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

JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Probability

If a random variable X follows the Binomial distribution

B (33, p)

such that

3 P (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

Answer: (A)

From

3 P (X = 0) = P (X = 1)

3 q^{33} = 33 p q^{32}

, giving

3 q = 33 p

, so

p = \frac{1}{12}

and

q = \frac{11}{12}

. Thus

\frac{q}{p} = 11

.

For binomial probabilities:

\frac{P ( X = 15 )}{P ( X = 18 )} = \frac{( 15 33 ) p ^{15} q ^{18}}{( 18 33 ) p ^{18} q ^{15}} = (\frac{q}{p})^{3} = 1 1^{3} = 1331

\frac{P ( X = 16 )}{P ( X = 17 )} = \frac{q}{p} = 11

Therefore

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

JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Single correct option