Question

The sum of the mean and variance of a binomial distribution is $15$ and the sum of their squares is $117$. Probability of atmost one success in case of these trials will be:

• A. $\frac{31}{3}{\left(\frac{1}{3}\right)}^{26}$
• B. $5{\left(\frac{2}{3}\right)}^{26}$
• C. $\frac{10}{3}{\left(\frac{2}{3}\right)}^{26}$
• D. $\frac{29}{3}{\left(\frac{2}{3}\right)}^{26}$

Answer 1

The correct option is D $\frac{29}{3}{\left(\frac{2}{3}\right)}^{26}$

Let $n$ and $p$ be parameters of the binomial distribution . Then,
$np+npq=15\cdots \left(i\right)$ and
$(np{)}^2+(npq{)}^2=117\cdots \left(ii\right)$
Equation $\left(ii\right)/(i{)}^2$
$\therefore \frac{n^2{p}^2(1+q^2)}{(np+npq{)}^2}=\frac{117}{{15}^2}$
$\frac{1+q^2}{(1+q{)}^2}=\frac{117}{225}$
$\Rightarrow 6q^2-13q+6=0$
$\Rightarrow q=\frac{2}{3},p=\frac{1}{3}$
$\therefore np+npq=15\Rightarrow n\times \frac{1}{3}+n\times \frac{2}{9}=15\Rightarrow n=27$
$P\left(\text{atmost one success}\right)=P\left(0\right)+P\left(1\right){=}^n{C}_0{p}^0{q}^n{+}^n{C}_1{p}^1{q}^{n-1}=q^{n-1}(q+np)$
$\Rightarrow {\left(\frac{2}{3}\right)}^{26}(\frac{2}{3}+\frac{27}{3})=\frac{29}{3}{\left(\frac{2}{3}\right)}^{26}$