Question

The mean and variance of binomial distribution are 4 and 2 respectively. Find the probability of two successes.

Answer 1

Mean = np = 4

Variance = npq = 2

∴ `(npq)/(np) = 2/4`

`implies` q = `1/2`,

∴ p = `1/2`

Now np = 4

`\implies n*1/2` = 4

`\implies` n = 8

∴ Probability of two successes,

P(X = 2) = `""^8C_2(1/2)^2(1/2)^6`

= `(8 xx 7)/2*1/4*1/2^8`

= `28/256`