Let's go through the problem.

The question states:

A pool of potential jurors consists of 20 men and 16 women. The Commissioner of Jurors randomly selects two names from the pool. What is the probability that the first is a man and the second is a woman if the two people are selected without replacement?

Solution:

1. Total number of jurors: $20 + 16 = 36$

2. Probability that the first person selected is a man: There are 20 men out of 36 people, so: $P(\text{First is a man}) = \frac{20}{36}$

3. Probability that the second person selected is a woman (given the first was a man): Since one man has already been selected, there are now 19 men and 16 women remaining, for a total of 35 people. $P(\text{Second is a woman | First was a man}) = \frac{16}{35}$

4. Combined Probability: Since these are sequential, independent events (with the pool changing after each selection due to "without replacement"), we multiply the probabilities: $P(\text{First is a man and second is a woman}) = \frac{20}{36} \times \frac{16}{35}$

5. Calculating: Simplify $\frac{20}{36}$ to $\frac{5}{9}$ and then calculate: $\frac{5}{9} \times \frac{16}{35} = \frac{80}{315}$

6. Final Answer: The answer is $\frac{80}{315}$, which does not seem to simplify further.

Thus, the closest answer to choose from the options given is likely to be similar to $\frac{80}{315}$, if it is among the choices listed.

Would you like further clarification or details on any part of the solution?

Related Questions

1. How would the probability change if the second selection was also a man?
2. What is the probability that both selected individuals are women?
3. If three names are selected, what is the probability that all are men?
4. How would the probability change if we were selecting with replacement instead?
5. What would the probability be if only one person was selected, and it had to be a woman?

Tip

In probability problems, carefully consider whether selections are made with or without replacement, as this significantly impacts the probabilities.