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:
![P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20h%28x%2CN%2Cn%2Ck%29%20%3D%20%5Cfrac%7BC_%7Bk%2Cx%7D%2AC_%7BN-k%2Cn-x%7D%7D%7BC_%7BN%2Cn%7D%7D)
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:
is the number of different combinations of x objects from a set of n elements, given by the following formula.
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
Group of 13 individuals:
This means that ![N = 13](https://tex.z-dn.net/?f=N%20%3D%2013)
6 candidates are selected:
This means that ![n = 6](https://tex.z-dn.net/?f=n%20%3D%206)
6 in trained internships:
This means that ![k = 6](https://tex.z-dn.net/?f=k%20%3D%206)
Find the probability that two or more internship trained candidates are hired.
This is:
![P(X \geq 2) = 1 - P(X < 2)](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%202%29%20%3D%201%20-%20P%28X%20%3C%202%29)
In which
![P(X < 2) = P(X = 0) + P(X = 1)](https://tex.z-dn.net/?f=P%28X%20%3C%202%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29)
So
![P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20h%28x%2CN%2Cn%2Ck%29%20%3D%20%5Cfrac%7BC_%7Bk%2Cx%7D%2AC_%7BN-k%2Cn-x%7D%7D%7BC_%7BN%2Cn%7D%7D)
![P(X = 0) = h(0,13,6,6) = \frac{C_{6,0}*C_{7,6}}{C_{13,6}} = 0.0041](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20h%280%2C13%2C6%2C6%29%20%3D%20%5Cfrac%7BC_%7B6%2C0%7D%2AC_%7B7%2C6%7D%7D%7BC_%7B13%2C6%7D%7D%20%3D%200.0041)
![P(X = 1) = h(1,13,6,6) = \frac{C_{6,1}*C_{7,5}}{C_{13,6}} = 0.0734](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20h%281%2C13%2C6%2C6%29%20%3D%20%5Cfrac%7BC_%7B6%2C1%7D%2AC_%7B7%2C5%7D%7D%7BC_%7B13%2C6%7D%7D%20%3D%200.0734)
![P(X < 2) = P(X = 0) + P(X = 1) = 0.0041 + 0.0734 = 0.0775](https://tex.z-dn.net/?f=P%28X%20%3C%202%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29%20%3D%200.0041%20%2B%200.0734%20%3D%200.0775)
![P(X \geq 2) = 1 - P(X < 2) = 1 - 0.0775 = 0.9225](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%202%29%20%3D%201%20-%20P%28X%20%3C%202%29%20%3D%201%20-%200.0775%20%3D%200.9225)
0.9225 = 92.25% probability that two or more internship trained candidates are hired.