The problem describes a scenario in a classroom with 24 students of different nationalities: 3 English, 2 French, 5 Chinese, 4 Korean, and 10 Khmer. We are asked to find the probabilities of three different events when 3 students are randomly selected to represent the class in a competition: A: All 3 students are from Europe (English or French). B: All 3 students are from Asia (Chinese, Korean, or Khmer), but none are Khmer. C: One student is Khmer, one is from Europe, and one is Asian but not Khmer.

1. Problem Description

The problem describes a scenario in a classroom with 24 students of different nationalities: 3 English, 2 French, 5 Chinese, 4 Korean, and 10 Khmer. We are asked to find the probabilities of three different events when 3 students are randomly selected to represent the class in a competition:

A: All 3 students are from Europe (English or French).

B: All 3 students are from Asia (Chinese, Korean, or Khmer), but none are Khmer.

C: One student is Khmer, one is from Europe, and one is Asian but not Khmer.

2. Solution Steps

First, let's determine the total number of ways to choose 3 students from the 24:

$Total = \binom{24}{3} = \frac{24!}{3!21!} = \frac{24 \times 23 \times 22}{3 \times 2 \times 1} = 4 \times 23 \times 22 = 2024$

A: All 3 students are from Europe (English or French). There are 3 English + 2 French = 5 European students.

$n(A) = \binom{5}{3} = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10$

$P(A) = \frac{n(A)}{Total} = \frac{10}{2024} = \frac{5}{1012}$

B: All 3 students are from Asia (Chinese, Korean, or Khmer), but none are Khmer. So, we are choosing from 5 Chinese + 4 Korean = 9 students.

$n(B) = \binom{9}{3} = \frac{9!}{3!6!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84$

$P(B) = \frac{n(B)}{Total} = \frac{84}{2024} = \frac{21}{506}$

C: One student is Khmer, one is from Europe, and one is Asian but not Khmer.

Number of Khmer students = 10

Number of European students = 5

Number of Asian (not Khmer) students = 5 Chinese + 4 Korean = 9

$n(C) = 10 \times 5 \times 9 = 450$

$P(C) = \frac{n(C)}{Total} = \frac{450}{2024} = \frac{225}{1012}$

3. Final Answer

A: $P(A) = \frac{5}{1012}$

B: $P(B) = \frac{21}{506}$

C: $P(C) = \frac{225}{1012}$