In a party there are 5 married couples (that is, 5 distinct pairs, each made of a husband and a wife). If 5 people are chosen at random from these 10 persons, what is the probability that the selected group contains at least two complete couples?

Introduction / Context:
This question combines combinatorics with probability and involves counting groups with a structural constraint: at least two complete couples must be present in the chosen group. Each couple consists of two specific people, and we select 5 people from 10. The challenge is to count the favorable combinations carefully without overcounting.

Given Data / Assumptions:

• There are 5 married couples, so 10 people in total.
• Each couple has exactly 2 members.
• We choose 5 people uniformly at random from these 10.
• A complete couple means both husband and wife are included in the selected group.
• We want the probability that there are at least two complete couples in the chosen group.

Concept / Approach:
The total number of ways to select any 5 people from 10 is 10C5. For the favorable cases, we must have at least 2 complete couples. Because we are choosing only 5 people, the only possible pattern that satisfies at least two couples is:

• Exactly two complete couples (4 people) and one additional person from a different couple.

There is no way to have three complete couples, as that would require 6 people. So we count selections with exactly two couples and one extra person, then divide by the total combinations.

Step-by-Step Solution:
Step 1: Compute total ways to choose any 5 people from 10: 10C5. 10C5 = 10 * 9 * 8 * 7 * 6 / (5 * 4 * 3 * 2 * 1) = 252. Step 2: Choose which 2 couples out of the 5 will both be fully present. Number of ways to choose 2 couples from 5 = 5C2 = 10. Step 3: After taking these 2 couples (4 people), we must choose 1 more person to make a group of 5. The remaining couples are 3 in number, giving 6 individuals available for the extra spot. Step 4: Number of ways to choose this extra person = 6C1 = 6. Step 5: Total favorable groups with at least two couples = 10 * 6 = 60. Step 6: Probability = favorable / total = 60 / 252. Step 7: Simplify 60 / 252 by dividing numerator and denominator by 12 to get 5 / 21.

Verification / Alternative check:
You can check the logic by reasoning about impossible patterns. Three couples would require 6 people, which is more than we select, so only the pattern exactly two couples plus one extra is valid. There is no other way to have at least two couples with a group size of 5. Checking the counts carefully shows that each selection is uniquely determined by choosing a pair of couples and then picking one additional person from the remaining couples. Because the total number of groups 10C5 is known, and our favorable count is 60, the probability 5/21 is consistent and complete.

Why Other Options Are Wrong:
Options 6/21 and 7/21 correspond to larger probabilities and would require more than 60 favorable combinations, which do not exist under the constraints. Option 5/42 is exactly half of the correct probability and might come from missing a factor of 2 or incorrectly counting couples. Option 11/21 is far too large, suggesting that a majority of all groups would contain at least two couples, which clearly is not true when you experiment with a few concrete groupings.

Common Pitfalls:
A frequent error is to try to count groups with at least two couples by adding several complicated cases or by attempting complementary counting without carefully defining the complement. Another pitfall is to forget that the extra person must not complete a third couple, which would violate the group size constraint. To avoid confusion, it is helpful to think in terms of couples first, then fill remaining spots with individuals from remaining couples, and finally confirm that there is no overlap between cases.

Final Answer:
The probability that the selected group of five people contains at least two complete couples is 5/21.