Total number of outcomes = 36
The possible doublets are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), and (6, 6).
Let p be the probability of success, therefore,
p = $\frac{6}{36}$ = $\frac{1}{6}$
So, q = 1 - p = 1 - $\frac{1}{6}$ = $\frac{5}{6}$
Since the dice is thrown 4 times, n=4
Let X denote the number of times of getting doublets in the experiment of throwing two dice simultaneously four times.
Therefore X can take the values 0, 1, 2, 3, or 4.
P (X = 0) = $\stackrel{4}{}{\mathit{C}}_0{\mathit{p}}^0{\mathit{q}}^4$ = ${\left(\frac{5}{6}\right)}^4$ = $\frac{625}{1296}$
P(X = 1) = $\stackrel{4}{}{\mathit{C}}_1{\mathit{p}}^1{\mathit{q}}^3$ = 4 $\left(\frac{1}{6}\right){\left(\frac{5}{6}\right)}^3$ = $\frac{500}{1296}$
P(X = 2) = $\stackrel{4}{}{\mathit{C}}_2{\mathit{p}}^2{\mathit{q}}^2$ = 6 ${\left(\frac{1}{6}\right)}^2{\left(\frac{5}{6}\right)}^2$ = $\frac{150}{1296}$
P (X = 3) = $\stackrel{4}{}{\mathit{C}}_3{\mathit{p}}^4{\mathit{q}}^0$ = ${\left(\frac{1}{6}\right)}^4$ = $\frac{1}{1296}$
Thus, the probability distribution is: