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

Question

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

Accepted Answer

Given: N denotes the sum obtained when two dice are rolled.

Find: The value of $4m - 3n$ , where the probability of $2N < N!$ is $\frac{m}{n}$ .

We are given that $2N < N!$ is satisfied for $N \geq 4$ . So the required probability is

P(N \geq 4) = 1 - P(N < 4)

For $N < 4$ :

$N = 1$ : not possible.
$N = 2$ : only outcome is $(1,1)$ , so probability is $\frac{1}{36}$ .
$N = 3$ : outcomes are $(1,2)$ and $(2,1)$ , so probability is $\frac{2}{36}$ .

Therefore,

P(N < 4) = P(N=2) + P(N=3) = \frac{1}{36} + \frac{2}{36} = \frac{3}{36}

Using the complement rule,

P(N \geq 4) = 1 - \frac{3}{36} = \frac{33}{36} = \frac{11}{12}

Hence,

m = 11, \quad n = 12

Now,

4m - 3n = 4(11) - 3(12) = 44 - 36 = 8

Therefore, the correct option is B.

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