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Question
Solve the following problem :
A computer installation has 3 terminals. The probability that any one terminal requires attention during a week is 0.1, independent of other terminals. Find the probabilities that 0
Solution
Let X denote the number of terminals that will require attention.
P(a terminal that will require attention during a week) = p = 0.1
∴ q = 1 – p = 1 – 0.1 = 0.9
Given, n = 3
∴ X ~ B(3, 0.1)
The p.m.f. of X is given by
P(X = x) = `""^3"C"_x (0.1)^x (0.9)^(3 - x),x` = 0, 1, 2, 3.
P(0 Terminal requires attention during a week)
= P(X = 0)
= `""^3"C"_0 (0.1)^0 (0.9)^3`
= (0.9)3.
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