Question

In an anthropology class there are 77 anthropology majors and 1414 non-anthropology majors. 55 students are randomly selected to present a topic. What is the probability that at least 11 of the 55 students selected is a non-anthropology major? Express your answer as a fraction or a decimal number rounded to four decimal places.

Answer 1

Answer:

P = 0.9989

Step-by-step explanation:

In order to do this, I will use the following numbers to make the calculations easier. In this case, We'll say that we have 7 majors and 14 non majors anthopology to present a topic.

This means that in the class we have 21 students.

Now, we choose 5 of them, and we want to know the probability that 1 of them is non major.

First, we need to calculate the number of ways we can select the students in all cases, and then, the probability.

First, we'll use the combination formula, to calculate the number of ways we can select the 5 students out of the 21. We use combination, because it does not matter the order that the students are selected.

C = m! / n!(m - n)!

Where:

m: number of students

n: number of selected students out of m.

With this expression we will calculate first, how many ways we can choose the 5 students out of 21:

C1 = 21! / 5!(21-5)! = 20,349

Now let's calculate the number of ways you can get the all 5 students are non majors:

C2 = 14! / 5!(14 - 5)! = 2002

Now we need to know the number of ways we can get 4 non majors and 1 major:

C3 = C3' * C3''

C3' represents the number of ways we can get 4 non majors and the C3'' represents the number of ways we can get 1 major.

C3' = 14! / 4!(14 - 4)! = 1,001

C3'' = 7! / 1!(7 - 1)! = 7

C3 = 1001 * 7 = 7,007 ways to get 4 non majors and 1 major

Now the way to get 3 non majors and 2 majors, we do the same thing we do to get 4 non majors and 1 major, but changing the numbers. Then the way to get 2 non majors and 3 majors, and finally 1 non major and 4 majors:

3 non majors and 2 majors:

C4 = C4' * C4'' = [14! / 3!(14 - 3)!] * [7! / 2!(7 - 2)!] = 7,644

2 non majors and 3 majors:

C5 = C5' * C5'' = [14! / 2!(14 - 2)!] * [7! / 3!(7 - 3)!] = 3,185

1 non major and 4 majors:

C6 = C6' * C6'' = [14! / 1!(14 - 1)!] * [7! / 4!(7 - 4)!] = 490

Finally to know the probability of getting 1 out of the 5 to be non major, we have to sum all the previous results, and divide them by the ways we can choose the 5 students (C1):

P = 2,002 + 7,007 + 7,644 + 3,185 + 490 / 20,349

P = 0.9989