Question

In southern California, a growing number of individuals pursuing teaching credentials are choosing paid internships over traditional student teaching programs. A group of thirteen candidates for six local teaching positions consisted of seven who had enrolled in paid internships and six who enrolled in traditional student teaching programs. All thirteen candidates appear to be equally qualified, so six are randomly selected to fill the open positions. Let Y be the number of internship trained candidates who are hired.

Find the probability that two or more internship trained candidates are hired.

Answer 1

0.9225 = 92.25% probability that two or more internship trained candidates are hired.

Explanation:

Candidates are chosen without replacement, which means that we use the hypergeometric distribution to solve this question.

Hypergeometric distribution:

The probability of x sucesses 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 sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

Group of 13 individuals:

This means that
`N = 13`

6 candidates are selected:

This means that
`n = 6`

6 in trained internships:

This means that
`k = 6`

Find the probability that two or more internship trained candidates are hired.

This is:

`P(X \geq 2) = 1 - P(X < 2)`

In which

`P(X < 2) = P(X = 0) + P(X = 1)`

So

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

`P(X = 0) = h(0,13,6,6) = (C_(6,0)*C_(7,6))/(C_(13,6)) = 0.0041`

`P(X = 1) = h(1,13,6,6) = (C_(6,1)*C_(7,5))/(C_(13,6)) = 0.0734`

`P(X < 2) = P(X = 0) + P(X = 1) = 0.0041 + 0.0734 = 0.0775`

`P(X \geq 2) = 1 - P(X < 2) = 1 - 0.0775 = 0.9225`

0.9225 = 92.25% probability that two or more internship trained candidates are hired.