Answer:
[tex]3.52 \times 10^{-9} = 3.52 \times 10^{-7}\%[/tex] probability that all the 15 students selected are girls
Step-by-step explanation:
The selection is from a sample without replacement, so 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]
All girls from the first group:
20 students, so [tex]N = 20[/tex]
10 girls, so [tex]k = 10[/tex]
5 students will be selected, so [tex]n = 5[/tex]
We want all of them to be girls, so we find P(X = 5).
[tex]P_1 = P(X = 5) = h(5,20,5,10) = \frac{C_{10,5}*C_{10,5}}{C_{20,5}} = 0.0163[/tex]
All girls from the second group:
20 students, so [tex]N = 20[/tex]
5 girls, so [tex]k = 5[/tex]
5 students will be selected, so [tex]n = 5[/tex]
We want all of them to be girls, so we find P(X = 5).
[tex]P_2 = P(X = 5) = h(5,20,5,5) = \frac{C_{5,5}*C_{15,5}}{C_{20,5}} = 0.00006[/tex]
All girls from the third group:
20 students, so [tex]N = 20[/tex]
8 girls, so [tex]k = 8[/tex]
5 students will be selected, so [tex]n = 5[/tex]
We want all of them to be girls, so we find P(X = 5).
[tex]P_3 = P(X = 5) = h(5,20,5,8) = \frac{C_{8,5}*C_{12,5}}{C_{20,5}} = 0.0036[/tex]
All 15 students are girls:
Groups are independent, so we multiply the probabilities:
[tex]P = P_1*P_2*P_3 = 0.0163*0.00006*0.0036 = 3.52 \times 10^{-9}[/tex]
[tex]3.52 \times 10^{-9} = 3.52 \times 10^{-7}\%[/tex] probability that all the 15 students selected are girls
