If the mean and variance of a binomial variable `X` are 2 and 1 respectively. Find `P(X >1)` .

Explanation

Mean `(mu) = np`

Variance `(sigma ^2) = np(1-p)`

`mu = 2 and Sigma^2 = 1 ldot ldots` Given

`therefore 2 * (1-p) = 1 Rightarrow 1-p = 1/2 Rightarrow p=1/2`

Since `sum P(X_i) = 1`

`therefore P(X ge 1) = 1 – P(X lt 1) = 1 – P(X=0) = 1 - \ ^4C_0 p^0 )1-p)^(4-0)`

`=1 – (4!)/(0!(4-0)!) * (1/2)^4 = 1 – 1/16 = 15/16`