The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively, then P(X=1), is
Answer & explanation
Correct answer: option 2
Let n and p be the parameters of the binomial distribution.
We have,
Mean = 4 and Variance = 2
$⇒np = 4 \, and \, npq = 2 $
$⇒p =\frac{1}{2}= q \, and \, n = 8 $
∴ Required probability $=P(X=1)={^8C}_1\left(\frac{1}{2}\right)^1\left(\frac{1}{2}\right)^7=\frac{1}{32}$