Question

If the sum of the mean and variance of a binomial distribution for $6$ trials be $10/3$, find the distribution.

Answer 1

Given for a binomial distribution, the sum of its mean and variance for 6 trials is $\frac{10}{3}$.

Let the parameters be $n,p$.

So, $np+npq=\frac{10}{3}$ where $q=1-p$

$\Rightarrow 6p(1+q)=\frac{10}{3}$

$\Rightarrow p(1+1-p)=\frac{5}{9}$

$\Rightarrow p(2-p)=\frac{5}{9}$

$\Rightarrow 9p^2-18p+5=0$

$\Rightarrow 9p^2-15p-3p+5=0$

$\Rightarrow 3p(3p-5)-1(3p-5)=0$

$\Rightarrow (3p-1)(3p-5)=0$

$\Rightarrow p=\frac{1}{3}$ or $p=\frac{5}{3}\to xp\le 1$

The binomial distribution is ,

$\therefore p(x-r)={}^6{C}_r{\left(\frac{1}{3}\right)}^r{\left(\frac{2}{3}\right)}^{6-r}$

where $r=0,1,2,.....6$.