Given that the two numbers appearing on throwing two dice are different. Find the probability of the event the sum of numbers on the dice is 4.

Solution: When two dice are thrown, there are $36$ possibilities.
However, we are given, every time both dice contain different numbers means doublet is not a possibility.
As, number of doublets are $6$, so, sample space, $S=36-6=30$
Number of possibilities of the event when sum of dice is 4, $\left(1,3\right),\left(3,1\right)$.
Here, we won${}^{\prime }$t take (2,2) as both dice contain different numbers.So,
$E=2$
So, $P\left(E\right)=\frac{2}{30}=\frac{1}{15}$