Math Problem Statement

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. 

Solution

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 kk objects from a set of nn objects is given by the combination formula:

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

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

(42)=4!2!(42)!=4×32×1=6\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:

(73)=7!3!(73)!=7×6×53×2×1=35\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:

Total ways=(42)×(73)=6×35=210\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:

(115)=11!5!(115)!=11×10×9×8×75×4×3×2×1=462\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(2 major, 3 electives)=2104620.454P(\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%.


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.

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Math Problem Analysis

Mathematical Concepts

Combinatorics
Probability Theory

Formulas

Combination formula: C(n, k) = n! / (k!(n-k)!)
Probability formula: P = Favorable outcomes / Total outcomes

Theorems

Basic Principle of Counting

Suitable Grade Level

Grades 10-12