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 $4m-3n$ is equal to

Answer 1

8

Answer 2

The sum $N$ of the numbers obtained when rolling two dice can range from $2$ to $12$. The total number of possible outcomes is $6 \times 6 = 36$.

The condition $2^N < N!$ needs to be checked. For $N=2$, $2^2=4$, $2!=2$. $4 \not< 2$. For $N=3$, $2^3=8$, $3!=6$. $8 \not< 6$. For $N=4$, $2^4=16$, $4!=24$. $16 < 24$. This inequality holds true for all $N \ge 4$.

The possible values of $N$ from rolling two dice that satisfy $N \ge 4$ are $\{4, 5, 6, 7, 8, 9, 10, 11, 12\}$. The number of ways to obtain each sum are:

Sum (N)	Ways
4	3
5	4
6	5
7	6
8	5
9	4
10	3
11	2
12	1

The total number of favorable outcomes is the sum of ways for $N=4$ to $N=12$: $3+4+5+6+5+4+3+2+1 = 33$. The probability is $\frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{33}{36}$. Simplifying the fraction, we get $\frac{11}{12}$. So, $m=11$ and $n=12$. These are coprime. The problem asks for the value of $4m-3n$. $4m - 3n = 4(11) - 3(12) = 44 - 36 = 8$.