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प्रश्न
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 3 or more, terminals will require attention during the next week.
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उत्तर
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(3 or more terminals will require attention)
= P(X ≥ 3)
= 1 - P(x < 3)
= 1 - [P(X = 0) + P(X = 1) + P(X = 2)]
= 1 - [p(0) + p(1) + p(2)]
= `1 - [(0.9)^40 + (0.9) + (0.45)(0.9)^8]`
`= 1 -[(0.9)^2 + (0.9)^4 + 0.45](0.9)^8`
= 1 - [0.81 + 0.9 + 0.45](0.9)8
= 1 - (2.16) × (0.9)8
Hence, the probability that 3 or more terminals require attention = 1 - (2.16) × (0.9)8
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