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Question #41032

Difficulty: 8/10 | 🟢 Free

The Question

Let $S = \{1, 2, 3, \dots, 20\}$ . A number $n$ is chosen at random from $S$ . Three bags $B_1, B_2,$ and $B_3$ are available. Bag $B_1$ is selected if $|n - 8| + |n - 14| = 6$ . Bag $B_2$ is selected if $|n - 8| + |n - 14| > 6$ and $|n - 11| \le 5$ . Bag $B_3$ is selected if $|n - 11| > 5$ . The contents of the bags are as follows: $B_1$ contains 2 Red and 3 Black balls, $B_2$ contains 3 Red and 2 Black balls, and $B_3$ contains 4 Red and 1 Black ball. A ball is drawn at random from the selected bag and is found to be Red. What is the probability that the chosen number $n$ is a prime number?

Options

A.
$\frac{27}{62}$
B.
$\frac{14}{31}$
C.
$\frac{25}{62}$
D.
$\frac{29}{62}$

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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

Conditional Probability Foundations|Bayes' Theorem Reversal|Modulus Function Symmetry

Why This Question Is Tricky

Common Mistakes

The problem fuses non-linear modulus inequalities with a Bayesian partition, creating a 'Symmetric Partition' trap.

Students often mechanically apply Bayes' theorem to find the posterior probability of the bags ( $P(B_i|R)$ ) instead of recognizing that the target event (choosing a prime) is a cross-sectional subset of the sample space $S$ .

Furthermore, the modulus equation $|n-8| + |n-14| = 6$ relies on the triangle inequality property $|x-a| + |x-b| = |a-b| \iff x \in [a,b]$ .

Failing to correctly decode these overlapping modulus domains leads to an incorrect partition of the sample space, corrupting both the total probability denominator and the cross-sectional numerator.

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