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

Question

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

Smart Achievers · Accepted Answer

(8)

$2^{N} < N! Which is true when N \geq 4$

$N = 1 (Not possible)$

$N = 2 i.e., (1, 1) (Not possible)$

$∴$ $required probability = \frac{36 - 3}{36} = \frac{33}{36} = \frac{11}{12}$

$∴$ $m = 11 and n = 12$

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

Solution

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

Ans.
(8)

$2^{N} < N! Which is true when N \geq 4$

$N = 1 (Not possible)$

$N = 2 i.e., (1, 1) (Not possible)$

$∴$ $required probability = \frac{36 - 3}{36} = \frac{33}{36} = \frac{11}{12}$

$∴$ $m = 11 and n = 12$

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

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Solution

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

Ans. (8) 2N<N! Which is true when N≥4 N=1 (Not possible) N=2 i.e., (1,1) (Not possible) ∴ required probability=36-336=3336=1112 ∴ m=11 and n=12 Now, 4m-3n=4(11)-3(12)=44-36=8

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

Ans.
(8)

$2^{N} < N! Which is true when N \geq 4$

$N = 1 (Not possible)$

$N = 2 i.e., (1, 1) (Not possible)$

$∴$ $required probability = \frac{36 - 3}{36} = \frac{33}{36} = \frac{11}{12}$

$∴$ $m = 11 and n = 12$

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