Answered

Get the answers you need at Westonci.ca, where our expert community is always ready to help with accurate information. Discover a wealth of knowledge from professionals across various disciplines on our user-friendly Q&A platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

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.

Sagot :

Answer:

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

Step-by-step 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:

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]

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:

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

Group of 13 individuals:

This means that [tex]N = 13[/tex]

6 candidates are selected:

This means that [tex]n = 6[/tex]

6 in trained internships:

This means that [tex]k = 6[/tex]

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

This is:

[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]

In which

[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex]

So

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]

[tex]P(X = 0) = h(0,13,6,6) = \frac{C_{6,0}*C_{7,6}}{C_{13,6}} = 0.0041[/tex]

[tex]P(X = 1) = h(1,13,6,6) = \frac{C_{6,1}*C_{7,5}}{C_{13,6}} = 0.0734[/tex]

[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.0041 + 0.0734 = 0.0775[/tex]

[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.0775 = 0.9225[/tex]

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

Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.