Maharashtra State BoardHSC Arts 12th Board Exam
Advertisement Remove all ads

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 0. - Mathematics and Statistics

Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
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 0.

Advertisement Remove all ads

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(no terminal will require attention) = P(X = 0)

`= "p"(0) = "^10C_0  (0.1)^0  (0.9)^(10 - 0)`

`= 1 xx 1 xx (0.9)^10 = (0.9)^10`

Hence, the probability that no terminal requires attention `(0.9)^10`

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 13.1 | Page 254
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×