JEE MAIN - Mathematics (2023 - 10th April Morning Shift - No. 8)
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 :
12
6
8
10
Explanation
$N$ denote the sum of the numbers obtained when two dice are rolled.
Such that $2^N < N$!
$$ \text { i.e., } 4 \leq N \leq 12 \text { i.e., } N \in\{4,5,6, \ldots 12\} $$
Now, $P(N=2)+P(N=3)=\frac{1}{36}+\frac{2}{36}=\frac{3}{36}=\frac{1}{12}$
So, required probability $=1-\frac{1}{12}=\frac{11}{12}=\frac{m}{n}$
$$ 4 m-3 n=4 \times 11-3 \times 12=44-36=8 $$
Such that $2^N < N$!
$$ \text { i.e., } 4 \leq N \leq 12 \text { i.e., } N \in\{4,5,6, \ldots 12\} $$
Now, $P(N=2)+P(N=3)=\frac{1}{36}+\frac{2}{36}=\frac{3}{36}=\frac{1}{12}$
So, required probability $=1-\frac{1}{12}=\frac{11}{12}=\frac{m}{n}$
$$ 4 m-3 n=4 \times 11-3 \times 12=44-36=8 $$
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