Question

Difference between mean and variance of a Binomial variate is ${}^{\prime }{1}^{\prime }$ and difference between their squares is ${}^{\prime }{11}^{\prime }.$ Find the probability of getting exactly three success

Answer 1

Mean=np & variance=npq
therefore, $np-npq=1$ (i)
$n^2{p}^2-n^2{p}^2{q}^2=11$ (ii)
Also, we know that $p+q=1$ (iii)
Divide equation (ii) by square of (i) and solve, we get, $q=\frac{5}{6},p\frac{1}{6}$ & $n=36$
Hence probability of ${}^{\prime }{3}^{\prime }$ success${=}^{36}{\text{C}}_3\times {\left(\frac{1}{6}\right)}^3\times {\left(\frac{5}{6}\right)}^{33}$