Question

If the mean and variance of a binomial variate X are 2 and 1 respectively, find P (X > 1).

Answer 1

$$\begin{gathered} \mathrm{Mean}=2,\mathrm{Variance}=1 \\ \therefore q=\frac{\mathrm{Variance}}{\mathrm{Mean}}=\frac{1}{2} \\ \text{and}p\mathit{}=1-\frac{1}{2}=\frac{1}{2} \\ n=\frac{\mathrm{Mean}}{p}=\frac{2}{{\displaystyle \frac{1}{2}}}=4 \\ \mathrm{The}\mathrm{binomial}\mathrm{distribution}\mathrm{is}\mathrm{given}\mathrm{by} \\ P(X=r)={}^{4}C_r{\left(\frac{1}{2}\right)}^r{\left(\frac{1}{2}\right)}^{4-r} \\ \therefore P(X=0)={}^{4}C_0{\left(\frac{1}{2}\right)}^0{\left(\frac{1}{2}\right)}^{4-0},r=0,1,2,3,4 \\ ={\left(\frac{1}{2}\right)}^4 \\ P(X>1)=1-P(X=0) \\ =1-{\left(\frac{1}{2}\right)}^4 \\ =\frac{15}{16} \end{gathered}$$