Two teachers are chosen randomly from a staff consisting 3 women and 2 men to attend a HIV/AIDS seminar. Calculate the probability that the two teachers chosen are:
(a) Of the same sex
(b) Of opposite sex
either 2 men or 2 women.
with no restriction:
number of choices = C(5,2) = 10
number choices of 2 women = C(2,2) = 1
number of choices of 2 men = C(3,2) = 3
prob of 2 of same sex = 4/10 = 2/5
prob of 2 of opposite sex = 1 - 2/5 = 3/5
To calculate the probabilities, we need to know the total number of possible outcomes and the number of favorable outcomes for each case.
(a) Probability that the two teachers chosen are of the same sex:
First, we need to determine the total number of possible outcomes. Since we are choosing 2 teachers from a staff of 3 women and 2 men, there are a total of 5 teachers to choose from. We can calculate this using combinations: C(5, 2) = 5! / (2!(5-2)!) = 10.
Next, we need to determine the number of favorable outcomes, which is the number of ways to choose 2 teachers of the same sex. There are 3 women, so we can choose 2 of them in C(3, 2) = 3! / (2!(3-2)!) = 3 ways. Similarly, there are 2 men, so we can choose 2 of them in C(2, 2) = 2! / (2!(2-2)!) = 1 way.
Therefore, the probability of choosing two teachers of the same sex is (3 + 1) / 10 = 4 / 10 = 2 / 5.
(b) Probability that the two teachers chosen are of opposite sex:
First, we need to determine the total number of possible outcomes, which is still 10 as calculated above.
Next, we need to determine the number of favorable outcomes, which is the number of ways to choose 1 woman and 1 man. There are 3 women, so we can choose 1 of them in C(3, 1) = 3! / (1!(3-1)!) = 3 ways. Similarly, there are 2 men, so we can choose 1 of them in C(2, 1) = 2! / (1!(2-1)!) = 2 ways.
Therefore, the probability of choosing two teachers of opposite sex is (3 * 2) / 10 = 6 / 10 = 3 / 5.