# Statistic class forum

Anonymous

### Question Description

Suppose that a class contains 15 boys and 30 girls and that ten students are to be selected at random for a particular assignment. Find the probability that exactly 3 boys will be chosen.

The question above represents a scenario that may require the application of the combination rule. The combination rule involves the situation in which one finds the most appropriate way that they can select r objects from a population of n subjects following no definite order. Through this approach, one can utilize the given information and generate the provided components to help in making a decision.

In this case, the formula to use is nCr=n! / (n-r)! r!

The selection will involve 3 boys and 7 girls.

In that case, (15C3* 30C7)/ 45C10

The combination of 15 and 3 will give;

15C3=15!/ ((15-3)!3! = 455. This case represents the various ways that a researcher can successfully select 3 boys from a population of 15 boys

30C7=30! / ((30-7)!7!)=2035800. This representation offers the various approaches that one can obtain 7 girls from a population containing 30 girls.

45C10=45! / ((45-10)! 10!)= 3190187800. This equation represents the multiple ways that one can select 10 students (both boys and girls) from a population containing 45 students.

In the attempt to obtain 3 boys from a selection containing 10 students, one needs to look for the product of the combination of 7 girls and 3 boys. The researcher may then need to look for the quotient of the above mixture and the combination of total students.

In this case, 455*2035800=926289000

By division we get, 926289000/ 3190187800= 0.29030

Thus the probability of obtaining 3 boys from a selection of 10 students= 0.2903 or 29.03

Follow up question from the instructor:

Did you use a calculator with a nCr button or Excel?

Why do you calculate (15C3* 30C7)/ 45C10 ? Why do you multiply and divide these combinations?

This question has not been answered.

Create a free account to get help with this and any other question!

Brown University

1271 Tutors

California Institute of Technology

2131 Tutors

Carnegie Mellon University

982 Tutors

Columbia University

1256 Tutors

Dartmouth University

2113 Tutors

Emory University

2279 Tutors

Harvard University

599 Tutors

Massachusetts Institute of Technology

2319 Tutors

New York University

1645 Tutors

Notre Dam University

1911 Tutors

Oklahoma University

2122 Tutors

Pennsylvania State University

932 Tutors

Princeton University

1211 Tutors

Stanford University

983 Tutors

University of California

1282 Tutors

Oxford University

123 Tutors

Yale University

2325 Tutors