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Question #41018

Difficulty: 8/10 | 🟢 Free

The Question

Urn $U_1$ contains $x-2$ red balls and $5$ black balls. Urn $U_2$ contains $x-2$ red balls and $1$ black ball, where $x \ge 2$ is an integer. An urn is selected by rolling a fair standard die. If the die shows a number less than 3, Urn $U_1$ is chosen; otherwise, Urn $U_2$ is chosen. A ball is drawn at random from the selected urn and is found to be red. Given that the drawn ball is red, the probability that it came from Urn $U_1$ is $\frac{3}{10}$ . Find the sum of all possible valid values of $x$ .

Options

A.
25
B.
27
C.
4
D.
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

Equiprobability Fallacy|Bayes' Theorem Reversal|The Zero-Factor Cancellation Trap

Why This Question Is Tricky

Common Mistakes

This problem simultaneously tests the 'Equiprobability Fallacy' and the 'Zero-Factor Cancellation Trap'.

Students who lazily assume the urns are equally likely ( $P(U_1)=P(U_2)=1/2$ ) will solve a different equation yielding $x=4$ .

Students who correctly identify the priors but blindly cross-multiply the Bayesian equation will obtain a quadratic with roots $x=2$ and $x=25$ .

The trap lies in failing to verify the roots against the domain of the conditional probability space: at $x=2$ , the event of drawing a red ball has a probability of strictly zero, rendering the Bayesian posterior $P(U_1|R)$ undefined.

Including this extraneous root leads to the trap answer of 27.

Quick Facts About This Question

Question Information

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