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 :
Solution
$N$ denote the sum of the numbers obtained when two dice are rolled.
<br/><br/>Such that $2^N < N$!
<br/><br/>$\text { i.e., } 4 \leq N \leq 12 \text { i.e., } N \in\{4,5,6, \ldots 12\}$
<br/><br/>Now, $P(N=2)+P(N=3)=\frac{1}{36}+\frac{2}{36}=\frac{3}{36}=\frac{1}{12}$
<br/><br/>So, required probability $=1-\frac{1}{12}=\frac{11}{12}=\frac{m}{n}$
<br/><br/>$4 m-3 n=4 \times 11-3 \times 12=44-36=8$
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
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