A committee consists of five men and four women. Two people are selected at random from the committee. What is the probability that one man and one woman will be selected?
9 people
probability of man first = 5/9
then prob of woman = 4/8
so 20/72 for man then woman
then
probability of woman first = 4/9
then prob of man = 5/8
so 20/72 again
add them
40/72 = .55
To find the probability of selecting one man and one woman from the committee, we need to determine the total number of possible outcomes and the number of favorable outcomes.
First, let's find the total number of possible outcomes. When two people are selected at random from the committee, it is similar to choosing two individuals from a group of nine people:
The total number of possible outcomes = C(9, 2) = 9! / (2! * 7!) = (9 * 8) / (2 * 1) = 36.
Next, let's determine the number of favorable outcomes, which involves selecting one man and one woman. The number of ways to choose one man from the five available men is C(5, 1) = 5. Similarly, the number of ways to select one woman from the four available women is C(4, 1) = 4.
The number of favorable outcomes = C(5, 1) * C(4, 1) = 5 * 4 = 20.
Therefore, the probability of selecting one man and one woman is:
Probability = favorable outcomes / total outcomes = 20 / 36 = 5 / 9 ≈ 0.556.
Hence, the probability that one man and one woman will be selected is approximately 0.556, or 55.6%.