Question

If the difference between the mean and variance of a binomial distribution for $5$ trials is $\frac{5}{9},$ then the distribution is

• A. ${(\frac{2}{3}+\frac{1}{3})}^5$
• B. ${(\frac{5}{9}+\frac{4}{9})}^5$
• C. ${(\frac{2}{5}+\frac{3}{5})}^5$
• D. ${(\frac{1}{4}+\frac{3}{4})}^5$

Answer 1

The correct option is A ${(\frac{2}{3}+\frac{1}{3})}^5$

We know. for binomial distribution
mean $=np,$ variance $=npq$
here $n=$ number of trials, $p=$ probability of success and
$q=$ probability of failure
$\Rightarrow np-npq=\frac{5}{9}\Rightarrow np(1-q)=\frac{5}{9}\Rightarrow 5p^2=\frac{5}{9}\Rightarrow p=\frac{1}{3},q=\frac{2}{3}$
then, the binomial distribution is ${(\frac{2}{3}+\frac{1}{3})}^5$