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Question #41406

Difficulty: 8/10 | 🟢 Free

The Question

A continuous random variable $X$ has a probability density function $f(x) = \frac{1}{2} f_1(x) + \frac{1}{2} f_2(x)$ , where $f_1(x) = 1$ and $f_2(x) = 2x$ for $x \in [0, 1]$ , and $f(x) = 0$ otherwise. Let $A$ be the event $X \in \left[0, \frac{1}{2}\right]$ and $B$ be the event $X \in \left[\frac{1}{4}, \frac{3}{4}\right]$ . What is the conditional expectation $E[X \mid A \cup B]$ ?

Options

A.
$\frac{3}{7}$
B.
$\frac{5}{12}$
C.
$\frac{5}{9}$
D.
$\frac{9}{28}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Linearity and Partitioning|Independent vs Disjoint Events|Conditional Probability Constraints

Why This Question Is Tricky

Common Mistakes

This problem tests the student's ability to correctly evaluate conditional expectations when the underlying probability density function is a mixture of distributions.

The core trap (Trap 359: Additivity Fallacy) lies in the partitioning of the expected value integral and the probability denominator.

Students often incorrectly assume linearity applies directly to the events, calculating $E[X \mid A \cup B] = \frac{\int_A x f(x) dx + \int_B x f(x) dx}{P(A) + P(B)}$ .

Because events $A$ and $B$ are not disjoint (they overlap on $[1/4, 1/2]$ ), this blind additivity double-counts the intersection, leading to the incorrect option 5/12.

The correct approach requires identifying the true measure-theoretic union $A \cup B = [0, 3/4]$ and integrating over this single, continuous sub-domain.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Integrals - Class 12
Concept	Linearity and Partitioning
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free