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) = ${\displaystyle {}^n{C}_x p^x q^{n-x}}$

i.e. p(x) = ${\displaystyle {}^{10}{C}_x {\left(0.1\right)}^x {\left(0.9\right)}^{10-x}}$, x = 0, 1, 2,...,10

P(2 terminals will require attention)

P(X = 2) = p(2) = ${\displaystyle {}^{10}{C}_2{\left(0.1\right)}^2{\left(0.9\right)}^{10-2}}$

${\displaystyle =\frac{10\times 9}{1\times 2}{\left(0.1\right)}^2{\left(0.9\right)}^8}$

${\displaystyle =45\left(0.01\right){\left(0.9\right)}^8}$

${\displaystyle =\left(0.45\right)\times {\left(0.9\right)}^8}$

Hence, the probability that 2 terminals require attention ${\displaystyle =\left(0.45\right)\times {\left(0.9\right)}^8}$