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

Let $X$ denote the number of doublets. Possible doublets are $(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)$
Clearly, $X$ can take the value $0,1,2,$ or $3$.

Probability of getting a doublet $=\frac{6}{36}=\frac{1}{6}$
Probability of not getting a doublet $=1-\frac{1}{6}=\frac{5}{6}$
Now,
$P(X=0)=P\left(\text{no doublet}\right)=\frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}=\frac{125}{216}$

$P(X=1)=P\left(\text{one doublet and two non-doublets}\right)=\frac{1}{6}\times \frac{5}{6}\times \frac{5}{6}+\frac{5}{6}\times \frac{1}{6}\times \frac{5}{6}+\frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}=\frac{75}{216}$

$P(X=2)=P\left(\text{two doublets and one non-doublet}\right)=\frac{1}{6}\times \frac{1}{6}\times \frac{5}{6}+\frac{1}{6}\times \frac{5}{6}\times \frac{1}{6}+\frac{5}{6}\times \frac{1}{6}\times \frac{1}{6}=\frac{15}{216}$

$P(X=3)=P\left(\text{three doublets}\right)=\frac{1}{6}\times \frac{1}{6}\times \frac{1}{6}=\frac{1}{216}$

Thus, the required probability distribution is

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