Let $X$ be a binomially distributed random variable with mean 4 and variance $\frac{4}{3}$. Then, $54 \,P(X \leq 2)$ is equal to :

<p>Mean $= 4 = \mu = np$</p> <p>Variance $= {\sigma ^2} = np(1 - P) = {4 \over 3}$</p> <p>$4(1 - P) = {4 \over 3}$</p> <p>$P = {2 \over 3}$</p> <p>$n \times {2 \over 3} = 4$</p> <p>$n = 6$</p> <p>$P(X = k) = {}^n{C_k}\,{P^k}{(1 - P)^{n - k}}$</p> <p>$P(X \le 2) = P(X = 0) + P(X = 1) + P(X = 2)$</p> <p>$$ = {}^6{C_0}{P^0}{(1 - P)^6} + {}^6{C_1}{P^1}{(1 - P)^5} + {}^6{C_2}{P^2}{(1 - P)^4}$$</p> <p>$$ = {}^6{C_0}{\left( {{1 \over 3}} \right)^6} + {}^6{C_1}\left( {{2 \over 3}} \right){\left( {{1 \over 3}} \right)^5} + {}^6{C_2}{\left( {{2 \over 3}} \right)^2}{\left( {{1 \over 3}} \right)^4}$$</p> <p>$= {\left( {{1 \over 3}} \right)^6}[1 + 12 + 60] = {{73} \over {{3^6}}}$</p> <p>$54P\,(X \le 2) = {{73} \over {{3^6}}} \times 54 = {{146} \over {27}}$</p>