Probability of Selecting Seniors and Juniors for a Conference: A Comprehensive Guide
When dealing with probability problems, particularly in the context of a student council committee, understanding the combination method is crucial for accurately calculating the likelihood of specific outcomes. This article explores a detailed solution to a common problem: selecting 2 seniors and 3 juniors from a pool of 8 seniors, 10 juniors, and 6 sophomores. We will delve into the mathematical steps involved and present the solution with clarity and precision.
Understanding the Problem
The problem statement clearly defines the scenario: a student council committee consists of 8 seniors, 10 juniors, and 6 sophomores. If 5 members are chosen at random to attend a conference, we are interested in the probability that exactly 2 seniors and 3 juniors are selected.
Mathematical Approach: Combinations
To solve this problem, we can use the combination method, which is a fundamental concept in probability and combinatorics. The combination formula is given by (C(n, k) frac{n!}{k!(n-k)!}), where (n) is the total number of items, (k) is the number of items to choose, and (n!) denotes factorial.
Step 1: Calculate the Number of Ways to Choose 2 Seniors
First, we need to find the number of ways to select 2 seniors from 8 seniors. Using the combination formula:
[C(8, 2) frac{8!}{2!(8-2)!} frac{8!}{2!6!} frac{8 times 7}{2 times 1} 28]Step 2: Calculate the Number of Ways to Choose 3 Juniors
Next, we calculate the number of ways to select 3 juniors from 10 juniors:
[C(10, 3) frac{10!}{3!(10-3)!} frac{10!}{3!7!} frac{10 times 9 times 8}{3 times 2 times 1} 120]Step 3: Calculate the Total Number of Ways to Choose 5 Members
Now, we determine the total number of ways to select 5 members from the 24 members of the committee. This calculation is as follows:
[C(24, 5) frac{24!}{5!(24-5)!} frac{24!}{5!19!} 42504]Step 4: Calculate the Probability
Finally, we calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes:
[text{Probability} frac{C(8, 2) times C(10, 3)}{C(24, 5)} frac{28 times 120}{42504} frac{3360}{42504} 0.0790 approx 7.90%]Conclusion
The probability that exactly 2 seniors and 3 juniors are chosen to attend the conference is approximately 7.90%. This solution provides a clear and step-by-step approach to solving such problems, emphasizing the importance of using the combination method and understanding the underlying mathematical principles.