Respuesta :
Answer:
23.63% chance that exactly three is elderly
Step-by-step explanation:
The people are chosen without replacement, which means that the hypergeometric distribution is used to solve this question.
Hypergeometric distribution:
The probability of x successes 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 successes.
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]
In this question:
101 total people means that [tex]N = 101[/tex]
Sample of 6 means that [tex]n = 6[/tex]
66 elderly means that [tex]k = 66[/tex]
If you talked to six random people there, what is the percent chance that exactly three is elderly?
This is P(X = 3). 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 = 3) = h(3,101,6,66) = \frac{C_{66,3}*C_{35,3}}{C_{101,6}} = 0.2363[/tex]
0.2363*100% = 23.63%
23.63% chance that exactly three is elderly
