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

Question

Let N denote the sum of the numbers obtained when two dice are rolled. If the probability that ${2^N} < N!$ is $${m \over n}$$, where m and n are coprime, then $4m-3n$ is equal to :

Accepted Answer

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

Such that ${2}^{N}<N$!

$4\le N\le 12N\in {4,5,6,\ldots 12}$

Now, $P(N=2)+P(N=3)=\frac{1}{36}+\frac{2}{36}=\frac{3}{36}=\frac{1}{12}$

So, required probability $=1-\frac{1}{12}=\frac{11}{12}=\frac{m}{n}$

$4m-3n=4	imes 11-3	imes 12=44-36=8$

Solutions · 1