Probability 5 Question 12
12. If the mean and the variance of a binomial variate $X$ are 2 and 1 respectively, then the probability that $X$ takes a value greater than one is equal to… .
(1991, 2M)
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Answer:
Correct Answer: 12. $\frac{11}{16}$
Solution:
- For Binomial distribution, mean $=n p$
and $\quad$ variance $=n p q$
$$ \begin{aligned} & \therefore \quad n p=2 \text { and } n p q=1 \quad \text { [given] } \\ & \Rightarrow \quad q=1 / 2 \text { and } p+q=1 \\ & \Rightarrow \quad p=1 / 2 \\ & \therefore \quad n=4, p=q=1 / 2 \\ & \text { Now, } \quad P(X>1)=1-{P(X=0)+P(X=1)} \\ & =1-{ }^{4} C _0 \frac{1}{2}^{0} \frac{1}{2}^{4}-{ }^{4} C _1 \frac{1}{2}^{1} \frac{1}{2}^{3} \\ & =1-\frac{1}{16}-\frac{4}{16}=\frac{11}{16} \end{aligned} $$
