If the sum and the product of mean and variance of a binomial distribution are 24 and 128 …

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 :

Accepted Answer

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

$$
\begin{aligned}
& \text { Mean }=n p \text { and variance }=n p q \text {. } \\
& \text { Here }\\
&  n p+n p q=24 \quad \dots(i)\\
& n p . n p q=128 \quad \dots(ii)\\
&\text { and } q=1-p \quad \dots(iii)
\end{aligned}
$$

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

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

$$
\begin{aligned}
& ={ }^{32} C_{1} \cdot\left(\frac{1}{2}\right)^{32}+{ }^{32} C_{2} \cdot\left(\frac{1}{2}\right)^{32} \\
& =\left(32+\frac{32 \times 31}{2}\right) \cdot \frac{1}{2^{32}} \\
& =\frac{33}{2^{28}}
\end{aligned}
$$

Solutions · 1