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 :

If $n$ is number of trails, $p$ is probability of success and $q$ is probability of unsuccess then, <br/><br/> $$ \begin{aligned} & \text { Mean }=n p \text { and variance }=n p q \text {. } \\\\ & \text { Here }\\\\ & n p+n p q=24 \quad \dots(i)\\\\ & n p . n p q=128 \quad \dots(ii)\\\\ &\text { and } q=1-p \quad \dots(iii) \end{aligned} $$ <br/><br/> from eq. (i), (ii) and (iii) : $p=q=\frac{1}{2}$ and $n=32$. <br/><br/> $\therefore$ Required probability $=p(X=1)+p(X=2)$ <br/><br/> $$ \begin{aligned} & ={ }^{32} C_{1} \cdot\left(\frac{1}{2}\right)^{32}+{ }^{32} C_{2} \cdot\left(\frac{1}{2}\right)^{32} \\\\ & =\left(32+\frac{32 \times 31}{2}\right) \cdot \frac{1}{2^{32}} \\\\ & =\frac{33}{2^{28}} \end{aligned} $$