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. (Round your answers to four decimal places.)
(a) What is the probability that all the cordless phones are among the first twelve to be serviced?

Ans: .0498 Correct: Your answer is correct.

(b) What is the probability that after servicing twelve of these phones, phones of only two of the three types remain to be serviced?

Ans: .0356 Incorrect: Your answer is incorrect.

(c) What is the probability that two phones of each type are among the first six serviced?

Ans: .0000000000005 Incorrect: Your answer is incorrect.

Answer 1

Answer:

Step-by-step explanation:

Given that 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.

a)  the probability that all the cordless phones are among the first twelve to be serviced

= chance of arranging all 6 cordless phone among the first 12

= $\frac{12C6}{18C6} \\=0.049774$

= 0.0498

b) the probability that after servicing twelve of these phones, phones of only two of the three types remain to be serviced

=chance of arranging 4 from each type in the first 12 and 2 of each type in the last 6

=$\frac{6C4 * 6C4 * 6C4}{18C12} \\=0.1818$

c) the probability that two phones of each type are among the first six serviced

= chance to get 2 each from each type

=$\frac{^6C2)^3}{18C6} \\=0.1818$