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Question #41719

Difficulty: 8/10 | 🟢 Free

The Question

Let $I_n$ be the set of all integers $k$ ( $0 \le k \le n$ ) for which the binomial coefficient $\binom{n}{k}$ attains its maximum value. Define $S_n = \sum_{k \in I_n} k$ . Two bags $B_1$ and $B_2$ are given. Bag $B_1$ contains 2 Red and 3 Black balls, while Bag $B_2$ contains 3 Red and 4 Black balls. A bag is selected at random such that the probability of choosing $B_1$ is proportional to $\sum_{m=1}^{7} S_{2m}$ and the probability of choosing $B_2$ is proportional to $\sum_{m=1}^{8} S_{2m-1}$ . A ball is drawn at random from the selected bag and is found to be Red. The probability that it came from Bag $B_1$ is:

Options

A.
$\frac{49}{169}$
B.
$\frac{14}{29}$
C.
$\frac{98}{233}$
D.
$\frac{49}{120}$

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

Bayes' Theorem Reversal|Sigma Notation Properties|Greatest Binomial Coefficient

Why This Question Is Tricky

Common Mistakes

This question tests the integration of combinatorics with conditional probability.

The core trap lies in the properties of the greatest binomial coefficient.

Students often incorrectly assume that the maximum binomial coefficient is always unique, taking only one central index for odd $n$ (e.g., just $(n-1)/2$ ).

This mistake halves the calculated weight for $B_2$ (yielding 28 or 36 instead of 64), which directly corrupts the prior probabilities $P(B_1)$ and $P(B_2)$ .

Consequently, the total probability denominator in Bayes' Theorem is poisoned, leading to incorrect trap answers like 14/29 or 98/233.

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