Respuesta :
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.
