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Question
A computer installation has 10 terminals. Independently, the probability that any one terminal will require attention during a week is 0.1. Find the probabilities that 2.
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Solution
Let X = number of terminals which required attention during a week.
p = probability that any terminal will require attention during a week
∴ p = 0.1
and q = 1 - p = 1 - 0.1 = 0.9
Given: n = 10
∴ X ~ B (10, 0.1)
The p.m.f. of X is given by
P(X = x) = `"^nC_x p^x q^(n - x)`
i.e. p(x) = `"^10C_x (0.1)^x (0.9)^(10 - x)`, x = 0, 1, 2,...,10
P(2 terminals will require attention)
P(X = 2) = p(2) = `"^10C_2 (0.1)^2 (0.9)^(10 - 2)`
`= (10 xx 9)/(1 xx 2) (0.1)^2 (0.9)^8`
`= 45(0.01)(0.9)^8`
`= (0.45) xx (0.9)^8`
Hence, the probability that 2 terminals require attention `= (0.45) xx (0.9)^8`
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