If the mean and variance of a binomial variate X are 2 and 1 respectively, then the probability that X takes a value greater than 1, is
Answer & explanation
Correct answer: option 4
Let n and p be the parameters of the binomial distribution. Then,
$np=2 $ and $ npq= 1$
$⇒ q=\frac{1}{2},p=\frac{1}{2}$ and n = 4 [Given]
$∴ P(X ≥ 1)= 1- P(X<1) = 1- P(X= 0)$
$⇒P(X ≥ 1)= 1 - {^nC}_0 \, p^0q^{n-0}= 1- {^4C}_0 × \left(\frac{1}{2}\right)^4=\frac{15}{16}$