概率题两种解法适用场景咨询及6对夫妻选4人概率求解

Hey there! Let's walk through this problem clearly, confirm your approach, and then break down when to use the two common methods for these types of probability questions.

Your Solution is Correct! #

First off, your reasoning for counting the valid selections is spot-on. Let's recap to make sure every step makes sense:

• Step 1: Choose the one married couple from the 6 available: ^6C_1 (which equals 6). This locks in the only pair of spouses in our 4-person group.
• Step 2: Pick 2 distinct couples from the remaining 5—we do this to avoid accidentally picking a second married pair. That's ^5C_2 (10 total ways).
• Step 3: From each of these 2 chosen couples, select one person (since we can't have both, or we'd end up with two married pairs). Each couple has 2 choices, so that's 2×2=4 ways.

Multiplying these together gives the total number of valid selections: 6 × 10 × 4 = 240.

To get the probability, divide this by the total number of ways to choose 4 people from 12: ^12C_4 = 495. So the probability is 240/495 = 16/33 ≈ 0.485.

Two Common Methods & Their Use Cases #

For these combinatorial probability problems, the two go-to approaches are direct counting (your method) and indirect counting (complementary probability). Here's when to use each:

1. Direct Counting

• Best for: When the target event (in this case, "exactly one married couple") can be broken down into clear, non-overlapping steps with no ambiguity.
• Pros: Logical and straightforward—each step directly builds the outcome you want, making it easy to verify if you're counting correctly.
• Gotcha: You need to be careful to avoid overcounting or including invalid cases. For example, if you'd tried to pick the initial couple then just choose 2 people from the remaining 10, you'd end up including cases where those 2 are another married pair (which we don't want). Your approach avoids this by first selecting separate couples, then picking one person from each.

2. Indirect Counting (Complementary Probability)

• Best for: When the target event has complex subcases, but the opposite event (the complement) is simpler to count. For this problem, the complement of "exactly one married couple" is "no married couples at all" + "exactly two married couples".
• Example for this problem:
  • Total selections: 495
  • Exactly two married couples: ^6C_2 = 15 (just pick 2 couples from the 6)
  • No married couples: ^6C_4 × 2^4 = 15 × 16 = 240 (pick 4 distinct couples, then one person from each)
  • Valid selections = Total - Complement = 495 - 15 - 240 = 240 (same as direct counting!)
• Pros: Reduces the problem to counting simpler, fewer cases, which minimizes the chance of making a mistake when the target event has multiple moving parts.

Quick Rule of Thumb #

• Use direct counting if you can clearly map out how to build the desired outcome step-by-step.
• Use indirect counting if the complement of your target event has fewer, easier-to-count scenarios.

内容的提问来源于stack exchange，提问作者Ishan Sharma