Question

If the mean and variance of a binomial variate X are 2 and 1 respectively, then the probability that X takes a value greater than 1 is
(a) 2/3
(b) 4/5
(c) 7/8
(d) 15/16

Answer 1

(d) 15/16

Mean =2 and variance =1

$$\begin{gathered} \Rightarrow np=2\mathrm{and}npq=1 \\ \Rightarrow q=\frac{1}{2} \\ \Rightarrow p=1-\frac{1}{2}=\frac{1}{2} \\ n=\frac{\mathrm{Mean}}{p} \\ \Rightarrow n=4 \\ \mathrm{Hence},\mathrm{the}\mathrm{distribution}\mathrm{is}\mathrm{given}\mathrm{by} \\ P\left(X=r\right)={}^{4}C_r{\left(\frac{1}{2}\right)}^r{\left(\frac{1}{2}\right)}^{4-r},r=0,1,2,3,4 \\ \therefore P(X\ge 1)=1-P(X=0) \\ =1-\frac{1}{2^4} \\ =\frac{15}{16} \end{gathered}$$