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

We need to find $$54P(X \leq 2)$$ where $$X \sim \text{Binomial}(n, p)$$ with mean $$4$$ and variance $$\dfrac{4}{3}$$.

Mean: $$np = 4$$

Variance: $$npq = \dfrac{4}{3}$$, where $$q = 1 - p$$.

Dividing: $$q = \dfrac{4/3}{4} = \dfrac{1}{3}$$, so $$p = \dfrac{2}{3}$$.

From $$np = 4$$: $$n = \dfrac{4}{2/3} = 6$$.

$$P(X = k) = \binom{6}{k}\left(\dfrac{2}{3}\right)^k\left(\dfrac{1}{3}\right)^{6-k}$$

$$P(X = 0) = \binom{6}{0}\left(\dfrac{1}{3}\right)^6 = \dfrac{1}{729}$$

$$P(X = 1) = \binom{6}{1}\left(\dfrac{2}{3}\right)\left(\dfrac{1}{3}\right)^5 = 6 \cdot \dfrac{2}{729} = \dfrac{12}{729}$$

$$P(X = 2) = \binom{6}{2}\left(\dfrac{2}{3}\right)^2\left(\dfrac{1}{3}\right)^4 = 15 \cdot \dfrac{4}{729} = \dfrac{60}{729}$$

$$P(X \leq 2) = \dfrac{1 + 12 + 60}{729} = \dfrac{73}{729}$$

$$54 \cdot \dfrac{73}{729} = \dfrac{54 \cdot 73}{729} = \dfrac{73}{729/54} = \dfrac{73}{13.5} = \dfrac{146}{27}$$

The correct answer is Option B: $$\dfrac{146}{27}$$.