The mean and variance of a binomial distribution are $\alpha$ and $\frac{\alpha}{3}$ respectively. If $P(X=1)=\frac{4}{243}$, then $P(X=4$ or $5)$ is equal to :

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Probability

The mean and variance of a binomial distribution are

α

and

\frac{α}{3}

respectively. If

P (X = 1) = \frac{4}{243}

, then

P (X = 4

5)

is equal to :

Answer: (C)

For a binomial distribution, mean

= n p = α

and variance

= n pq = \frac{α}{3}

. Dividing:

q = \frac{1}{3}

, so

p = \frac{2}{3}

. Given

P (X = 1) = (1 n) p q^{n - 1} = n \cdot \frac{2}{3} \cdot (\frac{1}{3})^{n - 1} = \frac{2 n}{3 ^{n}} = \frac{4}{243}

. Thus

\frac{n}{3 ^{n}} = \frac{2}{243} = \frac{2}{3 ^{5}}

, giving

n = 6

. Now

P (X = 4 or 5) = (4 6) (\frac{2}{3})^{4} (\frac{1}{3})^{2} + (5 6) (\frac{2}{3})^{5} (\frac{1}{3})^{1} = 15 \cdot \frac{16}{729} + 6 \cdot \frac{32}{729} = \frac{240 + 192}{729} = \frac{432}{729} = \frac{16}{27}

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Single correct option