Question

Find the binomial distribution when the sum of its mean and variance for 5 trials is 4.8.

Answer 1

Number of trials in the binomial distribution = 5

If p is the probability for success, then

np + npq = 4.8
Or 5p+5p (1$-$p) = 4.8

$$\begin{gathered} \Rightarrow 5p+5p-5p^2=4.8 \\ \text{Or}{p}^2-2p+0.96=0 \\ \mathrm{By}\mathrm{factorising},\mathrm{we}\mathrm{get} \\ (p-0.8)(p-1.2)=0 \\ \mathrm{As}p\mathrm{cannot}\mathrm{exceed}1, \\ p\mathit{}=0.8\mathrm{or}\frac{4}{5} \\ \text{and}q=1-p=\frac{1}{5} \\ \therefore P(X=r)={}^{5}C_r{\left(\frac{4}{5}\right)}^r{\left(\frac{1}{5}\right)}^{5-r},r=0,1,2,....5 \end{gathered}$$