The mean and variance of a binomial distribution are [] and [()/(3)] respectively.

Question

The mean and variance of a binomial distribution are $$\alpha$$ and $$\frac{\alpha}{3}$$ respectively. If $$\mathrm{P}(X=1)=\frac{4}{243}$$, then $$\mathrm{P}(X=4$$ or 5$)$ is equal to :

Accepted Answer

Given, mean $$ = np = \alpha $$.

and variance $$ = npq = {\alpha  \over 3}$$

$$ \Rightarrow q = {1 \over 3}$$ and $$p = {2 \over 3}$$

$$P(X = 1) = n.{p^1}.{q^{n - 1}} = {4 \over {243}}$$

$$ \Rightarrow n.{2 \over 3}.{\left( {{1 \over 3}} \right)^{n - 1}} = {4 \over {243}}$$

$$ \Rightarrow n = 6$$

$$P(X = 4\,\mathrm{or}\,5) = {}^6{C_4}\,.\,{\left( {{2 \over 3}} \right)^4}\,.\,{\left( {{1 \over 3}} \right)^2} + {}^6{C_5}\,.\,{\left( {{2 \over 5}} \right)^5}\,.\,{1 \over 3}$$

$$ = {{16} \over {27}}$$

Solutions · 1