There are 15 students that are assigned to three classes evenly, meaning each class has 5 students. Among the 15 students, 3 are outstanding students. (1) What is the probability that each class is assigned exactly one outstanding student? (2) What is the probability that all three outstanding students are assigned to the same class?

1. Problem Description

There are 15 students that are assigned to three classes evenly, meaning each class has 5 students. Among the 15 students, 3 are outstanding students.

(1) What is the probability that each class is assigned exactly one outstanding student?

(2) What is the probability that all three outstanding students are assigned to the same class?

2. Solution Steps

(1)

First we need to find the total number of ways to assign the 3 outstanding students to the 15 students. This is the same as choosing 3 spots out of 15 for the outstanding students. The number of ways is given by the combination formula:

$Total = {15 \choose 3} = \frac{15!}{3!(15-3)!} = \frac{15!}{3!12!} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 5 \times 7 \times 13 = 455$

Now, we need to find the number of ways to assign one outstanding student to each class.

Since there are 5 students in each class, the first outstanding student can be assigned to any of the 5 students in the first class, so there are 5 options. Similarly, the second outstanding student can be assigned to any of the 5 students in the second class, so there are 5 options. And the third outstanding student can be assigned to any of the 5 students in the third class, so there are 5 options. Also the order in which we pick the classes matters, so we have 3! ways to assign them. Therefore, the number of favorable outcomes is:

$Favorable = 5 \times 5 \times 5 \times 3! = 5^3 \times 1 = 125$

However this is not correct because it does not account for the other 12 students. Instead we should be focusing on where the outstanding students are placed.

There are 3 classes, each with 5 students. We want to choose one student from each class to be an outstanding student. So we have ${5 \choose 1}$ choices for each class. Since there are 3 classes, we have:

$5 \times 5 \times 5 = 125$ ways.

The total number of ways to choose 3 students from 15 is ${15 \choose 3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455$.

So the probability is:

$P(each \, class \, has \, 1) = \frac{125}{455} = \frac{25}{91}$

(2)

Now, we want to find the probability that all three outstanding students are assigned to the same class.

First, choose a class for them. There are 3 possible classes. Then we need to choose 3 students from the 5 in that class.

The number of ways to choose 3 students from 5 is ${5 \choose 3} = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10$.

So there are $3 \times 10 = 30$ ways for all three students to be in the same class.

The total number of ways to choose 3 students from 15 is ${15 \choose 3} = 455$.

So the probability is:

$P(all \, in \, same \, class) = \frac{30}{455} = \frac{6}{91}$

3. Final Answer

(1) $\frac{25}{91}$

(2) $\frac{6}{91}$