Question

If the mean and variance of a binomial variable $X$ are $2$ and $1$ respectively, then $P\left(X\ge 1\right)=$

• A. $\frac{2}{3}$
• B. $\frac{15}{16}$
• C. $\frac{7}{8}$
• D. $\frac{4}{5}$

Answer 1

The correct option is B $\frac{15}{16}$

For Binomial distribution, Mean$\left(\mu \right)=np$ and Variance$\left({\sigma }^2\right)=np(1-p)$

According to question, Mean = 2 and Variance = 1

Thus, $np=\mathrm{2.....}\left(1\right)np(1-p)=\mathrm{1.......}\left(2\right)$

Putting the value of np from first equation in the second equation.

$2\times (1-p)=1\Rightarrow 1-p=\frac{1}{2}\Rightarrow p=\frac{1}{2}$

Now, putting the value of p in equation(1), $n\times \frac{1}{2}=2\Rightarrow n=4$

$\because \Sigma P\left(X_i\right)=1$

$P(X\ge 1)=1-P(X<1)=1-P(X=0)=1-{}^4{C}_0{p}^0(1-p{)}^{4-0}$

$=1-\frac{4!}{0!(4-0)!}\times {\left(\frac{1}{2}\right)}^0\times {\left(\frac{1}{2}\right)}^4=1-\frac{1}{16}=\frac{15}{16}$