Twelvetelephones have just been received at an authorized service center. Fourof these telephones are cellular, four are cordless and the other fourare corded phones. Suppose that these components are randomly allocated the numbers 1, 2, ..., 12to establish the order in which they will be serviced. (a) What is the probability that all the cordless phones are among the first fourteen to be serviced? (b) What is the probability that after servicing fourteen of these phones, phones of only two of the three types remain to be serviced?

honestly, I'm confused

a) 0.1707

b) 0.4247

Step-by-step explanation:

This is a binomial distribution problem

There are 4 cellular phones, 4 cordless phones and 4 cellular phones.

a) Binomial distribution function is represented by

P (X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 8 phones to be serviced

x = Number of successes required = 4 cordless phones

p = probability of success = probability of a cordless phone amongst the phones to be serviced = 4/12 = (1/3)

q = probability of failure = probability of other phones = 1 - (1/3) = (2/3)

P (X = 4) = ⁸C₄ (1/3) ⁴ (2/3) ⁸⁻⁴ = 0.1707

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

This probability is equal to the sum of probabilities that

- exactly 4 cordless phones and others

- exactly 4 corded phones and others

- exactly 4 cellular phones and others

Are serviced,

Minus the probability that

- exactly 4 cordless phones and 4 corded phones

- exactly 4 cordless phones and 4 cellular phones

- exactly 4 cellular phones and 4 corded phones

Are serviced

Probability = 0.1707 + 0.1707 + 0.1707 - 3 (0.1707*0.1707) = 0.5121 - 0.087415 = 0.4247