Let N denote the sum of numbers obtained when two dice are rolled.

Question

Let $N$ denote the sum of numbers obtained when two dice are rolled. If the probability that $2^N<N!$ is $\dfrac{m}{n}$ (coprime), find $4m-3n$.

Accepted Answer

$2^N<N!$ holds for $N\ge4$.

$P(N=2)=\dfrac{1}{36}$, $P(N=3)=\dfrac{2}{36}$, so $P(N<4)=\dfrac{3}{36}$.

$$P(N\ge4) = 1-\frac{3}{36}=\frac{33}{36}=\frac{11}{12} \implies m=11,n=12$$

$4m-3n = 44-36=8$

Solutions · 1