Question

If $X$ follows Binomial distribution with mean $3$ and variance $2$, then $P(X\ge 8)$ is equal to :

• A. $\frac{17}{3^9}$
• B. $\frac{18}{3^9}$
• C. $\frac{19}{3^9}$
• D. $\frac{20}{3^9}$

Answer 1

Answer: (C)

As given, Mean $=np=3$ and Variance $=npq=2$
$\therefore \frac{npq}{np}=\frac{2}{3}$
$\Rightarrow q=\frac{2}{3}$ and $p=(1-q)=1-\frac{2}{3}-\frac{1}{3}$
$n\cdot \frac{1}{3}=3$
$\Rightarrow n=9$
$\therefore P(X\ge 8)=P(X=8)+P(X=9)$
$={}^9{C}_8{\left(\frac{1}{3}\right)}^8\times \frac{2}{3}+{}^9{C}_9{\left(\frac{1}{3}\right)}^9$
$=9{\left(\frac{1}{3}\right)}^8\frac{2}{3}+1{\left(\frac{1}{3}\right)}^9=\frac{18+1}{3^9}=\frac{19}{3^9}$