Question

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

Answer 1

Let $x=$ number of doublets

Expected value of $x$ wi be $0,1,2,3$

Set of all doublets available on a pair of dice in a throw $=(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)$

Number of all possible ways on throwing a pair of dice $=6\times 6=36$

Probability of getting a doublets on a pair of dice in a throw$=\frac{6}{36}=\frac{1}{6}$

Probability of getting doublets on a pair of dice throw$=1-\frac{1}{6}=\frac{5}{6}$

Now $P(X=0)=P$(No doublets)

$=\frac{5}{6}\times \frac{5}{6}\times v=\frac{125}{216}$

$P(X=1)=P$(one doublets, $2$ not)

$={{}^{3}C}_2\left(\frac{1}{6}\right){\left(\frac{5}{6}\right)}^2$

$=3\times \frac{1}{6}\times \frac{25}{36}=\frac{75}{216}$

$P(X=2)=P$ (2 doublets, 1 not)

$={{}^{3}C}_2{\left(\frac{1}{6}\right)}^2\times \frac{5}{6}=3\times \frac{1}{6}\times \frac{1}{6}\times \frac{5}{6}=\frac{15}{216}$

$P(X=3)=P$ (all doublets)

$={\left(\frac{1}{6}\right)}^3=\frac{1}{216}$

Required probability distribution is following

$X$ 0 1 2 3

$P\left(X\right)$ $\frac{125}{216}$ $\frac{75}{216}$ $\frac{15}{216}$$ $\frac{1}{216}$