If the mean and variance of a binomial distribution are 4 and 2 respectively, then the probability of 2 successes of that binomial variate X , is

Step-by-step solution

For a binomial distribution, Mean, =n =4 ... (i) Variance, ^2=n p q=2 ... (ii) On solving eqs. (i) and (ii), we get n=8, p = 1 2 , q= 1 2 Probability of getting 2 successes, p ( X =2)= ^8 C ₂ ( 1 2 )^2 ( 1 2 )^6= ^8 C ₂ 2^8 = 28 256 = 7 64