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Question #41808

Difficulty: 8/10 | 🟢 Free

The Question

Let ara_r be the coefficient of xrx^r in the polynomial P(x)=12((1+x)13+(1−x)13)P(x) = \frac{1}{2} \left( (1+x)^{13} + (1-x)^{13} \right). A set of 14 bags, labeled B0,B1,…,B13B_0, B_1, \dots, B_{13}, is given. A bag is selected such that the probability of selecting bag BrB_r is proportional to ara_r. Each bag BrB_r contains rr red balls and 13−r13-r black balls. A bag is chosen at random according to the given probabilities, and two balls are drawn from it, one after another, with replacement. Given that both drawn balls are red, what is the probability that they were drawn from bag B6B_6?

Options

  • A.

    297896\frac{297}{896}

  • B.

    2971792\frac{297}{1792}

  • C.

    165512\frac{165}{512}

  • D.

    1651024\frac{165}{1024}

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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|General Term Formulation|Polynomial Function Parity

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of binomial coefficient properties with Bayes' theorem.

The primary trap lies in the parity of the polynomial P(x)P(x).

Students who mechanically apply the binomial expansion without noting the +(1−x)13+ (1-x)^{13} term will assume ar=(13r)a_r = \binom{13}{r} for all rr.

This leads to an incorrect prior probability denominator of 2132^{13} instead of 2122^{12}.

Interestingly, the total probability of drawing two red balls remains 726\frac{7}{26} in both cases due to the symmetric properties of the second moment of the binomial distribution.

Consequently, missing the parity constraint exactly halves the final conditional probability, leading directly to the trap option 2971792\frac{297}{1792}.

Additional traps include calculating for drawing without replacement.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterProbability - Class 12
ConceptBayes' Theorem Reversal
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree