Question

Find the probability distribution of the number of doublets in three throws of a pair of dice and find its mean.

Answer 1

Total number of outcomes when two dice are thrown = 6 $\times$ 6, i.e. 36

Let X be the number of doublets in three throws of a pair of dice.
Then, X follows a binomial distribution with n = 3
$$\begin{gathered} p=P(\mathrm{Getting}\mathrm{a}\mathrm{doublet}\mathrm{in}\mathrm{three}\mathrm{throws})=\frac{6}{36}=\frac{1}{6}\mathrm{and}q=\frac{5}{6} \\ \therefore P(X=r)={}^{3}C_{\mathit{r}}{\left(\frac{1}{6}\right)}^{\mathrm{r}}{\left(\frac{\mathit{5}}{\mathit{6}}\right)}^{3-\mathit{\text{r}}},\mathit{}r=0,1,2,3 \\ \mathrm{Mean}\left(\mathit{n}\mathit{p}\right)=3\left(\frac{1}{6}\right)=\frac{1}{2} \end{gathered}$$
$$\begin{gathered} \text{The d}\mathrm{istribution}\mathrm{is}\mathrm{as}\mathrm{follows}: \\ X0123 \\ P\left(X\right){\left(\frac{5}{6}\right)}^{3}3{\left(\frac{1}{6}\right)}^1{\left(\frac{5}{6}\right)}^{3-1}3{\left(\frac{1}{6}\right)}^2{\left(\frac{5}{6}\right)}^{3-2}{\left(\frac{1}{6}\right)}^3 \\ \frac{125}{216}\frac{75}{216}\frac{15}{216}\frac{1}{216} \end{gathered}$$