Subject: Mathematics and Statistics
Topic: PROBABILITY
Language: English (U.S.)
Pages: 1
Instructions
Suppose that a class contains 15 boys and 30 girls, and that 10 students are to be selected at random for a special assignment. Find the probability that exactly 3 boys will be selected.

Probability

The solution to this problem is a hyper-geometric distribution. If three boys are selected for the special assignment then it means that seven girls will also be selected. This will make sure that the number of students selected for the special assignment is 10. The combination rule will be utilized to solve the equation. The combination rule can be stated as follows: nCr=n! / (n-r)! r!

The first step is identifying the combinations that will be involved in the selection of 10 students from the original 45.

C (45, 10) = 45! / ((45-10)! 10!)) = 3190187286

There are 15 boys in the classroom. This means that there are exactly C (15, 3) ways to select these boys at random.

C (15, 3) = 15! / ((15-3)! 12!)) = 455

The seven girls that will be chosen will be selected from a group of 30 girls in the classroom. We will need to determine the combination required to select seven girls from the group of 30.

C (30, 7) = 30! / ((30-7)! 7!)) =2035800

Based on the Classical Probability theory/ formulae, we will need to multiply the number of outcomes for each event. Therefore, the number of outcomes will be 455* 2035800= 926289000.

The last step is to divide the number of outcomes E by the total number of outcomes i.e. 926289000/ 3190187286= 0.29035568.

The answer to this hyper-geometric distribution is 0.2904.