Question

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

• A. $\frac{1}{2}$
  
• B. $\frac{29}{44}$
• C. $\frac{1}{4}$
• D. $\frac{3}{4}$

Answer 1

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

$Mean=np$
$=\frac{11}{4}$
$Variance=np(1-p)$
$=\frac{15}{16}$
Therefore
$\frac{11}{4}(1-p)=\frac{15}{16}$
$1-p=\frac{15}{44}$
$p=\frac{29}{44}$
$P$ being the probability of success.