Question

In southern California, a growing number of individuals pursuing teaching credentials are choosing paid internships over traditional student teaching programs. A group of eight candidates for three local teaching positions consisted of five who had enrolled in paid internships and three who enrolled in traditional student teaching programs. All eight candidates appear to be equally qualified, so three 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

The required probability = 0.7143

Explanation:

From the information given:

From a group of eight candidates

The no. of candidates that enrolled in internships = 5

The no. of candidates that enrolled in teaching = 3

Also, supposed all the eight candidates are equally qualified;

Then, Let assume that:

Y to represent the number of internship trainee candidates hired.

N to represent no. of candidates in a group = 8

r to represent those who enrolled in paid internship = 5

Now, N - r = 3 (for those who enrolled in traditional teaching program)

Suppose; n represent the positions for local teaching which is given as 3;

Then; selecting 3 from 8 whereby some enrolled in internships and some in traditional teaching programs;

Then, let assume Y is a random variable that follows a hypergeometric distribution; we have:

`p(Y = y) = \left \{ {{( \bigg (^r_y \bigg)\bigg (^(N-r)_(n-y) \bigg) )/( \bigg ( ^N_n \bigg) ) } _\atop { ^(0, otherwise) } } \right.`

`p(Y = y) = \left \{ {{( \bigg (^5_y \bigg)\bigg (^(3)_(3-y) \bigg) )/( \bigg ( ^8_3 \bigg) ) } } } \right, y= 0,1,2,3`

Thus, the probability that two or more internship trained candidates are hired can be computed as:

p(Y ≥ 2) = p(Y=2) + p(Y =3)

`p(Y \geq 2) = ( \bigg ( ^5_2\bigg) \bigg ( ^3_1 \bigg))/(\bigg (^8_3 \bigg)) + (\bigg (^5_3 \bigg) \bigg (^3_0 \bigg))/(\bigg ( ^8_3\bigg))`

`p(Y \geq 2) = (40)/(56)`

`\mathbf{p(Y \geq 2) = 0.7143}`