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Question #41718
The Question
Let and be the set of all functions from to . The set is partitioned into three subsets as follows:
: The set of all bijective functions. : The set of all functions that can be expressed as , where is surjective, is injective, and . : The set of all remaining functions in .
A biased die is rolled to select one of the subsets, such that the probability of selecting is proportional to . Then, a function is chosen uniformly at random from the selected subset. Given that the chosen function satisfies , what is the probability that it was chosen from ?
Options
- A.
- B.
- C.
- D.
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 and their composition .
If a student calculates by multiplying the number of surjective functions (36) by the number of injective functions (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 can be permuted in 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
| Subject | Mathematics |
| Chapter | Probability - Class 12 |
| Concept | Bayes' Theorem Reversal |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |