Question

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

Answer 1

Given that n = 6
The sum of mean and variance of a binomial distribution for 6 trials is $\frac{10}{3}$.
$$\begin{gathered} \Rightarrow 6p+6pq=\frac{10}{3} \\ \Rightarrow 18p+18p(1-p)=10 \\ \Rightarrow 18p^2-36p+10=0 \\ \Rightarrow (3p-1)(6p-10)=0 \\ \Rightarrow p=\frac{1}{3}\mathrm{or}\frac{5}{3} \\ p=\frac{5}{3}(\mathrm{Neglected}\mathrm{as}\mathrm{it}\mathrm{is}\mathrm{greater}\mathrm{than}1) \\ \therefore p=\frac{1}{3} \\ \Rightarrow q=1-p\mathit{}=\frac{2}{3} \\ \mathrm{Hence},\mathrm{the}\mathrm{distribution}\mathrm{is}\mathrm{given}\mathrm{by} \\ P(X=r)={}^{6}C_r{\left(\frac{1}{3}\right)}^r{\left(\frac{2}{3}\right)}^{6-r},r=0,1,2.....6 \end{gathered}$$