Six married couples are locked in a room, If two people are chosen at random, then the probability that one is male and the other is a female is

Options

A. `1/11`
B. `6/11`
C. `10/11`
D. `5/11`

Answer

To solve the problem of finding the probability that one person is male and the other is female when two people are chosen at random from six married couples, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the Total Number of People**:
- Since there are 6 married couples, the total number of people in the room is:
\[ 6 \text{ males} + 6 \text{ females} = 12 \text{ people} \]

2. **Calculate the Total Number of Ways to Choose 2 People**:
- The total number of ways to choose 2 people from 12 is given by the combination formula \( nCk \):
\[ \text{Total Outcomes} = \binom{12}{2} = \frac{12 \times 11}{2 \times 1} = 66 \]

3. **Calculate the Number of Favorable Outcomes (1 Male and 1 Female)**:
- To find the number of ways to choose 1 male and 1 female, we can use the combination formula:
- Choose 1 male from 6:
\[ \text{Ways to choose 1 male} = \binom{6}{1} = 6 \]
- Choose 1 female from 6:
\[ \text{Ways to choose 1 female} = \binom{6}{1} = 6 \]
- Therefore, the total number of favorable outcomes (1 male and 1 female) is:
\[ \text{Favorable Outcomes} = \binom{6}{1} \times \binom{6}{1} = 6 \times 6 = 36 \]

4. **Calculate the Probability**:
- The probability \( P \) that one person is male and the other is female is given by the ratio of favorable outcomes to total outcomes:
\[ P(\text{1 Male and 1 Female}) = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{36}{66} \]
- Simplifying the fraction:
\[ P(\text{1 Male and 1 Female}) = \frac{6}{11} \]

### Final Answer:
The probability that one person is male and the other is female is \( \frac{6}{11} \).