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Q.
The mean and variance of a binomial distribution are $\alpha$ and $\frac{\alpha}{3}$ respectively. If $P(X=1)=\frac{4}{243}$, then $P(X=4$ or 5$)$ is equal to :
$ n p =\alpha ....$ (1)
$ npq =\alpha / 3.....$(2)
From (1) & (2)
$q =1 / 3 \& p =2 / 3 $
$ { }^{ n } C _1 q ^{ n -1} p ^1=\frac{4}{243}$
$ \frac{ n }{3^{ n }}=\frac{2}{243}$
$ n =6$
$P (4 \text { or } 5)$
$={ }^6 C _4\left(\frac{2}{3}\right)^4\left(\frac{1}{3}\right)^2+{ }^6 C _5\left(\frac{2}{3}\right)^5 \cdot\left(\frac{1}{3}\right)^0 $
$ =\frac{16}{27}$