If the difference between mean and variance of a Binomial distribution is 1 and the difference of their squares is 5, then the probability of success is

$\frac{1}{3}$

The correct answer is Option (2) → $\frac{1}{3}$ **

Let a Binomial distribution have parameters $n$ and $p$.

Mean: $np$

Variance: $np(1-p)$

Given:

$np - np(1-p) = 1$

$\Rightarrow np[p] = 1$

$\Rightarrow np = \frac{1}{p}$ … (1)

Also given:

$(np)^{2} - (np(1-p))^{2} = 5$

$\frac{1}{p^{2}}(2p - p^{2}) = 5$

$\frac{2}{p} - 1 = 5$

$\frac{2}{p} = 6$

$p = \frac{1}{3}$

$p = \frac{1}{3}$