If the mean and variance of a binomial variable `X` are 2 and 1 respectively. Find `P(Xge1)` .

Explanation

For Binomial distribution, Mean `(mu) = np` and Variance `(sigma^2) = np(1-p)`

According to question, Mean` = 2` and Variance `= 1`

Thus, `np = 2 ldots (1) np(1-p)=1 ldots (2)`

Putting the value of np from first equation in the second equation.

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

Now, putting the value of p in equation `(1)`, `n x 1/2 = 2 Rightarrow n = 4`

`sum P (X_i) = 1`

`P(X ge 1) = 1 – P(X lt 1) = 1 – P(X=0) = 1 - \ ^4 c_0 p^0 (1-p)^(4-0)`

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