Let X be a binomially distributed random variable with mean 4 and variance [(4)/(3)].

Question

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 :

Accepted Answer

Mean $$ = 4 = \mu  = np$$

Variance $$ = {\sigma ^2} = np(1 - P) = {4 \over 3}$$

$$4(1 - P) = {4 \over 3}$$

$$P = {2 \over 3}$$

$$n \times {2 \over 3} = 4$$

$n = 6$

$$P(X = k) = {}^n{C_k}\,{P^k}{(1 - P)^{n - k}}$$

$$P(X \le 2) = P(X = 0) + P(X = 1) + P(X = 2)$$

$$ = {}^6{C_0}{P^0}{(1 - P)^6} + {}^6{C_1}{P^1}{(1 - P)^5} + {}^6{C_2}{P^2}{(1 - P)^4}$$

$$ = {}^6{C_0}{\left( {{1 \over 3}} \right)^6} + {}^6{C_1}\left( {{2 \over 3}} \right){\left( {{1 \over 3}} \right)^5} + {}^6{C_2}{\left( {{2 \over 3}} \right)^2}{\left( {{1 \over 3}} \right)^4}$$

$$ = {\left( {{1 \over 3}} \right)^6}[1 + 12 + 60] = {{73} \over {{3^6}}}$$

$$54P\,(X \le 2) = {{73} \over {{3^6}}} \times 54 = {{146} \over {27}}$$

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