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Question #41842

Difficulty: 8/10 | 🟢 Free

The Question

Let $N$ be the exponent of the prime $79$ in $10000!$ . Three distinct numbers are chosen at random without replacement from the set $S = \{1, 2, \dots, N\}$ . Let $P$ be the probability that the chosen subset contains at least one number $\le 30$ and at least one number $\ge 100$ . Which of the following equals $P$ ?

Options

A.
$1 - \frac{\binom{97}{3} + \binom{99}{3} - \binom{69}{3}}{\binom{127}{3}}$
B.
$1 - \frac{\binom{97}{3} + \binom{99}{3} + \binom{69}{3}}{\binom{127}{3}}$
C.
$\frac{\binom{30}{1}\binom{28}{1}\binom{69}{1}}{\binom{127}{3}}$
D.
$\frac{\binom{97}{3} + \binom{99}{3} - \binom{69}{3}}{\binom{127}{3}}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

De Morgan's Laws in Probability|Sampling With/Without Replacement|Exponent of Prime in n!

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to combine number theory (Legendre's Formula) with advanced combinatorial probability.

The primary trap lies in calculating $P(A \cap B)$ directly, which leads to missing multiple sub-cases (e.g., drawing two numbers $\le 30$ and one $\ge 100$ ).

Students must recognize that applying De Morgan's Law to shift to the union of the complements $1 - P(A' \cup B')$ drastically simplifies the problem.

Furthermore, the sample space size $N$ is hidden behind the exponent of a prime in a factorial, preventing trial-and-error and ensuring only those familiar with Legendre's formula can proceed.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 12
Concept	De Morgan's Laws in Probability
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free