Question

In a binomial distribution the sum and product of the mean and the variance are $\frac{25}{3}$ and $\frac{50}{3}$ respectively. Find the distribution.

Answer 1

$$\begin{gathered} \text{Given}: \\ \mathrm{Sum}\mathrm{of}\mathrm{the}\mathrm{mean}\mathrm{and}\mathrm{variance}=\frac{25}{3} \\ \Rightarrow np+npq=\frac{25}{3} \\ \Rightarrow np\left(1+q\right)=\frac{25}{3}...\left(1\right) \\ \mathrm{Product}\mathrm{of}\mathrm{the}\mathrm{mean}\mathrm{and}\mathrm{variance}=\frac{50}{3} \\ \Rightarrow np\left(npq\right)=\frac{50}{3}...\left(2\right) \\ \mathrm{Dividing}\mathrm{eq}\left(2\right)\mathrm{by}\mathrm{eq}\left(1\right),\mathrm{we}\mathrm{get} \\ \frac{\mathit{}np\mathit{\left(}npq\right)}{np(1+q)}=\frac{50}{3}\times \frac{3}{25} \\ \Rightarrow \frac{npq}{1+q}=2 \\ \Rightarrow npq=2(1+q) \\ \Rightarrow np(1-p)=2(2-\mathrm{p}) \\ \Rightarrow np=\frac{2(2-p)}{(1-p)} \\ \mathrm{Substituting}\mathrm{this}\mathrm{value}\mathrm{in}\mathit{}np+npq\mathit{}=\frac{25}{3},\mathrm{we}\text{get} \\ \frac{2(2-p)}{(1-p)}(2-p)=\frac{25}{3} \\ \Rightarrow 6(4-4p+p^2)=25-25p \\ \Rightarrow 6p^2+p-1=0 \\ \Rightarrow (3p-1)(2p+1)=0 \\ \Rightarrow p=\frac{1}{3}\mathrm{or}\frac{-1}{2}. \\ \mathrm{As}p\mathrm{cannot}\mathrm{be}\mathrm{negative},\text{the}\mathrm{only}\mathrm{answer}\mathrm{for}\mathit{}p\text{is}\frac{1}{3}. \end{gathered}$$
$$\begin{gathered} q=1-p=\frac{2}{3} \\ \Rightarrow np+npq\mathit{}=\frac{25}{3} \\ \Rightarrow n\left(\frac{\mathit{1}}{\mathit{3}}\right)\left(1+\frac{2}{3}\right)=\frac{25}{3} \\ \Rightarrow n=15 \\ \therefore P(X=r)={}^{15}C_{\mathit{r}}{\left(\frac{1}{3}\right)}^r{\left(\frac{2}{3}\right)}^{15-r},\mathit{}r=0,1,2......15 \end{gathered}$$