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Question #40683

Difficulty: 8/10 | 🟢 Free

The Question

An urn contains 15 balls numbered $1, 2, 3, \dots, 15$ . Four balls are drawn simultaneously from the urn. Given that the sum of the numbers on the drawn balls is 34, let $p$ be the probability that no three of the drawn numbers form an arithmetic progression. Then $p$ is equal to

Options

A.
$\frac{46}{67}$
B.
$\frac{42}{67}$
C.
$\frac{53}{67}$
D.
$\frac{45}{67}$

Check Your Answer

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

Selection of AP Terms|Combinatorial Probability|De Morgan’s Laws in Probability

Why This Question Is Tricky

Common Mistakes

This question tests combinatorial probability under strict Diophantine constraints, specifically targeting the 'Lethal Collision' trap in Inclusion-Exclusion.

Students must use De Morgan's Laws to find the probability of the complement (no 3 terms form an AP).

The primary trap is assuming that 3-term APs in a 4-element sorted subset must be contiguous ( $T_1$ or $T_2$ ), completely missing the non-contiguous APs ( $T_3$ and $T_4$ ).

Furthermore, even if all 4 AP configurations are identified, students often fail to subtract the intersections (e.g., 4-term APs and sequences like $a, a+d, a+2d, a+4d$ ), leading to a double-counting error.

The constant sum constraint $x_1+x_2+x_3+x_4=34$ elegantly reduces the AP condition to simple linear equations (e.g., $3x_2 + x_4 = 34$ ), rewarding algebraic substitution over brute-force counting.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Sequences and Series - Class 11
Concept	Selection of AP Terms
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free