Respuesta :
Answer:
a) 0.3056 = 30.56% probability that 3 females and 2 males are selected.
b) 0.0306 = 3.06% probability that all five students selected are males.
c) 0.0131 = 1.31% probability that all five students selected are females.
d) 0.9869 = 98.69% probability that at least one male is selected.
Step-by-step explanation:
The students are chosen from the sample 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.
In this question:
15 + 13 = 28 students, which means that [tex]N = 28[/tex]
5 are selected, which means that [tex]n = 5[/tex]
13 females, which means that [tex]k = 13[/tex]
a. 3 females and 2 males are selected?
3 females, so this is P(X = 3).
[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(3,28,5,13) = \frac{C_{13,3}*C_{15,2}}{C_{28,5}} = 0.3056[/tex]
0.3056 = 30.56% probability that 3 females and 2 males are selected.
b.all five students selected are males?
0 females, so this is P(X = 0).
[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(0,28,5,13) = \frac{C_{13,0}*C_{15,5}}{C_{28,5}} = 0.0306[/tex]
0.0306 = 3.06% probability that all five students selected are males.
c. all five students selected are females?
This is P(X = 5). 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 = 5) = h(5,28,5,13) = \frac{C_{13,5}*C_{15,0}}{C_{28,5}} = 0.0131[/tex]
0.0131 = 1.31% probability that all five students selected are females.
d.at least one male is selected?
Less than five females, so:
[tex]P(X < 5) = 1 - P(X = 5) = 1 - 0.0131 = 0.9869[/tex]
0.9869 = 98.69% probability that at least one male is selected.
