In a certain TV game show, exactly 1 prize was placed behind each of 6 curtains, which were labeled A,...

1. Translate the problem requirements: We need to find the probability that after swapping prizes between positions A,B,C and D,E,F, the same two curtain positions still contain winning prizes
2. Identify all possible initial winning configurations: Determine how many ways 2 winning prizes can be distributed among 6 curtains initially
3. Apply the rearrangement rule systematically: For each initial configuration, determine where the winning prizes end up after the A↔D, B↔E, C↔F swap
4. Count favorable outcomes and calculate probability: Identify which initial configurations result in the same curtain positions having winning prizes after rearrangement

Execution of Strategic Approach

1. Translate the problem requirements

Let's understand what's happening in plain English. We have 6 curtains (A, B, C, D, E, F) with prizes behind them - 4 consolation prizes and 2 winning prizes. Initially, the prizes are randomly placed.

Then there's a specific rearrangement: whatever prize was behind curtain A moves to curtain D, whatever was behind B moves to E, whatever was behind C moves to F. At the same time, whatever was behind D moves to A, whatever was behind E moves to B, and whatever was behind F moves to C.

We want to find the probability that the same two curtain positions still have winning prizes after this swap.

2. Identify all possible initial winning configurations

Since we have 2 winning prizes among 6 curtains, we need to count how many ways we can choose 2 curtains out of 6 to place our winning prizes.

Let's think about this systematically. We can pick any 2 curtains from the 6 available curtains. This gives us all possible pairs:

• Pairs involving A: (A,B), (A,C), (A,D), (A,E), (A,F)
• Pairs involving B (not already counted): (B,C), (B,D), (B,E), (B,F)
• Pairs involving C (not already counted): (C,D), (C,E), (C,F)
• Pairs involving D (not already counted): (D,E), (D,F)
• Pairs involving E (not already counted): (E,F)

Counting these up: 5 + 4 + 3 + 2 + 1 = 15 total possible initial configurations.

In mathematical notation: \(\mathrm{C(6,2)} = 15\)

3. Apply the rearrangement rule systematically

Now let's see what happens to each initial configuration after the rearrangement. Remember: \(\mathrm{A↔D, B↔E, C↔F}\).

Let's work through each of our 15 initial configurations:

• (A,B) → (D,E) - winning prizes move from A,B to D,E
• (A,C) → (D,F) - winning prizes move from A,C to D,F
• (A,D) → (D,A) - this is the same as (A,D)! ✓
• (A,E) → (D,B) - winning prizes move from A,E to D,B
• (A,F) → (D,C) - winning prizes move from A,F to D,C
• (B,C) → (E,F) - winning prizes move from B,C to E,F
• (B,D) → (E,A) - winning prizes move from B,D to E,A
• (B,E) → (E,B) - this is the same as (B,E)! ✓
• (B,F) → (E,C) - winning prizes move from B,F to E,C
• (C,D) → (F,A) - winning prizes move from C,D to F,A
• (C,E) → (F,B) - winning prizes move from C,E to F,B
• (C,F) → (F,C) - this is the same as (C,F)! ✓
• (D,E) → (A,B) - winning prizes move from D,E to A,B
• (D,F) → (A,C) - winning prizes move from D,F to A,C
• (E,F) → (B,C) - winning prizes move from E,F to B,C

4. Count favorable outcomes and calculate probability

Looking at our systematic analysis above, we found exactly 3 configurations where the winning prizes end up in the same curtain positions:

• (A,D) stays as (A,D)
• (B,E) stays as (B,E)
• (C,F) stays as (C,F)

This makes perfect sense! These are exactly the configurations where one winning prize is in the 'left group' (A, B, or C) and the other is in the corresponding position in the 'right group' (D, E, or F respectively). When we swap between corresponding positions, these pairs stay together.

So we have 3 favorable outcomes out of 15 total possible outcomes.

Probability = Number of favorable outcomes / Total number of outcomes

Probability = \(\frac{3}{15} = \frac{1}{5}\)

Final Answer

The probability that the same 2 curtains have winning prizes before and after the rearrangement is 1/5.

This corresponds to answer choice C.

Common Faltering Points

Errors while devising the approach

Faltering Point 1: Misunderstanding the rearrangement pattern
Students often struggle to correctly interpret what "the prizes that had been behind Curtains A, B, and C were now behind Curtains D, E, and F, respectively" means. They might think this is a one-way movement rather than understanding it's a complete swap between corresponding positions (A↔D, B↔E, C↔F). This leads to incorrect mapping of where prizes end up after rearrangement.

Faltering Point 2: Confusing "same curtains" with "same prizes"
The question asks for the probability that the same 2 curtains have winning prizes, but students might mistakenly think about whether the same physical prizes end up in their original positions. The key insight is that we care about which curtain positions contain winning prizes, not which specific prize objects are where.

Faltering Point 3: Incorrect problem setup for probability calculation
Students might approach this as a complex conditional probability problem or try to account for the "randomness" of the initial arrangement in their calculations, when actually they should recognize that all possible initial configurations of 2 winning prizes among 6 curtains are equally likely, making this a straightforward counting problem.

Errors while executing the approach

Faltering Point 1: Incomplete enumeration of cases
When listing all possible initial configurations, students often miss some combinations or double-count others. The systematic approach of \(\mathrm{C(6,2)} = 15\) total cases requires careful organization to ensure no pairs are missed. Students might list only 10-12 cases instead of all 15.

Faltering Point 2: Incorrect application of the swap rule
When checking what happens to each initial configuration after rearrangement, students might incorrectly apply the A↔D, B↔E, C↔F rule. For example, they might think (A,D) becomes (D,D) or (A,A) instead of recognizing that A→D and D→A simultaneously, so (A,D) remains (A,D).

Faltering Point 3: Missing the pattern in favorable outcomes
Students might laboriously check each case without recognizing that only pairs consisting of one curtain from {A,B,C} and its corresponding partner from {D,E,F} will remain unchanged. This makes them more prone to calculation errors and less likely to verify their answer makes logical sense.

Errors while selecting the answer

Faltering Point 1: Arithmetic errors in fraction reduction
Even after correctly identifying 3 favorable outcomes out of 15 total outcomes, students might make simple arithmetic mistakes when reducing \(\frac{3}{15}\) to \(\frac{1}{5}\), or might leave their answer as \(\frac{3}{15}\) without simplifying it to match the given answer choices.