Question

Find the probability distribution of the number of doublets in 4 throws of a pair of dice.

Answer 1

Let X be the number of doublets in 4 throws of a pair of dice.

X follows a binomial distribution with n =4,
$$\begin{gathered} p=\text{N}o\mathrm{of}\mathrm{getting}(1,1)(2,2)...(6,6)=\frac{6}{36}=\frac{1}{6} \\ q=1-p=\frac{5}{6} \\ P(X=r)={}^{4}C_r{\left(\frac{1}{6}\right)}^r{\left(\frac{5}{6}\right)}^{4-r},\mathit{}r=0,1,2,3,4 \\ P(X=0)={}^{4}C_0{\left(\frac{1}{6}\right)}^0{\left(\frac{5}{6}\right)}^{4-0} \\ P(X=1)={}^{4}C_1{\left(\frac{1}{6}\right)}^1{\left(\frac{5}{6}\right)}^{4-1} \\ P(\mathrm{X}=2)={}^{4}C_2{\left(\frac{1}{6}\right)}^2{\left(\frac{5}{6}\right)}^{4-2} \\ P(X=3)={}^{4}C_3{\left(\frac{1}{6}\right)}^3{\left(\frac{5}{6}\right)}^{4-3} \\ P(X=4)={}^{4}C_4{\left(\frac{1}{6}\right)}^4{\left(\frac{5}{6}\right)}^{4-4} \\ \text{The d}\mathrm{istribution}\mathrm{is}\mathrm{as}\mathrm{follows}. \\ X01234 \\ P\left(X\right)\frac{625}{1296}\frac{500}{1296}\frac{150}{1296}\frac{20}{1296}\frac{1}{1296} \end{gathered}$$