Question

In a binomial distribution mean and variance are $\frac{11}{4}$ and $\frac{15}{16}$ respectively, then the probability of success is

• A. $\frac{1}{11}$
• B. $\frac{15}{44}$
• C. $\frac{29}{44}$
• D. $\frac{11}{60}$

Answer 1

The correct option is C $\frac{29}{44}$

Given : mean$\left(\mu \right)=np=\frac{11}{4},p=$ probability of success
variance $={\sigma }^2=npq=\frac{15}{16}$
from above, we get $q=\frac{15}{44}$
So, $p=1-q=\frac{29}{44}$