Before You Begin

Question #41800

Difficulty: 8/10 | 🟢 Free

The Question

Let $S$ be the sample space of all functions $f: A \to A$ , where $A = \{1, 2, 3, 4\}$ . A function is chosen at random from $S$ with uniform probability. Define an equivalence relation $\sim$ on $S$ by $f \sim g \iff \text{Range}(f) = \text{Range}(g)$ . Let $C_f$ denote the equivalence class containing $f$ . Let $E$ be the event that the chosen function satisfies $f(f(x)) = f(x)$ for all $x \in A$ . Let $F$ be the event that the chosen function belongs to an equivalence class of size exactly $16$ . Find the conditional probability $P(E \mid F)$ .

Options

A.
$\frac{1}{4}$
B.
$\frac{2}{7}$
C.
0
D.
\text{Undefined}

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

Conditional Probability Constraints|Equivalence Class Partitioning|The Empty Set Trap

Why This Question Is Tricky

Common Mistakes

This question targets the 'Empty Set Trap' and the 'Denominator Zero' fallacy in conditional probability.

Students often mechanically apply the conditional probability formula $P(E|F) = P(E \cap F)/P(F)$ without verifying if the conditioning event $F$ is actually possible.

Here, $F$ relies on an equivalence class having exactly 16 elements.

A classic combinatorial error is assuming the number of functions with a range of size 2 is $2^4 = 16$ , forgetting to subtract the non-surjective cases.

Students making this error will incorrectly assume $F$ is non-empty and calculate a spurious probability of $1/4$ (since there are 24 idempotent functions out of the flawed 96).

Careful verification reveals $F = \emptyset$ , making the conditional probability undefined, thereby separating rote learners from those who rigorously check constraints.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 12
Concept	Conditional Probability Constraints
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free