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 $=2$
$\Rightarrow n p=4$ and $n p q=2$
So, $\frac{n p q}{n p}=\frac{2}{4} $
$\Rightarrow q=\frac{1}{2}$
Then, $p=1-q=1-\frac{1}{2}=\frac{1}{2}$
Mean $=n p=4$
$\Rightarrow n \times \frac{1}{2}=4$
$ \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
$P(X \geq 7)=P(X=7)+P(X=8)$
$=\left({ }^{8} C_{7}+{ }^{8} C_{8}\right)\left(\frac{1}{2}\right)^{8}$
$=\left(\frac{8 !}{7 ! ! !}+\frac{8 !}{8 ! 0 !}\right)\left(\frac{1}{2}\right)^{8}$
$=(8+1)\left(\frac{1}{2}\right)^{8}=\frac{9}{256}$