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.

Answer 1

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`