Before You Begin

Question #41830

Difficulty: 8/10 | 🟢 Free

The Question

Let $S = \{0, 1, 2, \dots, 61\}$ . A subset $A$ of $S$ is chosen uniformly at random. Let $X = \sum_{r \in A} \binom{61}{r}$ . The probability that $|X - 2^{60}| \le 2^{60} - 1892$ and $X \equiv 0 \pmod{61}$ is:

Options

A.
$\frac{2^{56} - 1}{2^{58}}$
B.
$\frac{2^{58} - 3}{2^{60}}$
C.
$\frac{2^{58} - 5}{2^{60}}$
D.
$\frac{2^{57} - 1}{2^{60}}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Complementary Counting|Properties of Binomial Coefficients|Modulus Inequality Logic

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of modular arithmetic on binomial coefficients with complementary counting in probability.

The primary trap lies in the asymmetric boundary conditions of the absolute value inequality.

Students often correctly deduce that $X \equiv 0 \pmod{61}$ forces $0 \notin A$ and $61 \notin A$ due to the prime property of 61.

However, when applying complementary counting for $|X - 2^{60}| > 2^{60} - 1892$ , they frequently assume symmetry between the lower bound ( $X < 1892$ ) and the upper bound ( $X > 2^{61} - 1892$ ).

The trap is that the complement set $B = A^c$ inherently contains the elements 0 and 61, reducing the available sum threshold from 1892 to 1890.

This subtle shift eliminates valid combinations (like $1830 + 61 = 1891$ ), breaking the symmetry (10 ways vs 6 ways) and invalidating standard binomial distribution approximations.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 12
Concept	Complementary Counting
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free