Chapter Solutions
Solution 1
This problem is an example of the relative frequency type of probability, because the probability
of an event happening is based on the number of times the particular event happened in the past
relative to the total number of observations.
a. The probability that exactly two cars are sold is:
P(2cars) 
Number of days two cars were sold 10

 0.17
Total number of days
60
b. The probability of selling three or more cars is found by using a special rule of addition
given in formula [4-2]. Let X represent the number of cars sold. ( is read “greater than or
equal to.” The notation > would be just greater than.) Then:
P( X  3)  P(3)  P( 4)  P(5 or more) 
20 15 5 40
 

 0.67
60 60 60 60
Interpreting, three cars or more are sold 67 percent of the days.
c. The probability of selling at least one car is determined by adding the probabilities of
selling one, two, three, four, and five or more cars. Again let X be the number of cars
sold, then
P( X  1)  P(1)  P( 2)  P(3)  P( 4)  P(5 or more)

5 10 20 15 5 55





 0.9166  0.92
60 60 60 60 60 60
The same result can also be found by using the complement rule. Obtain the probability of the
occurrence of a particular event by computing the probability it did not occur and then subtracting
that value from 1.0. In this example, the probability of not selling any cars is 5  60 = 0.08, then
(1  0.08) = 0.92.
P( X  1)  10
.  P( 0)
 10
. 
5
 (1  0.08)  0.92
60
Solution 2
If we combine the probabilities (0.45, 0.60, and 0.20), they exceed 1.00. The group that reads
both papers, of course, is being counted twice and must be subtracted to arrive at the answer.
Letting T represent the Morning Times, and D the Evening Dispatch, and using the general rule
of addition, formula [4-4]:
P(T or D )  P(T )  P( D )  P(T and D )
 0.45  0.60  0.20
 0.85
Thus, 85 percent of the households in the community read at least one paper.
Solution 3
These events are independent since the probability that one arrow hits the bull’s-eye does not
depend on whether the other hits it. The special rule of multiplication, formula [4-7], is used to
find the joint probability. B1 represents the first arrow, B2 the second arrow, and B3 the third
arrow.
P(all3 hitbull's-eye )  P( B1 ) P( B2 ) P( B3 )
 (0.70)(0.70)(0.70)
 0.343
Hence the probability that all three hit the bull’s-eye is 0.343.
The probability that at least one arrow hits the bull’s-eye is found by combining the complement
rule and the multiplication rule. To explain: The probability of a miss with the first arrow is 0.30,
found by P(M1) = 1  0.70. The probability for M2 and M3 is also 0.30. The multiplication rule is
used to obtain the probability that all three miss. Let X be the number of hits.
P( X  0 )  1  P( 0 )
 1  [ P( M1 )][ P( M 2 )][ P( M 3 )]
 1  [( 0.30 )( 0.30 )( 0.30 )]
 1  0.027  0.973
Thus, the likelihood that at least one of the arrows hits the bull’s-eye is 0.973.
Solution 4
a. The selections of the two carburetors are not independent events because the selection of the
first affects the second outcome. Let G1 represent the first “good” carburetor and G2 the
second “good” one.
P(G1 and G2) = P(G1) x P(G2|G1)
3 out of 4 are good
2 out of 3 are good
3 IF
2I
F
G
H4J
KG
H3J
K 0.50
Hence, the probability that neither of the two selected carburetors is defective is 0.50.
b. The probability that the defective carburetor is found requires the general rule of
multiplication and the general rule of addition.
In this case the defect may be detected either in the first test or in the second one. The general
rule of multiplication is used. Let D1 represent a defect on the first test and D2 on the second
test. The probability is:
P( find thedefect )  P( G1 )  P( D2 | G1 )  P( D1 )  P( G2 D1 )

3 IF
1I F
1 IF
3I
F
 GJ
G
J
G
J
G
H4KH3KH4KH3J
K 0.50
To explain further, the probability that the first carburetor tested is good is P(D1) = ¾. If the first
one selected is good, then to meet the requirements of the problem the second one sampled must
be defective. This conditional probability is P(D2|G1) = 1/3. The joint probability of these two
events is 3/12 or ¼. The defective part could be found on the first test P(D1 ). Since there is one
defect among the four carburetors, the probability that it will be found on the first test is ¼. If the
defect is found on the first test, then the three remaining parts are good. Hence the conditional
probability of selecting a good carburetor on the second trial is 1.0 P(G2|D1). The joint probability
of a defective part being followed by a good part is ¼, found by P(D1) x P(G2|D1) = (1/4)(3/3) =
¼. The sum of these two outcomes is 0.50.
Solution 5
a. If we combine the events “Less than $20” (B1) and “Cash payment” (A1), then those who paid
cash for a purchase of less than $20 are counted twice. That is, these two events are not
mutually exclusive. Therefore, the general rule of addition formula [4-4] is used.
P( A1 or B1 )  P( A1 )  P( B1 )  P( A1 and B1 )
30 35 15
50




 0.36
140 140 140 140
The probability of selecting a customer who made a cash payment or purchased an item for
less than $20 is 0.36.
b. Conditional probability is used to find the probability of selecting someone who paid by
check (A2) and who made a purchase of over $50 (B3)
There are two qualifications: “paid by check” and “made a purchase of over $50.” Referring
to the table, 20 out of 140 customers meet both qualifications, therefore, 20  140 = 0.14.
This probability could also be computed in a three-step process.
1.
The probability of selecting those who paid by check (A2) is 60  140 = 0.43.
2.
Of the 60 persons who paid by check, 20 made a purchase of over $50. Therefore P(B3|A2)
= 20  60 = 0.33
3.
These two events are then combined using the general rule of multiplication, formula [5-6].
P( A2 and B3 )  P( A2 ) P( B3 A2 )
 ( 0.43)( 0.33)  014
.
Solution 6
Using the multiplication formula, there are five cheeses (c), four meats (m), and three rolls (r).
The total number of possible sandwiches is 60 found by:
cmr = (5)(4)(3) = 60
Solution 7
This is an example of a permutation because a different assignment of the scholarships means
another arrangement. Jones could be awarded the $2,000 scholarship, Sinski the $2,400
scholarship, and Peters the $3,000 scholarship. Or, Sinski could be awarded the $2,000, Seiple the
$2,400 one, and Orts the $3,000 scholarship, and so on.
n
Pr 
n!
12!
12  11  10  9!


 1,320
n  r ! 12  3 !
9!
b gb g
Where:
n is the total number of applicants
r is number of scholarships.
Solution 8
This is an example of a combination because the order in which the players are selected is not
important. Jocko, Camden, Urfer, Smith, and Marchal are the same starting lineup as Smith,
Camden, Marchal, Jocko, and Urfer, and so on.
n Cr

n!
12!
12  11  10  9  8  7! 95,040



 792
r ! n  r ! 5! 12  5 !
5  4  3  2  1  7!
120
b g b g
Where:
n is the total number of available players.
r is the number in the starting lineup.