Let N denote the sum of the numbers obtained when two dice are rolled. If the probability that 2 N < N !…

Given that,

$2^N<N!$

$N$ is the sum of numbers of two dice.

$\Rightarrow 2\le N\le 12$

Let us check the given condition $2^N<N!$

$N=1$ (not possible) $\to 0$

$N=2$ (not possible) $\to 1$

$N=3$ (not possible) $\to 2$

$N=4$ (possible)

We need to find the probability for $N\ge 4$.

$\Rightarrow P\left(N\ge 4\right)=1-P\left(N=2\right)-P\left(N=3\right)$

$=1-\frac{1}{36}-\frac{2}{36}$

$=\frac{11}{12}=\frac{m}{n}$

Now let us find $4m-3n$.

$=4\times 11-3\times 12=8$.

Hence, the required answer is $8$.