Question

If the mean and variance of a binomial distribution are 4 and 3, respectively, find the probability of no success.

Answer 1

Mean (np)= 4

Variance (npq )= 3

.$$\begin{gathered} \Rightarrow q=\frac{3}{4} \\ \mathrm{Hence},p=1-\frac{3}{4}=\frac{1}{4} \\ \text{and}n\mathit{}=\frac{\mathrm{Mean}}{p}=4\times 4=16 \\ \mathrm{Therefore},\mathrm{the}\mathrm{binomial}\mathrm{distribution}\mathrm{is}\mathrm{given}\mathrm{by} \\ \mathrm{P}(\mathrm{X}=r)={}^{16}C_r{\left(\frac{1}{4}\right)}^r{\left(\frac{3}{4}\right)}^{16-r},r=0,1,2....r \\ \mathrm{Probability}\mathrm{of}\mathrm{no}\mathrm{success}={}^{16}C_0{\left(\frac{1}{4}\right)}^0{\left(\frac{3}{4}\right)}^{16-0}={\left(\frac{3}{4}\right)}^{16} \end{gathered}$$