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(A) Let $N$ be the sum of the numbers on two dice. The possible values for $N$ are $2, 3, 4, \dots, 12$.We need to find the probability that $2^{N} <

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$8$

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(A) Let $N$ be the sum of the numbers on two dice. The possible values for $N$ are $2, 3, 4, \dots, 12$.We need to find the probability that $2^{N} Let us check the condition $2^{N} For $N=2: 2^2 = 4, 2! = 2$. $4 For $N=3: 2^3 = 8, 3! = 6$. $8 For $N=4: 2^4 = 16, 4! = 24$. $16 For $N=5: 2^5 = 32, 5! = 120$. $32 For $N \geq 4$,the condition $2^N Thus,we need to find the probability that $N \geq 4$.$P(N \geq 4) = 1 - P(N The total number of outcomes is $6 	imes 6 = 36$.$P(N=2) = \frac{1}{36}$ (outcomes: $(1,1)$).$P(N=3) = \frac{2}{36}$ (outcomes: $(1,2), (2,1)$).$P(N Therefore,$P(N \geq 4) = 1 - \frac{1}{12} = \frac{11}{12}$.Here,$m = 11$ and $n = 12$. Since $11$ and $12$ are coprime,we calculate $4m - 3n = 4(11) - 3(12) = 44 - 36 = 8$.

Let $N$ denote 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 $4m - 3n$ is equal to $......$.

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