Question

A box contains 10 red balls and a single black ball. Balls are drawn (one at the time) without replacing until the black ball is drawn. Let X = number of balls drawn before the black ball is drawn. Find the p.m.f, the expected value and the standard deviation of X.

Answer 1

Given

Number of red balls =10

Number of black balls = 1

n=total number of balls = 10+1 = 11

P=p(ball is black) = 1/11 = 0.0909 = 0.091

X= number of black balls are drawn before the red ball is drawn

This is same as number of failures before the first success

Drawing black ball is success here.

And

P=p(success) =p(black ball) = 0.091

All the draws are independent of each other.

Here

X follows Geometric distribution with P=0.091

The p.m.f. Is

P(X=x) =P*(1-P) ^x x=0,1,2,3,...

We know that FOR Geometric distribution

Expected value = (1-P)/P =9.989

Standard deviation = ​​​​​​

= = 10.4771