Question

Suppose X has a binomial distribution with n = 6 and $p=\frac{1}{2}.$ Show that X = 3 is the most likely outcome.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}n=6\mathrm{and}p=\frac{1}{2} \\ \therefore \mathit{}q=1-p=\frac{1}{2} \\ \mathrm{Hence},\mathrm{the}\mathrm{distribution}\mathrm{is}\mathrm{given}\mathrm{by} \\ P(X=r)={}^{6}C_r{\left(\frac{\mathit{1}}{\mathit{2}}\right)}^r{\left(\frac{\mathit{1}}{\mathit{2}}\right)}^{6-r},r=0,1,2,3,4,5,6 \\ ={}^{6}C_r{\left(\frac{1}{2}\right)}^6 \end{gathered}$$

$$\begin{gathered} P(X=r)=\frac{{}^{6}C_r}{2^6} \\ \mathrm{By}\mathrm{substituting}r\mathit{}=0,1,2,3,4,5\text{and}6,\mathrm{we}\mathrm{get}\mathrm{the}\mathrm{following}\mathrm{distribution}\mathrm{for}X. \\ X0123456 \\ P\left(X\right)\frac{1}{64}\frac{6}{64}\frac{15}{64}\frac{20}{64}\frac{15}{64}\frac{6}{64}\frac{1}{64} \end{gathered}$$

Comparing the probabilities, we get that X = 3 is the most likely outcome.