Question

If the sum and the product of mean and variance of a binomial distribution are $24$ and $128$ respectively, then the probability of one or two successes is:

Answer 1

Sol. If $n$ is number of trails, $p$ is probability of success and $q$ is probability of unsuccess then,

Mean $=np$ and variance $=npq$.

$\begin{array}{cc}\text{Here}np+npq=24& \dots \text{(i)}\\ np\cdot npq=128& \dots \text{(ii)}\\ \text{and}q=1-p& \dots \text{(iii)}\end{array}$

from eq. $\left(i\right),\left(ii\right)$ and $\left(iii\right):p=q=\frac{1}{2}$ and $n=32$.

$\therefore \text{Required probability}=p(X=1)+p(X=2)$

$={}^{32}{C}_1\cdot {\left(\frac{1}{2}\right)}^{32}+{}^{32}{C}_2\cdot {\left(\frac{1}{2}\right)}^{32}$

$=(32+\frac{32\times 31}{2})\cdot \frac{1}{2^{32}}$

$=\frac{33}{2^{28}}$