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PA() = 1
216

Example 10.2
Find the probability of getting only two sixes in a roll of three dice.

Solution: The event in this case consists of all elementary occurrences having
the following forms:

(a, 6, 6), (6, b, 6), (6, 6, c)

where a = 1, …, 5; b = 1, …, 5; and c = 1, …, 5. Therefore, the event A consists
of elements corresponding to 15 elementary occurrences, and its probability is

PA() = 15
216

Example 10.3
Find the probability that, if three individuals are asked to guess a number
from 1 to 10, their guesses will be different numbers.

Solution: There are 1000 distinct equiprobable 3-tuplets (a, b, c), where each
component of the 3-tuplet can have any value from 1 to 10. The event A
occurs when all components have unequal values. Therefore, while a can
have any of 10 possible values, b can have only 9, and c can have only 8.
Therefore, n(A) = 8 × 9 × 10, and the probability for the event A is

×× 10
PA() = 89 = . 072
1000

Example 10.4
An inspector checks a batch of 100 microprocessors, 5 of which are defective.
He examines ten items selected at random. If none of the ten items is defec-
tive, he accepts the batch. What is the probability that he will accept the batch?

Solution: The number of ways of selecting 10 items from a batch of 100 items is:

N = 100! = 100! = C 100
10 100!( − 10)! 10 90! ! 10

where C n is the binomial coefficient and represents the number of combina-
k
tions of n objects taken k at a time without regard to order. It is equal to
n!
kn k!( − ! ) . All these combinations are equally probable.

© 2001 by CRC Press LLC