A pair of dice is rolled 4 times. The probability of getting doublet exactly 2 times is 
 

Given n=4

The number of possible doublets

$=\left\{ (1, 1), (2, 2) (3, 3), (4, 4), (5, 5), (6, 6) \right\}$

P (getting a double on a roll of two dice)$=\frac{6}{36}$

$\Rightarrow \mathrm{P}=\frac{1}{6}$

P (not getting a doublet)$=\mathrm{q}=1-\mathrm{p}=\frac{5}{6}$

where X follows a binomial distribution 

Now 

          $\begin{array}{rcl}\mathrm{P}(X& =& 2)={}^4{\mathrm{C}}_2 {\mathrm{p}}^2 {\mathrm{q}}^{4-2}\\ & =& 6\times {\left(\frac{1}{6}\right)}^2 {\left(\frac{5}{6}\right)}^2\\ & =& 6\times \frac{1}{36}\times \frac{25}{36}\\ & =& \frac{25}{216}\end{array}$