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
Answer & explanation
Correct answer: option 2
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:
$(np)^{2} - (np(1-p))^{2} = 5$
$\frac{1}{p^{2}}(2p - p^{2}) = 5$
$\frac{2}{p} - 1 = 5$
$\frac{2}{p} = 6$
$p = \frac{1}{3}$
$p = \frac{1}{3}$