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Question #41809

Difficulty: 8/10 | 🟢 Free

The Question

Let $E_1$ and $E_2$ be two mutually exclusive and exhaustive events such that the prior probability $P(E_1)$ is given by $P(E_1) = \frac{6}{13} \lim_{n \to \infty} \frac{1}{n^2} \sum_{k=1}^n \sqrt{ 9k^2 - 6kn + n^2 }$ A box $B_1$ contains 4 red and 1 black balls, while box $B_2$ contains 3 red and 5 black balls. A box is selected such that the probability of selecting $B_1$ is $P(E_1)$ and the probability of selecting $B_2$ is $P(E_2)$ . A ball is drawn at random from the selected box and is found to be red. The probability that it was drawn from box $B_2$ is:

Options

A.
$\frac{3}{7}$
B.
$\frac{4}{7}$
C.
$\frac{25}{41}$
D.
$\frac{15}{47}$

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

Bayes' Theorem Reversal|Integral as the Limit of a Sum|Square Root and Modulus Logic

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of definite integration (as a limit of a sum) and Bayes' Theorem, specifically targeting the common student oversight of the identity $\sqrt{x^2} = |x|$ .

The integrand masks a piecewise function behind a square root.

If a student blindly simplifies $\sqrt{(3x-1)^2}$ to $3x-1$ without applying the modulus, they will evaluate the integral to $1/2$ instead of the correct $5/6$ .

This cascades into incorrect prior probabilities ( $3/13$ and $10/13$ ), leading to a flawed application of the Law of Total Probability and Bayes' Theorem, ultimately yielding the trap answer $25/41$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 12
Concept	Bayes' Theorem Reversal
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free