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Question #41071

Difficulty: 8/10 | 🟢 Free

The Question

Let $A, B,$ and $C$ be three independent events such that $P(A) = p$ , $P(B) = 2p - 1$ , and $P(C) = 1 - \frac{p}{2}$ for some $p \in \mathbb{R}$ . The minimum possible value of $P(A \cup B \cup C)$ is

Options

A.
$\frac{23}{27}$
B.
$\frac{7}{8}$
C.
$\frac{1}{8}$
D.
$\frac{3}{4}$

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

Complementary Counting|De Morgan's Laws in Probability|Domain and Range Nuance

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to integrate calculus-based optimization with the foundational axioms of probability.

The primary trap lies in the domain constraints of the probability variables.

Students often blindly differentiate the complementary probability function $f(p) = p(1-p)^2$ and find a local maximum at $p = 1/3$ , which leads directly to the trap option 23/27.

However, they fail to realize that $p = 1/3$ results in $P(B) = -1/3$ , which is an impossible event.

By enforcing the axiom $0 \le P(E) \le 1$ as a system of inequalities, the valid domain is restricted to $p \in [0.5, 1]$ , forcing the true maximum of the complement to occur at the boundary $p = 0.5$ .

Quick Facts About This Question

Question Information

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