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

(N/A) Solution: Let $X$ denote the number of doublets. The possible doublets are $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$.
Clearly,$X$ can take the values $0, 1, 2,$ or $3$.
Probability of getting a doublet $p = \frac{6}{36} = \frac{1}{6}$.
Probability of not getting a doublet $q = 1 - \frac{1}{6} = \frac{5}{6}$.
Using the binomial distribution formula $P(X=k) = \binom{n}{k} p^k q^{n-k}$ where $n=3$:
$P(X=0) = \binom{3}{0} (\frac{1}{6})^0 (\frac{5}{6})^3 = 1 \times 1 \times \frac{125}{216} = \frac{125}{216}$.
$P(X=1) = \binom{3}{1} (\frac{1}{6})^1 (\frac{5}{6})^2 = 3 \times \frac{1}{6} \times \frac{25}{36} = \frac{75}{216}$.
$P(X=2) = \binom{3}{2} (\frac{1}{6})^2 (\frac{5}{6})^1 = 3 \times \frac{1}{36} \times \frac{5}{6} = \frac{15}{216}$.
$P(X=3) = \binom{3}{3} (\frac{1}{6})^3 (\frac{5}{6})^0 = 1 \times \frac{1}{216} \times 1 = \frac{1}{216}$.
Thus,the required probability distribution is:

$X$	$P(X)$
$0$	$\frac{125}{216}$
$1$	$\frac{75}{216}$
$2$	$\frac{15}{216}$
$3$	$\frac{1}{216}$

Verification: $\sum P(X) = \frac{125+75+15+1}{216} = \frac{216}{216} = 1$.