Start Learning Free

Before You Begin

Question #41718

Difficulty: 8/10 | 🟢 Free

The Question

Let A={1,2,3,4}A = \{1, 2, 3, 4\} and SS be the set of all functions from AA to AA. The set SS is partitioned into three subsets S1,S2,S3S_1, S_2, S_3 as follows:

S1S_1: The set of all bijective functions. S2S_2: The set of all functions h∈Sh \in S that can be expressed as h=f∘gh = f \circ g, where g:A→Bg: A \to B is surjective, f:B→Af: B \to A is injective, and B={1,2,3}B = \{1, 2, 3\}. S3S_3: The set of all remaining functions in SS.

A biased die is rolled to select one of the subsets, such that the probability of selecting SiS_i is proportional to ii. Then, a function hh is chosen uniformly at random from the selected subset. Given that the chosen function satisfies h(1)=h(2)h(1) = h(2), what is the probability that it was chosen from S2S_2?

Options

  • A.

    1156\frac{11}{56}

  • B.

    38\frac{3}{8}

  • C.

    1141\frac{11}{41}

  • D.

    1128\frac{11}{28}

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|Composite Injectivity/Surjectivity|Composition Inverse Identity Traps

Why This Question Is Tricky

Common Mistakes

This question tests the 'Composition Inverse Identity Trap' within a Bayesian probability framework.

Students often incorrectly assume a one-to-one correspondence between the pairs of functions (f,g)(f, g) and their composition h=f∘gh = f \circ g.

If a student calculates ∣S2∣|S_2| by multiplying the number of surjective functions gg (36) by the number of injective functions ff (24), they obtain 864, which absurdly exceeds the total number of functions (256).

This forces the realization that the composition mapping is many-to-one (specifically 6-to-1, as the intermediate set BB can be permuted in 3!3! ways).

The problem then requires rigorous combinatorial counting of domain partitions to find the conditional probabilities for Bayes' Theorem, heavily penalizing any structural assumptions about inverse mappings.

Quick Facts About This Question

Question Information

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