There are 15 students that are randomly and evenly assigned to three classes, which means there are 5 students in each class. Among the 15 students, 3 are outstanding. (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 randomly and evenly assigned to three classes, which means there are 5 students in each class. Among the 15 students, 3 are outstanding.

(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) Probability that each class is assigned exactly one outstanding student.

Total number of ways to assign 3 outstanding students to 15 slots, with 5 slots in each class:

The number of ways to choose 3 students out of 15 is given by the combination formula:

$C(n, k) = \frac{n!}{k!(n-k)!}$

The total number of ways to assign the 3 outstanding students to the 15 students is $C(15, 3) = \frac{15!}{3!12!} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 5 \times 7 \times 13 = 455$.

Number of ways such that each class has one outstanding student:

First outstanding student can be placed in any of the three classes. So, we can assign one outstanding student to a class with 5 slots. This gives us 5 options for the first outstanding student.

Second outstanding student must be assigned to one of the remaining two classes, so we have 5 options.

Third outstanding student must be assigned to the last remaining class, so we have 5 options.

So the number of ways to assign one outstanding student to each class is $5 \times 5 \times 5 = 125$.

The probability is the ratio of the number of favorable outcomes to the total number of possible outcomes.

$P(\text{each class has one outstanding student}) = \frac{\text{Number of ways to assign one student to each class}}{\text{Total number of ways to assign 3 outstanding students}}$

$P = \frac{125}{455} = \frac{25}{91}$.

(2) Probability that all three outstanding students are assigned to the same class.

Number of ways to assign all three students to one class:

We can choose one class out of the three classes, so there are 3 options.

Once a class has been chosen, we need to select 3 students from the 5 students in that class. However, in this problem, all three are the outstanding students and we want them in one class. Hence, we choose the 3 outstanding students to be in any of the 3 classes. This can be done in $3 \times 1 = 3$ ways. The selection of each student can only be done in one way since all three are chosen. The placement can be in any of the 3 classes.

Number of ways to select three students to be in the same class is given as $\binom{5}{3}$.

There are 3 classes to chose from. Then number of ways to pick 3 outstanding students assigned to a specific class is 1, so overall $\binom{5}{3}=1$. Thus, there are 3 ways to assign the 3 outstanding students to a class.

Number of ways to pick a class and select the 3 outstanding students for it = $3 \times 1 = 3$.

$P(\text{all three students in the same class}) = \frac{\text{Number of ways to have all three outstanding students in the same class}}{\text{Total number of ways to assign 3 outstanding students}}$

$P = \frac{3}{455}$.

3. Final Answer

(1) The probability that each class is assigned exactly one outstanding student is $\frac{25}{91}$.

(2) The probability that all three outstanding students are assigned to the same class is $\frac{3}{455}$.