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.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\frac{15}{16}$
mean = np = 2 ...(1)
variance = npq = 1 ...(2)
eq. (2)/ eq. (1)
$⇒q=\frac{1}{2}⇒p=\frac{1}{2}$
$⇒n=4$
$P(X≥1)=1-P(X=0)$
$1-{^4C}_0×(\frac{1}{2})^0(\frac{1}{2})^4=\frac{15}{16}$