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 2.

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(2 terminals will require attention)

P(X = 2) = p(2) = `"^10C_2 (0.1)^2 (0.9)^(10 - 2)`

`= (10 xx 9)/(1 xx 2) (0.1)^2 (0.9)^8`

`= 45(0.01)(0.9)^8`

`= (0.45) xx (0.9)^8`

Hence, the probability that 2 terminals require attention `= (0.45) xx (0.9)^8`