A student is in the process of choosing classes for the next semester. They may choose from 11 classes on the advisors list of courses available. Four courses for the major and seven elective courses. Can any class be repeated? Find the probability of choosing 2 major classes and 3 electives. 


To find the probability of choosing 2 major classes and 3 electives from the list, we need to assume that classes cannot be repeated (as is typical in course selection). We will use **combinations** to calculate the number of ways to choose the courses from the available major and elective classes. 

### Steps:

1. **Total number of major courses:** 4  
2. **Total number of elective courses:** 7

We are asked to find the probability of choosing 2 major courses and 3 elective courses. 

#### 1. Number of ways to choose 2 major courses from 4:
The number of ways to choose \(k\) objects from a set of \(n\) objects is given by the combination formula:

\[
\binom{n}{k} = \frac{n!}{k!(n-k)!}
\]

Thus, the number of ways to choose 2 major courses from 4 is:

\[
\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6
\]

#### 2. Number of ways to choose 3 elective courses from 7:

\[
\binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35
\]

#### 3. Total number of ways to choose 2 major and 3 elective courses:
Since the selection of major courses and elective courses are independent events, the total number of ways to make the selection is the product of the combinations:

\[
\text{Total ways} = \binom{4}{2} \times \binom{7}{3} = 6 \times 35 = 210
\]

#### 4. Total number of ways to choose any 5 classes from 11 (major + elective):

The total number of ways to choose 5 classes from the 11 available (without regard to major or elective) is:

\[
\binom{11}{5} = \frac{11!}{5!(11-5)!} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462
\]

#### 5. Probability of choosing 2 major and 3 elective courses:
The probability is the ratio of favorable outcomes to total outcomes:

\[
P(\text{2 major, 3 electives}) = \frac{210}{462} \approx 0.454
\]

Thus, the probability of choosing 2 major courses and 3 electives is approximately **0.454** or **45.4%**.

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Would you like more details on any of the steps or concepts? Here are some related questions to explore further:

1. How do combinations differ from permutations?
2. What is the difference between independent and dependent events in probability?
3. How would the result change if the student could repeat courses?
4. How can the binomial coefficient formula be applied to real-world problems?
5. Can the same probability approach be used for situations with different categories, such as job applications or project assignments?

**Tip:** When solving probability problems, clearly distinguish between independent and dependent events. This helps determine whether to multiply or add probabilities.

A student is in the process of choosing classes for the next semester. They may choose from 11 classes on the advisors list of courses available. Four courses for the major and seven elective courses. Can any class be repeated? Find the probability of choosing 2 major classes and 3 electives.

Calculate the probability of selecting 2 major and 3 elective courses from a set of 11 total classes using combinations. Learn the step-by-step process of applying the combination formula and finding the probability, suitable for students in Grades 10-12.

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