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Q.
If $X$ follows Binomial distribution with mean $3$ and variance $2$, then $P ( X \geq 8)$ is equal to :
Probability - Part 2
Solution:
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 \geq 8)= P ( X =8)+ P ( X =9)$
$={ }^{9} C _{8}\left(\frac{1}{3}\right)^{8} \times \frac{2}{3}+{ }^{9} C _{9}\left(\frac{1}{3}\right)^{9}$
$=9\left(\frac{1}{3}\right)^{8} \frac{2}{3}+1\left(\frac{1}{3}\right)^{9}=\frac{18+1}{3^{9}}=\frac{19}{3^{9}}$