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Question #41033

Difficulty: 8/10 | 🟢 Free

The Question

Six distinct balls are distributed randomly into four distinct boxes. Let $E$ be the event that no box is empty, and $F$ be the event that Box 1 and Box 2 contain an equal number of balls. The conditional probability $P(F|E)$ is:

Options

A.
$\frac{5}{13}$
B.
$\frac{43}{78}$
C.
$\frac{7}{26}$
D.
$\frac{15}{52}$

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

Conditional Probability Foundations|Combinatorial Counting (P&C Integration)|Cardinality Inclusion-Exclusion

Why This Question Is Tricky

Common Mistakes

This question tests the application of the Inclusion-Exclusion Principle within a conditional probability framework.

The primary trap lies in calculating the cardinality of the intersection $F \cap E$ .

Students often correctly use Inclusion-Exclusion for the reduced sample space $n(E)$ but fail to apply it to the restricted subspace for $n(F \cap E)$ .

By merely assigning equal balls to Box 1 and Box 2 and distributing the rest freely, they overcount cases where Box 3 or Box 4 remain empty (yielding an incorrect intersection size of 860), leading to the trap probability of 43/78.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 11
Concept	Conditional Probability Foundations
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free