14. In a statistics class with 15 males and 13 females, five students are selected to put problems on the board. What is the probability that: a. 3 females and 2 males are selected? b. all five students selected are males? c. all five students selected are females? d. at least one male is selected?

a) 0.3056 = 30.56% probability that 3 females and 2 males are selected.

b) 0.0306 = 3.06% probability that all five students selected are males.

c) 0.0131 = 1.31% probability that all five students selected are females.

d) 0.9869 = 98.69% probability that at least one male is selected.

Step-by-step explanation:

The students are chosen from the sample without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

In this question:

15 + 13 = 28 students, which means that

5 are selected, which means that

13 females, which means that

a. 3 females and 2 males are selected?

3 females, so this is P(X = 3).

0.3056 = 30.56% probability that 3 females and 2 males are selected.

b.all five students selected are males?

0 females, so this is P(X = 0).

0.0306 = 3.06% probability that all five students selected are males.

c. all five students selected are females?

This is P(X = 5). So

0.0131 = 1.31% probability that all five students selected are females.

d.at least one male is selected?

Less than five females, so:

0.9869 = 98.69% probability that at least one male is selected.

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