Question

If mean and variance of a binomial variate X are 2 and 1 respectively, then the probability that X takes a value atleast one is

• A. $\frac{1}{16}$
• B. $\frac{3}{16}$
• C. $\frac{5}{16}$
• D. $\frac{15}{16}$

Answer 1

Answer: (D)

$\frac{15}{16}$

Answer 2

Given mean, $np=2$ and variance, $npq=1$
$\therefore$ $2(q)\,=2\,\,\Rightarrow q=\frac{1}{2}$
$\Rightarrow$ $p=\frac{1}{2}$ and $n=4$
$\therefore$ $P(X\ge 1)=1-P(X=0)$
$=1{{-}^{4}}{{C}_{0}}{{\left( \frac{1}{2} \right)}^{0}}{{\left( \frac{1}{2} \right)}^{4}}$
$=1-1\times \frac{1}{16}=\frac{15}{16}$