Question

The mean and variance of a binomial distribution are $\frac{4}{3}$ and $\frac{8}{9}$ respectively. Find P (X ≥ 1).

Answer 1

$$\begin{gathered} \mathrm{Mean}\left(np\right)=\frac{4}{3}\mathrm{and}\mathrm{Variance}\left(npq\right)=\frac{8}{9} \\ \therefore q=\frac{2}{3} \\ \text{and}p=1-\frac{2}{3}=\frac{1}{3} \\ \mathrm{Therefore},\mathit{}n\mathit{}=\frac{\mathrm{Mean}}{p}=4 \end{gathered}$$
$$\begin{gathered} \mathrm{Hence},\mathrm{the}\mathrm{distribution}\mathrm{is}\mathrm{given}\mathrm{by} \\ P(X=r)={}^{4}C_{\mathit{r}}{\left(\frac{\mathit{1}}{\mathit{3}}\right)}^r{\left(\frac{\mathit{2}}{\mathit{3}}\right)}^{4-r},r=0,1,2,3,4 \\ P(X\ge 1)=1-\left[P\right(X=0\left)\right] \\ =1-\left[{\left(\frac{2}{3}\right)}^4\right] \\ =\frac{81-16}{81} \\ =\frac{65}{81} \end{gathered}$$