Question

The mean variance of a binomial variable $X$ are $2$ and $1$ respectively. The probability that $X$ takes values greater than $1$, is

• A. $\frac{5}{16}$
• B. $\frac{9}{16}$
• C. $\frac{11}{16}$
• D. None of these

Answer 1

Answer: (C)

$\frac{11}{16}$

Given: mean $np=2$ ...(1)

and variance $npq=144$ ...(2)

Dividing (2) by (1), then $q=\frac{1}{2}$

$\therefore p=1-q=\frac{1}{2}$

From (1), $n\times \frac{1}{2}=2\Rightarrow n=4$

The binomial distribution is ${(\frac{1}{2}+\frac{1}{2})}^4$

Now, $P(X>1)=P(X=2)+P(X=3)+P(X=4)$

${=}^4{C}_2{\left(\frac{1}{2}\right)}^2\times {\left(\frac{1}{2}\right)}^2{+}^4{C}_3{\left(\frac{1}{2}\right)}^1\times {\left(\frac{1}{2}\right)}^3{+}^4{C}_4{\left(\frac{1}{2}\right)}^4$

$=\frac{6+4+1}{16}=\frac{11}{16}$