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.

#### Solution

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}^{ }