# 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. - 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 2.

#### 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(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

Concept: Binomial Distribution
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#### APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board
Chapter 8 Binomial Distribution
Miscellaneous exercise 8 | Q 13.3 | Page 254