It is given that mean and variance of a binomial variable X are 2 and 1 respectively. Find the probability that X takes value greater than or equal to 1.

$\frac{15}{16}$

The correct answer is Option (2) → $\frac{15}{16}$

mean = np = 2 ...(1)

variance = npq = 1  ...(2)

eq. (2)/ eq. (1)

$⇒q=\frac{1}{2}⇒p=\frac{1}{2}$

$⇒n=4$

$P(X≥1)=1-P(X=0)$

$1-{^4C}_0×(\frac{1}{2})^0(\frac{1}{2})^4=\frac{15}{16}$