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Question #41806

Difficulty: 8/10 | 🟢 Free

The Question

A biased coin with probability of heads $p \in (0, 1)$ is tossed repeatedly until a head appears. Let $X$ be the total number of tosses required. Let $A$ be the event that $X$ is a multiple of 3, and $B$ be the event that $X \ge 4$ . If $P(A \cup B) = \frac{25}{36}$ , find the probability that $X$ is an even number.

Options

A.
$\frac{5}{11}$
B.
$\frac{4}{9}$
C.
$\frac{6}{11}$
D.
$\frac{5}{12}$

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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

Independent vs Disjoint Events|Complementary Counting|Equiprobability Fallacy

Why This Question Is Tricky

Common Mistakes

This problem tests the application of complementary counting in infinite sample spaces while setting a trap based on the Equiprobability Fallacy.

Students often attempt to use the additive rule $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ .

In doing so, they may incorrectly assume $P(A) = 1/3$ (believing multiples of 3 are uniformly distributed, ignoring the geometric decay of probabilities).

Even if they correctly identify the geometric series for $P(A)$ , calculating the intersection $P(A \cap B)$ requires recognizing the memoryless property (independence) or summing another complex series.

By forcing the complementary counting approach $1 - P(A^c \cap B^c)$ , the infinite series collapses into a simple finite set $\{1, 2\}$ , bypassing the algebraic nightmare and exposing students who rely on heuristic shortcutting over formal set-theoretic definitions.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 12
Concept	Independent vs Disjoint Events
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free