JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Probability

Let X be a binomially distributed random variable with mean 4 and variance

\frac{4}{3}

. Then

54 P (X \leq 2)

is equal to

Answer: (B)

Solution

For a binomial distribution, mean

n p = 4

and variance

n pq = \frac{4}{3}

. Thus

q = \frac{1}{3}

and

p = \frac{2}{3}

, giving

n = 6

. Now

P (X \leq 2) = P (X = 0) + P (X = 1) + P (X = 2) = (0 6) (\frac{2}{3})^{0} (\frac{1}{3})^{6} + (1 6) (\frac{2}{3})^{1} (\frac{1}{3})^{5} + (2 6) (\frac{2}{3})^{2} (\frac{1}{3})^{4}

. Computing:

= \frac{1}{729} + \frac{12}{729} + \frac{60}{729} = \frac{73}{729}

. Thus

54 P (X \leq 2) = 54 \cdot \frac{73}{729} = \frac{146}{27}

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Single correct option