Let $$X$$ be a random variable having binomial distribution $$B(7, p)$$. If $$P(X = 3) = 5P(X = 4)$$, then the sum of the mean and the variance of $$X$$ is

$$X \sim B(7, p)$$ and $$P(X = 3) = 5 \cdot P(X = 4)$$. Find mean + variance.

First, we write the binomial probabilities: $$P(X = 3) = \binom{7}{3} p^3 (1-p)^4 = 35 p^3 (1-p)^4$$ and $$P(X = 4) = \binom{7}{4} p^4 (1-p)^3 = 35 p^4 (1-p)^3$$.

Since $$P(X = 3) = 5 \times P(X = 4)$$, we have $$35 p^3 (1-p)^4 = 5 \times 35 p^4 (1-p)^3$$. Dividing both sides by $$35 p^3 (1-p)^3$$ gives $$1 - p = 5p$$, which leads to $$1 = 6p$$ and hence $$p = \frac{1}{6}$$.

Now we calculate the mean and variance. Using $$np$$ for the mean yields $$7 \times \frac{1}{6} = \frac{7}{6}$$, and using $$np(1-p)$$ for the variance gives $$7 \times \frac{1}{6} \times \frac{5}{6} = \frac{35}{36}$$.

Finally, the sum of the mean and variance is $$\frac{7}{6} + \frac{35}{36} = \frac{42}{36} + \frac{35}{36} = \frac{77}{36}$$.

The answer is Option B: $$\frac{77}{36}$$.