Answer:
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
