JEE Main · 2004 · 24 Apr 2004 · Offline · Probability

The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is

Answer: (D)

Solution

For a binomial distribution the mean is

n p = 4

and the variance is

n p (1 - p) = 2

. Dividing,

1 - p = \frac{1}{2}

, so

p = \frac{1}{2}

and then

n = \frac{4}{p} = 8

. The probability of exactly two successes is

(2 8) (\frac{1}{2})^{2} (\frac{1}{2})^{6} = 28 \cdot \frac{1}{256} = \frac{28}{256}

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option