Full Question
Eighteen telephones have just been received at an authorized service center. Six of these telephones are cellular, six are cordless, and the other six are corded phones. Suppose that these components are randomly allocated the numbers 1, 2, . . . , 18 to establish the order in which they will be serviced.
What is the probability that after servicing twelve of these phones, phones of only two of the three types remain to be serviced?
What is the probability that two phones of each type are among the first six serviced?
Answer:
a. 0.149
b. 0.182
Step-by-step explanation:
Given
Number of telephone= 18
Number of cellular= 6
Number of cordless = 6
Number of corded = 6
a.
There are 18C6 ways of choosing 6 phones
18C6 = 18564
From the Question, there are 3 types of telephone (cordless, Corded and cellular)
There are 3C2 ways of choosing 2 out of 3 types of television
3C2 = 3
There are 12C6 ways of choosing last 6 phones from just 2 types (2 types = 6 + 6 = 12)
12C6 = 924
There are 2 * 6C6 * 6C0 ways of choosing none from any of these two types of phones
2 * 6C6 * 6C0 = 2 * 1 * 1 = 2.
So, the probability that after servicing twelve of these phones, phones of only two of the three types remain to be serviced is
3 * (924 - 2) / 18564
= 3 * 922/18564
= 2766/18564
= 0.149
b)
There are 6C2 * 6C2 * 6C2 ways of choosing 2 cellular, 2 cordless, 2 corded phones
= (6C2)³
= 3375
So, the probability that two phones of each type are among the first six serviced is
= 3375/18564
= 0.182