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 $=np=2$ and
variance $npq=1$
$\Rightarrow 2q=1$
$\Rightarrow q=\frac{1}{2}$
$\Rightarrow p=1-q=\frac{1}{2}$
$\Rightarrow n=4$
$\therefore P(x\ge 1)=1-{}^n{C}_0{P}^0(q{)}^n$
$=1-{}^4{C}_0{\left(\frac{1}{2}\right)}^0{\left(\frac{1}{2}\right)}^4$
$=1-\frac{1}{16}=\frac{15}{6}$