




36 36 36 18
(c) You cannot roll a sum of 7 and a sum of 11 at the same time. These are mutually exclusive events.
6
2
8 2



P(sum is 7 or 11) = P(sum is 7) + P(sum is 11) =
36 36 36 9
13. (a) No, the draws are not independent. The key idea is “without replacement” because the probability of
the second card drawn depends on the first card drawn. Let the card draws be represented by an (x, y)
ordered pair. For example, (K, 6) means the first card drawn was a king and the second card drawn was
a 6. Here the order of the cards is important.
16
4
 4  4 
(b) P(ace on first draw and king on second draw) = P(ace, king)     

 52  51  2,652 663
There are four aces and fpour kings in the deck. Once the first card is drawn and not replaced, there are
only 51 cards left to draw from, but all the kings are available.
16
4
 4  4 
(c) P(king, ace)     

 52  51  2652 663
(d) P(ace and king in either order)
= P[(ace, king) or (king, ace)]
= P(ace, king) + P(king, ace)
because these two outcomes are mutually exclusive
16
16
32
8
=



2, 652 2, 652 2, 652 663
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Part IV: Complete Solutions, Chapter 4
303
14. (a) No, the draws are not independent. The key idea is “without replacement” because the probability of
the second card drawn depends on the first card drawn. Let the card draws be represented by an (x, y)
ordered pair. For example, (K, 6) means the first card drawn was a king and the second card drawn was
a 6. Here the order of the cards is important.
(b) P(3, 10)  P[(3 on 1st) and (10 on 2nd, given 3 on 1st)]
 P(3 on 1st)  P(10 on 2nd, given 3 on 1st)
16
4
 4  4 
    

 0.006
52
51
2,
652
663
  
(c) P(10, 3)  P[(10 on 1st) and (3 on 2nd, given 10 on 1st)]
 P(10 on 1st)  P(3 on 2nd, given 10 on 1st)
16
4
 4  4 
    

 0.006
 52  51  2, 652 663
(d) P[(3, 10) or (10, 3)] = P(3, 10) + P(10, 3) because these two outcomes are mutually exclusive.
4
4
8


 0.012
=
663 663 663
15. (a) Yes, the draws are independent. The key idea is “with replacement.” When the first card drawn is
replaced, the sample space is the same for the second card as it was for the first card. In fact, it is
possible to draw the same card twice. Let the card draws be represented by an (x, y) ordered pair; for
example, (K, 6) means a king was drawn, replaced, and then the second card, a 6, was drawn.
(b) P(A, K)  P(A)  P(K) because they are independent.
16
1
 4  4 
    

52
52
2,
704
169
  
(c) P(K, A)  P(K)  P(A) because they are independent.
16
1
 4  4 
    

 52  52  2, 704 169
(d) P[(A, K) or (K, A)] = P(A, K) + P(K, A) because the two outcomes are mutually exclusive.
1
1
2


=
169 169 169
16. (a) Yes, the draws are independent. The key idea is “with replacement.” When the first card drawn is
replaced, the sample space is the same for the second card as it was for the first card. In fact, it is
possible to draw the same card twice. Let the card draws be represented by an (x, y) ordered pair; for
example, (K, 6) means a king was drawn, replaced, and then the second card, a 6, was drawn.
(b) P(3, 10)  P(3)  P(10) because draws are independent.
16
1
 4  4 
    

 0.0059
 52  52  2, 704 169
(c) P(10, 3)  P(10)  P(3) because of independence.
16
1
 4  4 
    

 0.0059
 52  52  2, 704 169
(d) P[(3, 10) or (10, 3)] = P(3, 10) + P(10, 3) because the two outcomes are mutually exclusive.
1
1
2


 0.0118
=
169 169 169
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Part IV: Complete Solutions, Chapter 4
304
17. (a)
(b)
(c)
(d)
P(6 years old or older) = 27% + 14% + 22% = 63%
P(12 years old or younger) = 1 – P(13 years old or older) = 100% – 22% = 78%
P(Between 6 and 12 years old) = 27% + 14% = 41%
P(Between 2 and 9 years old) = 22% + 27% = 49%
The 13-and-older category may include children up to 17 or 18 years old. This is a larger category.
18. Let S denote “senior.” Let F denote “got the flu.” We are given the following probabilities:
P(F | S) = 0.14
P(F | Sc) = 0.24
P(S) = 0.125
P(Sc) = 0.875
(a) P(S and F) = P(S) × P(F | S) = (0.125) × (0.14) = 0.0175
(b) P(Sc and F) = P(Sc) × P(F | Sc) = (0.875) × (0.24) = 0.21
(c) Here, P(S) = 0.95, so P(Sc) = 1 – 0.95 = 0.05
(a) P(S and F) = P(S) × P(F | S) = (0.95) × (0.14) = 0.133
(b) P(Sc and F) = P(Sc) × P(F | Sc) = (0.05) × (0.24) = 0.012
(d) Here, P(S) = P(Sc) = 0.50.
(a) P(S and F) = P(S) × P(F | S) = (0.50) × (0.14) = 0.07
(b) P(Sc and F) = P(Sc) × P(F | Sc) = (0.50) × (0.24) = 0.12
19. Let T denote “telling the truth.” Let L denote “machine catches a person lying.” We are given the following
probabilities:
P(L | Tc) = 0.72
P(L | T) = 0.07
(a) Given P(T) = 0.90. Then
P(T and L) = P(T) × P(L | T) = (0.90) × (0.07) = 0.063
(b) Given P(Tc) = 0.10. Then
P(Tc and L) = P(Tc) × P(L | Tc) = (0.10) × (0.72) = 0.072
(c) Given P(T) = P(Tc) = 0.50. Then
P(T and L) = (0.50) × (0.07) = 0.035
P(Tc and L) = (0.50) × (0.72) = 0.36
(d) Given P(T) = 0.15 and P(Tc) = 0.85. Then
P(T and L) = (0.15) × (0.07) = 0.0105
P(Tc and L) = (0.85) × (0.72) = 0.612
20. (a) We want to solve for P(Tc). There are two possibilities when the polygraph says that the person is
lying: Either the polygraph is right, or the polygraph is wrong. If the polygraph is right, the polygraph
results show “lying,” and the person is not telling the truth; i.e., P(L and not T). If the polygraph is
wrong, then the polygraph results show “lying,” but in fact, the person is telling the truth; i.e., P(L and
T).
P(L) = P(L and Tc) + P(L and T) =
[P(Tc) × P(L | Tc)] + [P(T) × P(L | T)] =
[P(Tc) × P(L | Tc)] + {[1 – P(Tc)] × P(L | T)}
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Part IV: Complete Solutions, Chapter 4
305
We are told that P(L) = 0.30, so
0.30 = [P(Tc) × P(L | Tc)] + {[1 – P(Tc)] × P(L | T)}
= [P(Tc) × 0.72 ] + {[1 – P(Tc)] × 0.07}
= (0.72) × P(Tc) + {0.07 – [0.07 × P(Tc)]}
0.23 = P(Tc) × (0.72 – 0.07) = P(Tc) × (0.65)
(**)
0.23/0.65 = P(Tc) = 0.354 = 35.4%
(b) Here, P(L) = 70% = 0.70. Replace the 0.30 with 0.70 in (**) and solve.
P(Tc) = 0.63/0.65 = 0.969
686
1,160
270
P(S | A) 
580
416
P(S | Pa) 
580
(b) No, they are not independent. P(S | Pa) ≠ P(S) based on the previous part.
21. (a) P(S) 
(c) P(A and S) = 270/1,160 using the table.
P(Pa and S) = 416/1,160 using the table.
474
1,160
310
P(N | A) 
580
(e) No, they are not independent. P(N | A) ≠ P(N) based on the preceding part.
(d) P(N) 
(f)
P( A or S )  P( A)  P( S )  P( A and S )
580
686
270
996




1,160 1,160 1,160 1,160
110
130
20
(b) P(– |condition present) 
130
50
(c) P(– | condition absent) =
70
20
(d) P(+ | condition absent) 
70
(e) P(condition present and +) = P(condition present) × P(+ | condition present)
 130  110  110



 200  130  200
(f) P(condition present and –) = P(condition present) × P(– | condition present)
 130  20  20



 200  130  200
22. (a) P(+ | condition present) 
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Part IV: Complete Solutions, Chapter 4
306
23. Let C denote the presence of the condition and not C denote absence of the condition.
72
(a) P(+ | C) 
154
82
(b) P(– | C) 
154
79
(c) P(– | not C) 
116
37
(d) P(+ | not C) 
116
 154  72  72
(e) P(C and +) = P(C) × P(+ | C)  


 270  154  270
 154  82  82
(f) P(C and –) = P(C) × P(– | C)  


 270  154  270
24. (a) P(10 to 14 years) =
291
2008
(b) P(10 to 14 years | East) =
(c) P(at least 10 years) =
77
452
291  535 826

2008
2008
(d) P(at least 10 years | East) =
45  86 131

373
373
(e) P(West | less than 1 year) =
41
157
(f) P(South | less than 1 year) =
53
157
(g) P(1 or more years | East) = 1 – P(less than 1 year | East) = 1 
32
420

452 452
(h) P(1 or more years | West) = 1 – P(less than 1 year | West) = 1 
41 332

373 373
(i) We can check if P(East) = P(15 or more years | East). If these probabilities are equal, then the events
are independent.
P(East) 
452
 0.225
2008
P (15 years  | East) 
118
 0.261
452
Since the probabilities are not equal, the events are not independent.
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Part IV: Complete Solutions, Chapter 4
307
25. Given: Let A be the event that a new store grosses > $940,000 in year 1; then Ac is the event the new store
grosses ≤ $940,000 the first year.
Let B be the event that the store grosses > $940,000 in the second year; then Bc is the event the store
grosses ≤ $940,000 in the second year of operation.
2-Year Results
A and B
A and Bc
Ac and B
Ac and Bc
Translations
Profitable both years
Profitable first but not second year
Profitable second but not first year
Not profitable either year
P(A) = 65% (show profit in first year)
P(Ac) = 35%
P(B) = 71% (show profit in second year)
P(Bc) = 29%
P(close) = P(Ac and Bc)
P(B, given A) = 87%
(a)
(b)
(c)
(d)
(e)
(f)
P(A) = 65% = 0.65
P(B) = 71% = 0.71
P(B | A) = 87% = 0.87
P(A and B) = P(A) × P(B | A) = (0.65)(0.87) = 0.5655 ≈ 0.57
P(A or B) = P(A) + P(B) – P(A and B) = 0.65 + 0.71 – 0.57 = 0.79
P(not closed) = P(show a profit in year 1 or year 2 or both) = 0.79
P(closed) = 1 – P(not closed) = 1 – 0.79 = 0.21
26. P(female) = 85%, so P(male) = 15%
P(BSN | female) = 70%
P(BSN | male) = 90%
(a)
(b)
(c)
(d)
(e)
P(BSN | female) = 70% = 0.70
P(BSN and female) = P(female) × P(BSN | female) = (0.85) × (0.70) = 0.595
P(BSN | male) = 90% = 0.90
P(BSN and male) = P(male) × P(BSN | male) = (0.15)(0.90) = 0.135
Of the graduates, some are female and some are male. We can add the mutually exclusive
probabilities.
P(BSN) = [P(BSN | female) × P(female)] + [P(BSN | male) × P(male)] = [(0.70) × (0.85)] + [(0.90) ×
(0.15)] = 0.73
(f) The phrase “will graduate and is female” describes the proportion of all students who are female and
will graduate. The phrase “will graduate, given female” describes the proportion of the females who
will graduate. Observe from parts (a) and (b) that the probabilities are indeed different.
27. Let TB denote that the person has tuberculosis.
Let + denote the test for tuberculosis indicates the presence of the disease.
Let – denote the test for tuberculosis indicates the absence of the disease.
We are given the following probabilities:
P(+ | TB) = 0.82 (sensitivity of the test)
P(+ | TBc) = 0.09 (false-positive rate)
P(TB) = 0.04
(a) P(TB and +) = P(+ | TB) × P(TB) = (0.82) × (0.04) = 0.0328
(b) P(TBc) = 1 – P(TB) = 1 – 0.04 = 0.96
(c) P(TBc and +) = P(+ | TBc) × P(TBc) = (0.09) × (0.96) = 0.0864
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Part IV: Complete Solutions, Chapter 4
308
28. Known: Let A be the event the client relapses in phase I.
Let B be the event the client relapses in phase II.
Let C be the event that the client has no relapse in phase I; i.e., C = not A.
Let D be the event that the client has no relapse in phase II; i.e., D = not B.
P(A) = 0.27, so P(Ac) = P(C) = 1 – 0.27 = 0.73
P(B) = 0.23, so P(Bc) = P(D) = 1 – 0.23 = 0.77
P(Bc | Ac) = 0.95 = P(D | C) = 0.95
P(B | A) = 0.70
Possible Outcomes
A, B
Ac, B (= C, B)
A, Bc (= A, D)
Ac, Bc (= C, D)
Translation
Relapse in I, relapse in II
No relapse in I, relapse in II
Relapse in I, no relapse in II
No relapse in I no relapse in II
(a) P(A) = 0.27, P(B) = 0.23, P(C) = 0.73, P(D) = 0.77
(b) P(B | A) = 0.70, P(D | C) = 0.95
(c) P(A and B) = P(A) × P(B | A) = (0.27) × (0.70) = 0.189
P(C and D) = P(C) × P(D | C) = (0.73) × (0.95) = 0.6935
(d) P(A or B) = P(A) + P(B) – P(A and B) = 0.27 + 0.23 – 0.189 = 0.311
(e) P(C and D) = 0.69
(f) P(A and B) = 0.189
(g) Translate as the inclusive or. P(A or B) = 0.31.
Section 4.3
1.
The permutations rule counts the number of different arrangements, or r items out of n distinct items. Here,
the ordering matters. The combinations rule counts the number of groups of r items out of n distinct items.
Here, the ordering does not matter. For a permutation, ABC is different from ACB. For a combination,
ABC and ACB are the same item. The number of permutations is larger than the number of combinations.
2.
A tree diagram lists all possible events. The user of the diagram can trace the sequential event from the start
to the end by following a distinct path along the branches. Counting the number of final branches gives the
total number of outcomes.
3.
(a) Use the combinations rule because we are concerned only with the groups of size five.
(b) Use the permutations rules because we are concerned with the number of different arrangements of
size five.
4.
Both methods are correct because you are counting the number of possible arrangements of five items
taken five at a time.
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Part IV: Complete Solutions, Chapter 4
5.
309
(a)
(b) HHT, HTH, THH. There are three outcomes.
(c) There are eight possible outcomes, and three outcomes have exactly two heads.
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3
.
8
Part IV: Complete Solutions, Chapter 4
310
6.
(a)
(b) H5, H6. There are two outcomes.
(c) There are 12 possible outcomes, and 2 outcomes meet the requirements.
2 1
 .
12 6
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Part IV: Complete Solutions, Chapter 4
7.
311
(a)
(b) Let P(x, y) be the probability of choosing an x-colored ball on the first draw and a y-colored ball on the
second draw. Notice that the probabilities add to 1.
 3  2  6 1
P ( B, R )      

 6   5  30 5
1
 3  2  6 1
 2  1  2
P ( B, B )      

P( R, R)     

 6   5  30 5
 6  5  30 15
1
 3  1  3
 2  3  6 1
P ( B, Y )      

P( R, B)     

 6   5  30 10
 6  5  30 5
1
1
 1  2  2
 2  1  2
P (Y , R )      

P( R, Y )     

6
5
30
15
  
 6  5  30 15
1
 1  3  3
P (Y , B )      

 6   5  30 10
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312
8.
Part IV: Complete Solutions, Chapter 4
(a) For clarity, only a partial tree diagram is provided. Each of the branches for B, C, and D would
continue in the same manner as the fully expanded A branch.
 1  1  1  1
(b) If the outcomes are equally likely, then P(all 3 correct)      
.
 4  4  4  64
9.
Using the provided hint, we multiply. There are 4 × 3 × 2 × 1 = 4! = 24 possible wiring configurations.
10. Using the multiplication rule, we multiply. There are 4! = 24 possible ways to visit the four cities. This
problem is exactly like Problem 9.
11. There are four fertilizers, three temperature zones for each fertilizer, and three water treatments for every
fertilizer–temperature zone combination. She needs to test 4 × 3 × 3 = 36 plots.
12. (a) The die rolls are independent, so multiply the six outcomes for the first die and the six outcomes for
the second die. There are 6 × 6 = 36 possible outcomes.
(b) There are three possible even outcomes per die. There are 3 × 3 = 9 outcomes.
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Part IV: Complete Solutions, Chapter 4
313
9 1
  0.25
36 4
number of favorable outcomes
Using P(event) 
total number of outcomes
(c) P(even, even) 
Problems 13, 14, 15, and 16 deal with permutations.
n!
Use Pn, r 
to count the number of ways r objects can be selected from n objects when ordering
(n  r )!
matters.
13. P5, 2 : n  5, r  2
P5, 2 
5!
5  4  3  2 1

 20
(5  2)!
3!
14. P8,3 : n  8, r  3
P8,3 
8!
8  7  6  5  4  3  2 1 8  7  6  5!


 336
(8  3)!
5!
5!
15. P7,7 : n  r  7
P7, 7 
7!
7!
  7!  5, 040 (recall 0!  1)
(7  7)! 0!
In general, Pn, n 
n!
n! n!
   n! .
(n  n)! 0! 1
16. P9,9 : n  r  9
P9,9 
9!
9! 9!
   362,880
(9  9)! 0! 1
Problems 17, 18, 19, and 20 deal with combinations.
n!
Use Cn, r 
to count the number of ways r objects can be selected from n objects when ordering
r !(n  r )!
is irrelevant.
17. C5, 2 : n  5, r  2
C5, 2 
5!
5! 5  4  3  2 1 20



 10
2!(5  2)! 2!3! 2 1 3  2 1 2
18. C8,3 : n  8, r  3
C8,3 
8!
8! 8  7  6  5!


 56
3!(8  3)! 3!5! 3  2 1  5!
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Part IV: Complete Solutions, Chapter 4
314
19. C7,7 : n  r  7
7!
7!
7!


 1 (recall 0!  1)
7!(7  7)! 7!0! 7!(1)
C7, 7 
In general, Cn, n 
n!
n!
n!


 1. There is only one way to choose n objects without regard
n !(n  n)! n !0! n !(1)
to order.
20. C8,8 : n  r  8
8!
8!
8!


 1 (recall 0!  1)
8!(8  8)! 8!0! 8!(1)
C8,8 
21. Since the order matters (first is day supervisor, second is night supervisor, and third is coordinator), this is a
permutation of 15 nurse candidates to fill three positions.
P15,3 
15!
15! 15 14 13 12!


 2, 730
(15  3)! 12!
12!
22. Order matters here because the order of the finalists selected determines the prize awarded.
P10,3 
10!
10! 10  9  8  7!


 720
(10  3)! 7!
7!
23. Order matters because the resulting sequence determines who wins first, second, and third place.
P5,3 
5!
5! 120
 
 60
(5  3)! 2!
2
24. The order of the software packages selected is irrelevant, so use the combinations method.
C10,3 
10!
10! 10  9  8  7! 720



 120
3!(10  3)! 3!7!
3!7!
6
25. The order of trainee selection is irrelevant, so use the combinations method.
C15,5 
15!
15! 15 14 13 12 1110! 15 14 13 12 11



 3, 003
5!(15  5)! 5!10!
5!10!
5  4  3  2 1
26. The order of the problems selected is irrelevant, so use the combinations method.
12!
12! 12 11 10  9  8  7! 12 11 10  9  8



 792
5!(12  5)! 5!7!
5!7!
5  4  3  2 1
(b) Jerry must have completed the same five problems as the professor selected to grade.
1
 0.001
P(Jerry chose the right problems) 
792
7!
7! 7  6  5! 7  6 42




 21
(c) Silvia did seven problems, so she completed C7,5 
5!(7  5)! 5!2!
5!2!
2 1 2
possible subsets.
(a) C12,5 
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Part IV: Complete Solutions, Chapter 4
P(Silvia picked the correct set of graded problems) 
315
21
 0.027
792
Silvia increased her chances by a factor of 21 compared with Jerry.
27. (a) Six applicants are selected from among 12 without regard to order. C12, 6 
12! 479, 001, 600

 924.
6!6!
(720)2
(b) This problem is asking, “In how many ways can six women be selected from seven applicants?”
C7,6 
7!
7
6!  1!
number of favorable outcomes
total number of outcomes
7
1

 0.008
P(all hired are women) 
924 132
(c) P(event A) 
Chapter 4 Review
1.
(a)
(b)
(c )
(d)
(e)
The individual does not own a cell phone.
The individual owns both a cell phone and a laptop computer.
The individual owns either a cell phone or a laptop computer or both.
A laptop owner who owns a cell phone.
A cell phone owner who owns a laptop.
2.
(a) Only if events A and B are mutually exclusive. Then P(A and B) = 0 and P(A or B) = P(A) + P(B).
(b) Yes, see above.
3.
(a) No, unless events A and B are independent. If they are not, we need either P(A | B) or P(B | A) to
compute P(A and B).
(b) Yes, now we can compute P(A and B) = P(A) × P(B).
4.
The information yields P(B | A) = 2. Probabilities must be between 0 and 1 inclusive. Also, P(A and B)
cannot be greater than P(A) or P(B) individually.
5.
P(asked) = 24% = 0.24
P(received | asked) = 45% = 0.45
P(asked and received) = P(asked) × P(received | asked) = (0.24) × (0.45) = 0.108 = 10.8%
6.
P(asked) = 20% = 0.20
P(received | asked) = 59% = 0.59
P(asked and received) = P(asked) × P(received | asked) = (0.20) × (0.59) = 0.118 = 11.8%
7.
(a) Throw a large number of similar thumbtacks or one thumbtack a large number of times, and record the
relative frequency of the outcomes. Assume that the thumbtack falls either flat side down or tilted. To
estimate the probability the tack lands on its flat side, find the relative frequency of this occurrence,
dividing the number of times this occurred by the total number of thumbtack tosses.
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Part IV: Complete Solutions, Chapter 4
316
(b) The sample space consists of two outcomes: flat side down and tilted.
340
 0.68
(c) P(flat side down) 
500
8.
(a)
(b)
(c)
(d)
P(tilted)  1  0.68  0.32
470
P( N ) 
 0.470
1000
390
P( M ) 
 0.390
1000
140
P( S ) 
 0.140
1000
420
 0.840
P(N | W) 
500
20
 0.040
P(S | W) 
500
50
 0.100
P(N | A) 
500
120
 0.240
P(S | A) 
500
P(N and W) = P(W) × P(N | W) = (0.50) × (0.84) = 0.42
P(M and W) = P(W) × P(M | W) = (0.50) – (0.12) = 0.06
(e) P( N or M )  P ( N )  P ( M ) if mutually exclusive
 470   390  860

 0.860


 1, 000   1, 000  1, 000
No reaction is mutually exclusive from a mild reaction; they cannot occur at the same time.
(f) If N and W were independent, P(N and W) = P(N) · P(W) = (0.470) × (0.500) = 0.235.
However, from (d), we have P(N and W) = 0.420. They are not independent.
9.
(a) Possible values for x are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12.
(b) Below, the values for x are listed, along with the combinations required.
2
3
4
5
6
7
8
9
10
11
12
1 and 1
1 and 2, or 2 and 1
1 and 3, 2 and 2, 3 and 1
1 and 4, 2 and 3, 3 and 2, 4 and 1
1 and 5, 2 and 4, 3 and 3, 4 and 2, 5 and 1
1 and 6, 2 and 5, 3 and 4, 4 and 3, 5 and 2, 6 and 1
2 and 6, 3 and 5, 4 and 4, 5 and 3, 6 and 2
3 and 6, 4 and 5, 5 and 4, 6 and 3
4 and 6, 5 and 5, 6 and 4
5 and 6, 6 and 5
6 and 6
1 way
2 ways
3 ways
4 ways
5 ways
6 ways
5 ways
4 ways
3 ways
2 ways
1 way
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Part IV: Complete Solutions, Chapter 4
x
2
3
4
5
6
7
8
9
10
11
12
P(x)
1
 0.028
36
2
 0.056
36
3
 0.083
36
4
 0.111
36
5
 0.139
36
6
 0.167
36
5
 0.139
36
4
 0.111
36
3
 0.083
36
2
 0.056
36
1
 0.028
36
Where there are (6)(6) = 36
possible, equally likely outcomes.
(The sums, however, are not
equally
likely).
10. P(pass 101) = 0.77
P(pass 102 | pass 101) = 0.90
P(pass 101 and pass 102) = P(pass 101) × P(pass 102 | pass 101) = (0.77) × (0.90) = 0.693
11. C8, 2 
8!
8  7  6! 56


 28
2!6! (2 1)6! 2
12. (a) P7, 2 
7!
7!
  7(6)  42
(7  2)! 5!
7!
76

 21
2!5!
2
3!
3!
 6
(c) P3,3 
(3  3)! 0!
(b) C7, 2 
(d) C4, 4 
4!
4!

1
4!(4  4)! 4!0!
13. Five multiple choice questions, each with flurossible (A, B, C, or D).
There are 4 × 4 × 4 × 4 × 4 = 1,024 possible sequences, such as A, D, B, B.
1
 0.00098
P(getting the correct sequence) 
1024
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317
318
Part IV: Complete Solutions, Chapter 4
14.
15. There are 10 possible numbers per turn of dial and, we turn the dial three times.
There are 10 × 10 × 10 = 1,000 possible combinations.
16. The combination uses the three numbers 2, 9, and 5, in an ordered sequence.
3!
3  2 1

 6.
The number of sequences is P3,3 
(3  3)!
0!
The possible combinations are 259, 295, 529, 592, 925, and 952.
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