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

Answer 1

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

$np-np(1-p)=\frac{5}{9}$
$np\left(p\right)=\frac{5}{9}$
$n\left(p^2\right)=5(\frac{1}{3}{)}^2$
By comparing we get
$p=\frac{1}{3}$ and $n=5$
Where, p is the probability of success and n is the no. of trials.