# A computer installation has 10 terminals. Independently, the probability that any one terminal will require attention during a week is 0.1. - Mathematics and Statistics

Sum

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

Concept: Binomial Distribution
Is there an error in this question or solution?

#### APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board
Chapter 8 Binomial Distribution
Miscellaneous exercise 8 | Q 13.4 | Page 254