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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 1.
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(1 terminal will require attention)
P(X = 1) = p(1) = `"^10C_1 (0.1)^1 (0.9)^(10 - 1)`
`= 10(0.1)(0.9)^9`
`= (1.0)(0.9)^9`
`= (0.9)^9`
Hence, the probability that 1 terminal requires attention `= (0.9)^9`
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