Probability of Selecting Students with Passed Math Test using Hypergeometric Distribution

Imagine we have a class of 25 students, and 15 of them have passed the Math test. Now, if 3 students are selected randomly, what is the probability that exactly 2 of the selected students have passed the test? This question can be answered using probability theory and, specifically, the hypergeometric distribution. Let's break down the problem and explore the solution in detail.

Understanding the Hypergeometric Distribution

The hypergeometric distribution is a discrete probability distribution that describes the probability of k successes in n draws, without replacement, from a finite population of size N that contains exactly K successes. In this case, the population is the class of 25 students, and the number of successes (students who passed the Math test) is 15. We want to find the probability of selecting exactly 2 students who passed out of 3 randomly selected students.

The Basic Approach

Let's start by finding the probability of picking 2 passers and 1 non-passer in a specific order. The steps for this calculation are as follows:

1. **Probability of picking 1 student who had passed:** [ P(text{pass}) frac{15}{25} 0.6 ]2. **Probability of picking another student who had passed from the remaining class:** [ P(text{pass}) frac{14}{24} 0.5833 ]3. **Probability of picking a non-passer from the remaining class:** [ P(text{not pass}) frac{10}{23} 0.4348 ]

The overall probability of this specific order (pass, pass, not pass) is the product of these probabilities:

[ P(text{pass, pass, not pass}) 0.6 times 0.5833 times 0.4348 approx 0.15217 ]

Considering All Possible Orders

Since the order in which we pick the students does not matter, we need to consider all the different ways we can pick 2 passers and 1 non-passer. The number of different orders in which this can happen is given by the combination ( binom{3}{2} times binom{1}{1} ), which is simply 3 (i.e., 15, 14, 10; 15, 10, 14; 10, 15, 14; and so on). The probability of any one of these individual orders is 0.15217. Therefore, the total probability is:

[ P(text{2 passers and 1 non-passer}) binom{3}{2} times 0.15217 3 times 0.15217 approx 0.4565 ]

Using the Hypergeometric Distribution

To confirm our result, we can use the hypergeometric distribution directly. The probability mass function (PMF) of the hypergeometric distribution is given by:

[ P(X k) frac{binom{K}{k} binom{N - K}{n - k}}{binom{N}{n}} ]

Where ( N ) is the total number of students, ( K ) is the number of passers, ( n ) is the number of students we are drawing, and ( k ) is the number of passers we are drawing. In our case:

- ( N 25 )- ( K 15 )- ( n 3 )- ( k 2 )

Plugging these values into the formula, we get:

[ P(X 2) frac{binom{15}{2} binom{10}{1}}{binom{25}{3}} frac{frac{15!}{2!(15-2)!} times frac{10!}{1!(10-1)!}}{frac{25!}{3!(25-3)!}} ]

Using a calculator or Excel function, we can find:

[ P(X 2) text{HYPGEOM.DIST(2, 3, 15, 25, FALSE)} approx 0.4565 ]

This confirms our earlier result, showing that the probability of selecting exactly 2 students who passed the Math test out of 3 randomly selected students is approximately 0.4565.

Conclusion

The hypergeometric distribution provides a powerful tool for solving such probability problems. By considering the specific order and all possible orders, we can accurately calculate the probability of an event occurring under these conditions. Using the Excel function HYPGEOM.DIST, we can verify our calculations and ensure the accuracy of the result.