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Question #41074

Difficulty: 8/10 | 🟢 Free

The Question

Two players, A and B, each independently toss a fair coin $50$ times. Let $E_1$ be the event that player A obtains strictly more heads than player B, and $E_2$ be the event that player A obtains at least as many heads as player B. The conditional probability $P(E_1 \mid E_2)$ is given by:

Options

A.
$\frac{2^{100} - \binom{100}{50}}{2^{100} + \binom{100}{50}}$
B.
$\frac{2^{101} - \binom{100}{50}}{2^{101} + \binom{100}{50}}$
C.
$\frac{2^{99} - \binom{100}{50}}{2^{99} + \binom{100}{50}}$
D.
$\frac{2^{50} - \binom{50}{25}}{2^{50} + \binom{50}{25}}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Conditional Probability Foundations|Binomial Distribution Tail-Ends|Properties of Binomial Coefficients

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of conditional probability and binomial coefficient identities.

The primary trap lies in attempting to manually expand the tail-end probabilities $P(X > Y)$ and $P(X \ge Y)$ as double summations, which is computationally unfeasible.

Students must recognize the symmetry of the binomial distribution ( $P(X > Y) = P(X < Y)$ ) and apply the combinatorial identity $\sum \binom{n}{k}^2 = \binom{2n}{n}$ to evaluate $P(X = Y)$ .

Failing to leverage these properties or making algebraic errors in the complement rule leads directly to the distractor options.

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