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

Answer: (C)

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