Question

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 or equal to one is :

• A. $\frac{1}{16}$
• B. $\frac{9}{16}$
• C. $\frac{3}{4}$
• D. $\frac{15}{16}$

Answer 1

Answer: (D)

In binomial distribution mean $=n p=2$ and
variance $n p q=1$
$\Rightarrow 2 q=1$
$\Rightarrow q=\frac{1}{2}$
$\Rightarrow p=1-q=\frac{1}{2}$
$\Rightarrow n=4$
$\therefore P(x \geq 1)=1-{ }^{n} C_{0} P^{0}(q)^{n}$
$=1-{ }^{4} C_{0}\left(\frac{1}{2}\right)^{0}\left(\frac{1}{2}\right)^{4}$
$=1-\frac{1}{16}=\frac{15}{6}$