Question

The mean and variance of a binomial variate with parameters n and p are 16 and 8, respectively. Find P (X = 0), P (X = 1) and P (X ≥ 2).

Answer 1

Given: mean =16 and variance = 8
Let n and p be the parameters of the distribution.

That is, np = 16 and npq = 8

$$\begin{gathered} q=\frac{npq}{np}=\frac{1}{2} \\ \text{and}p\mathit{}=1-q=\frac{1}{2} \\ np=16 \\ \Rightarrow n=32 \\ \therefore P(X=r)={}^{32}C_r{\left(\frac{\mathit{1}}{\mathit{2}}\right)}^r{\left(\frac{1}{2}\right)}^{32-r},r=0,1,2....32, \\ \Rightarrow P(X=0)={\left(\frac{1}{2}\right)}^{32} \\ P(X=1)=32{\left(\frac{1}{2}\right)}^{32}={\left(\frac{1}{2}\right)}^2 \\ P(X\ge 2)=1-\mathrm{P}(X=0)-P(X=1) \\ =1-{\left(\frac{1}{2}\right)}^{32}-{\left(\frac{1}{2}\right)}^{27} \\ =1-\left(\frac{1+32}{2^{32}}\right) \\ =1-\frac{33}{2^{32}} \end{gathered}$$