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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 0.
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(no terminal will require attention) = P(X = 0)
`= "p"(0) = "^10C_0 (0.1)^0 (0.9)^(10 - 0)`
`= 1 xx 1 xx (0.9)^10 = (0.9)^10`
Hence, the probability that no terminal requires attention `(0.9)^10`
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