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Question #41075

Difficulty: 8/10 | 🟢 Free

The Question

Let $A$ and $B$ be two events such that $P(A) = \frac{1}{2}$ and $P(B) = \frac{1}{3}$ . For a continuous parameter $x \in \left[0, \frac{\pi}{3}\right]$ , the conditional probability $P(A|B)$ is given by a function $y(x)$ which satisfies the differential equation: $\frac{dy}{dx} = y \tan x - y^2 \sec x$ It is known that $y(x)$ represents a valid probability for all $x \in \left[0, \frac{\pi}{3}\right]$ . Let $M$ be the minimum possible value of $P(A \cup B)$ at $x = \frac{\pi}{3}$ over all such valid functions $y(x)$ . The value of $M$ is:

Options

A.
$\frac{7 - 2\sqrt{3}}{6}$
B.
$\frac{1}{2}$
C.
$\frac{3 + 2\sqrt{3}}{6}$
D.
$\frac{11 - 6\sqrt{3}}{6}$

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|Arbitrary Constants vs Parameters|Variable Separable Constraints

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of differential equations and probability axioms.

The core trap lies in treating the integration constant $C$ as a free parameter.

Students often solve the Bernoulli equation correctly but fail to apply the boundary condition $0 \le P(A|B) \le 1$ across the entire domain $x \in [0, \frac{\pi}{3}]$ .

By forcing $y(x) \le 1$ , a domain constraint $C \ge 1$ is established.

Failing to maximize the constraint function at $x=0$ leads to incorrect bounds for $C$ (such as evaluating at the endpoint $x=\pi/3$ ), resulting in the lethal distractor $1/2$ .

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