Question

An unbiased coin is tossed 4 times. Find the mean and variance of the number of heads obtained. [CBSE 2015]

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ p=\mathrm{probability}\mathrm{of}\mathrm{getting}\mathrm{a}\mathrm{head}\mathrm{in}\mathrm{a}\mathrm{toss}=\frac{1}{2}, \\ q=\mathrm{probability}\mathrm{of}\mathrm{getting}\mathrm{a}\mathrm{tail}\mathrm{in}\mathrm{a}\mathrm{toss}=\frac{1}{2} \\ \mathrm{Let}X\mathrm{denote}\mathrm{a}\mathrm{success}\mathrm{of}\mathrm{getting}\mathrm{a}\mathrm{head}\mathrm{in}\mathrm{a}\mathrm{toss}.\mathrm{Then}, \\ X\mathrm{follows}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{parameters}n=4\mathrm{and}p=\frac{1}{2} \\ \therefore \mathrm{Mean},\mathrm{E}\left(X\right)=np=4\times \frac{1}{2}=2 \\ \mathrm{Also},\mathrm{variance},\mathrm{Var}\left(X\right)=npq=4\times \frac{1}{2}\times \frac{1}{2}=1 \end{gathered}$$