Probability of Selecting Almost 2 Rotten Eggs: A Detailed Analysis

In this article, we delve into a specific problem where we have to calculate the probability of selecting almost 2 rotten eggs from a set of 12 mixed eggs. This problem involves combinatorial mathematics and probability theory, which are fundamental in various fields, including statistics, data analysis, and data science. We will break down the solution step by step to provide a clear and comprehensive understanding.

Introduction to the Problem

Consider a fictional scenario where there are 12 eggs in a refrigerator, out of which 2 are rotten and 10 are good. We want to make a dessert by randomly selecting 4 eggs. The objective is to determine the probability that the selection includes either 1 or 2 rotten eggs. To achieve this, we will follow a series of steps, utilizing the principles of combinatorial mathematics and probability theory.

Step 1: Calculate the Total Number of Ways to Choose 4 Eggs from 12

The first step is to determine the total number of ways to choose 4 eggs from 12. This can be calculated using the combination formula:

(12)12!4!#x2212;8!12!4!8!#x2248;495

This formula, C(12,4)495, gives the total number of combinations of choosing 4 eggs from 12.

Step 2: Calculate the Number of Ways to Select 1 Rotten and 3 Good Eggs

Next, we calculate the number of ways to select 1 rotten egg from the 2 rotten eggs and 3 good eggs from the 10 good eggs:

C(2,1)2 C(10,3)10!3!7!120

Therefore, the total number of ways to choose 1 rotten and 3 good eggs is:

C(2,1)#x22C5;C(10,3)2#x22C5;120240

Step 3: Calculate the Number of Ways to Select 2 Rotten and 2 Good Eggs

Similarly, we calculate the number of ways to select 2 rotten eggs from the 2 rotten eggs and 2 good eggs from the 10 good eggs:

C(2,2)1 C(10,2)10!2!8!10!2!8!10!2!8!45

Thus, the total number of ways to choose 2 rotten and 2 good eggs is:

C(2,2)#x22C5;C(10,2)1#x22C5;4545

Step 4: Combine the Probabilities

Now, we combine the probabilities of selecting 1 rotten and 3 good eggs and 2 rotten and 2 good eggs to find the total number of favorable outcomes:

Total favorable outcomesC(2,1)#x22C5;C(10,3) C(2,2)#x22C5;C(10,2)240 45285

Step 5: Calculate the Final Probability

The probability of selecting almost 2 rotten eggs (1 or 2) is given by the ratio of the favorable outcomes to the total outcomes:

P(almost2rotten)favorable outcomes285total outcomes495285495

To simplify this fraction, we find the greatest common divisor (GCD) of 285 and 495, which is 15. Dividing both the numerator and the denominator by 15, we get:

P(almost2rotten)285#x0305;15495#x0305;151933

Therefore, the probability that almost 2 eggs are rotten (either 1 or 2 rotten) when selecting 4 eggs from 12 is:

boxed({1933})

Conclusion

This article provides a detailed step-by-step solution to the problem of calculating the probability that almost 2 eggs are rotten when selecting 4 eggs from 12, with 2 rotten and 10 good eggs. By using combinatorial methods and probability theory, we were able to determine that the probability is 19/33. This problem and its solution demonstrate how combinatorial mathematics can be applied in various real-world scenarios, particularly in data analysis and decision-making processes.