The mean and the variance of a binomial distribution are 4 and 2 respectively. Then, the probability of 2 successes is
Answer & explanation
Correct answer: option 1
Let n and p bet the parameters for the distribution.
We have,
Mean = 4 and Variance = 2
$⇒ np = 4 $ and $ npq = 2 ⇒p = 1 = \frac{1}{2}$ and $ n = 8 $
Let X denote the number of successes. Then,
$P(X=r)={^8C}_r \left(\frac{1}{2}\right)^r \left(\frac{1}{2}\right)^{8-1}={^8C}_r \left(\frac{1}{2}\right)^8$
∴ Required probability $= P(X=2)= {^8C}_r \left(\frac{1}{2}\right)^8= \frac{28}{256}$