If a random variable $$X$$ follows the Binomial distribution $$B(33, p)$$ such that $$3P(X = 0) = P(X = 1)$$, then the value of $$\frac{P(X = 15)}{P(X = 18)} - \frac{P(X = 16)}{P(X = 17)}$$ is equal to

A random variable $$X$$ follows $$B(33, p)$$ with $$3P(X=0) = P(X=1)$$.

Find p.

$$P(X=0) = q^{33}$$, $$P(X=1) = 33pq^{32}$$.

$$3q^{33} = 33pq^{32} \implies 3q = 33p \implies q = 11p$$

Since $$p + q = 1$$: $$p + 11p = 1 \implies p = \frac{1}{12},\, q = \frac{11}{12}$$

Compute the ratio $$\frac{P(X=r)}{P(X=r+1)}$$.

$$\frac{P(X=r)}{P(X=r+1)} = \frac{\binom{33}{r}}{\binom{33}{r+1}} \cdot \frac{q}{p} = \frac{r+1}{33-r} \cdot 11$$

Compute $$\frac{P(X=15)}{P(X=18)}$$.

$$\frac{P(X=15)}{P(X=18)} = \frac{P(X=15)}{P(X=16)} \cdot \frac{P(X=16)}{P(X=17)} \cdot \frac{P(X=17)}{P(X=18)}$$

$$= \frac{16 \cdot 11}{18} \cdot \frac{17 \cdot 11}{17} \cdot \frac{18 \cdot 11}{16}$$

$$= \frac{88}{9} \cdot 11 \cdot \frac{99}{8}$$

$$= \frac{88 \times 11 \times 99}{9 \times 8} = \frac{88 \times 11 \times 11}{8} = \frac{10648}{8} = 1331$$

Compute $$\frac{P(X=16)}{P(X=17)}$$.

$$\frac{P(X=16)}{P(X=17)} = \frac{17 \times 11}{17} = 11$$

Final answer.

$$\frac{P(X=15)}{P(X=18)} - \frac{P(X=16)}{P(X=17)} = 1331 - 11 = 1320$$

Answer: Option A (1320)