We know that the mean and variance of a binomial distribution with number of trials as ‘n’ and the probability of success as ‘p’ are Mean = np Variance = np(1-p) Therefore applying the given conditions yields us np=2 np(1−p)=1 2(1−p)=1 1−p=
1
2
p=
1
2
Therefore the number of trials will be 4 P(X=0) =4C0(1−P)4 =