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

Q: Why do we stop checking after N=4?

Because N! grows factorially while 2^N grows exponentially. Once N! overtakes 2^N, it never looks back. For N=4, 16 4, the factorial growth will always outpace the exponential growth.

Q: Is it wrong to list all 36 outcomes?

It is not wrong, but it is inefficient. In competitive exams like JEE, time is your most valuable currency. Using the complement rule allows you to solve the problem by calculating only the cases where the condition fails, which is much faster.

Q: What if I forgot the complement rule?

You would be forced to sum the probabilities for N=4, 5, 6, 7, 8, 9, 10, 11, and 12. That is 9 separate calculations instead of just 2. The complement rule is a powerful tool to simplify your workflow and reduce the chance of arithmetic errors.

Question

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

Options:

(A) 6

(B) 12

(C) 10

(D) 8

Sigma Percentile · Accepted Answer

### Analyzing the Setup

When we roll two fair six-sided dice, we are dealing with a sample space of $36$ distinct, equally likely outcomes. Our random variable $N$ represents the sum of these two dice, where $N \in \{2, 3, 4, \dots, 12\}$.

We are tasked with finding the probability that the condition $2^N < N!$ is satisfied. This problem explores the fundamental tension between exponential growth and factorial growth.

### The Search for the Crossover

Let us test the smallest values of $N$ to identify where the inequality holds:

For $N=2$:
$$2^2 = 4, \quad 2! = 2 \quad \Rightarrow \quad 4 < 2 	ext{ is False.}$$

For $N=3$:
$$2^3 = 8, \quad 3! = 6 \quad \Rightarrow \quad 8 < 6 	ext{ is False.}$$

For $N=4$:
$$2^4 = 16, \quad 4! = 24 \quad \Rightarrow \quad 16 < 24 	ext{ is True.}$$

Since the factorial function $N!$ grows significantly faster than the exponential function $2^N$ for all $N \ge 4$, the condition $2^N < N!$ holds true for all sums $N \in \{4, 5, 6, 7, 8, 9, 10, 11, 12\}$.

### The Elegance of the Complement

Rather than calculating the probability for each favorable sum individually, we utilize the complement rule:
$$P(N \ge 4) = 1 - P(N < 4)$$

The unwanted outcomes are $N=2$ and $N=3$. The number of ways to obtain these sums are:
- For $N=2$: Only $(1, 1)$, which is $1$ outcome.
- For $N=3$: $(1, 2)$ and $(2, 1)$, which are $2$ outcomes.

The probability of the unwanted outcomes is:
$$P(N < 4) = \frac{1 + 2}{36} = \frac{3}{36} = \frac{1}{12}$$

### The Final Calculation

With the unwanted probability determined, the favorable probability is:
$$P(N \ge 4) = 1 - \frac{1}{12} = \frac{11}{12}$$

We are given that this probability is $\frac{m}{n}$ where $m$ and $n$ are coprime. Thus, $m=11$ and $n=12$.

Finally, we calculate the value of $4m - 3n$:
$$4(11) - 3(12) = 44 - 36 = 8$$

The final answer is **8**.

Animated Solution for Mathematics - Probability: 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 $4 m - 3 n$ is equal to

Select Answer:

Visualized Solution

$N = Sum of Two Dice$

$2^{N} < N!$

$Testing N = 2$

$Testing N = 3$

$Testing N = 4$

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

$P (N < 4) = \frac{3}{36}$

$P (N \geq 4) = \frac{11}{12}$

$4 m - 3 n = 8$

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

The Search for the Crossover

The Elegance of the Complement

The Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N=Sum of Two Dice

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2N<N!

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Testing N=2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Testing N=3

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Testing N=4

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(N≥4)=1−P(N<4)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(N<4)=363​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(N≥4)=1211​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4m−3n=8