Questions: Suppose there are 22 men and 68 women in the local Rotary club. Two members are chosen at random each year to serve on the hospitality committee. What is the probability of choosing two members at random and the first is a woman and the second is a man? Round your answer to four decimal places, if necessary.

Solution Steps

To find the probability of choosing two members at random where the first is a woman and the second is a man, we can use the following approach:

1. Calculate the total number of members.
2. Calculate the probability of selecting a woman first.
3. Calculate the probability of selecting a man second, given that a woman was selected first.
4. Multiply these probabilities to get the final answer.

The total number of members in the Rotary club is given by: \[ \text{Total members} = 22 + 68 = 90 \]

The probability of selecting a woman first is: \[ P(\text{Woman first}) = \frac{68}{90} \approx 0.7556 \]

Given that a woman was selected first, the remaining number of members is \(90 - 1 = 89\). The probability of selecting a man second is: \[ P(\text{Man second} \mid \text{Woman first}) = \frac{22}{89} \approx 0.2472 \]

The final probability of selecting a woman first and a man second is the product of the two probabilities: \[ P(\text{Woman first and Man second}) = P(\text{Woman first}) \times P(\text{Man second} \mid \text{Woman first}) \approx 0.7556 \times 0.2472 \approx 0.1868 \]

Final Answer