Question

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?

Answer 1

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.

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:

`P(X = x) = h(x,N,n,k) = (C_(k,x)*C_(N-k,n-x))/(C_(N,n))`

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
`N = 28`

5 are selected, which means that
`n = 5`

13 females, which means that
`k = 13`

a. 3 females and 2 males are selected?

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

`P(X = x) = h(x,N,n,k) = (C_(k,x)*C_(N-k,n-x))/(C_(N,n))`

`P(X = 3) = h(3,28,5,13) = (C_(13,3)*C_(15,2))/(C_(28,5)) = 0.3056`

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).

`P(X = x) = h(x,N,n,k) = (C_(k,x)*C_(N-k,n-x))/(C_(N,n))`

`P(X = 0) = h(0,28,5,13) = (C_(13,0)*C_(15,5))/(C_(28,5)) = 0.0306`

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

`P(X = x) = h(x,N,n,k) = (C_(k,x)*C_(N-k,n-x))/(C_(N,n))`

`P(X = 5) = h(5,28,5,13) = (C_(13,5)*C_(15,0))/(C_(28,5)) = 0.0131`

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

d.at least one male is selected?

Less than five females, so:

`P(X < 5) = 1 - P(X = 5) = 1 - 0.0131 = 0.9869`

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