Question

If the mean and variance of a binomial distribution are $4$ and $2$, respectively. Then, the probability of atleast $7$ successes is

• A. $\frac{3}{214}$
• B. $\frac{4}{173}$
• C. $\frac{9}{256}$
• D. $\frac{7}{231}$

Answer 1

The correct option is C $\frac{9}{256}$

Here, $mean=4$ and $variance=2$
$\Rightarrow np=4$ and $npq=2$
So, $\frac{npq}{np}=\frac{2}{4}\Rightarrow q=\frac{1}{2}$
Then, $p=1-q=1-\frac{1}{2}=\frac{1}{2}$
$Mean=np=4$
$\Rightarrow n\times \frac{1}{2}=4\Rightarrow n=8$
$\therefore P(X=r)={}^n{C}_r{p}^r{q}^{n-r}$
$={}^8{C}_r{\left(\frac{1}{2}\right)}^8[\because p=q=\frac{1}{2}]$
The required probability of atleast $7$ successes is
$P(X\ge 7)=P(X=7)+P(X=8)$
$={(}^8{C}_7+{}^8{C}_8){\left(\frac{1}{2}\right)}^8$
$=(\frac{8!}{7!1!}+\frac{8!}{8!0!}){\left(\frac{1}{2}\right)}^8$
$=(8+1){\left(\frac{1}{2}\right)}^8=\frac{9}{256}$

Hence, option C is correct.