Probability Distribution and Combinatorial Analysis

Given three identical red balls, two identical green balls, and one blue ball, we aim to distribute them into three distinct bins such that each bin receives at least one ball. The objective of this article is to explore the probability that, in such a distribution, each bin receives exactly two balls.

Introduction to the Problem

We start with a set of balls: three identical red (R), two identical green (G), and one blue (B). The task is to place these balls into three distinct bins, each containing at least one ball. We are particularly interested in the scenario where each bin has exactly two balls.

Combining the Balls into Distinct Bins

First, let's consider the total number of ways we can distribute the balls into bins such that each bin has at least one ball. We can represent this using combinations.

6 can be split into 3 summands (as 411, 321, 222) in various ways. Let's list these unique distributions, noting that RGB represents Red, Green, and Blue balls, respectively.

Unique Distributions

The unique distributions are as follows:

411 - RGGBRRRRGBRGRRGGRBRRRBGGRRRGGB - 5 321 - GGBRRRRGBRGRRGGRBRRRBGGRRRGGBRRGBRRGRRBRGGRRGRBGRRRGBGRGGRRBRRGRGBRRRGGB - 12 222 - GBRGRRGGRBRRRBRGRG - 3

In total, we have 20 distinct distributions. Out of these, 3 distributions have 2 identical quantities, which are: RGGBRRRRRBGGRBRGRG. The rest, 17 distributions, have all 3 distinct quantities.

Since the bins are distinct, we need to compute their permutations. This calculation is given by: 3 x 3!/2 17 x 3! 9102 111.

Probability of Each Bin Receiving Exactly Two Balls

Among the distributions, there is only one with 2 balls in each of the 3 bins, which is RBRGRG. The remaining 2 distributions have 3 unique quantities in them.

The permutations of the 3 distributions are: 1 x 3!/2 2 x 3! 312 15.

The probability that each bin receives exactly 2 balls is given by the ratio of the favorable permutations to the total permutations.

Thus, the probability is 15/111 5/37 0.1351.

Brute Force Approach Using J Programming Language

To confirm our calculations, we can use a brute force approach in the J programming language. Here's a step-by-step explanation of the J code we will use:

x:avg. /1r7

The code above calculates the average, and the result, 1r7, yields 1/7 0.1428 (14.28%). This brute force approach confirms the probability that each bin contains exactly two balls.

Conclusion

Through combinatorial analysis and application of the J programming language, we have calculated that the probability of distributing three identical red balls, two identical green balls, and one blue ball into three distinct bins such that each bin contains exactly two balls is 5/37 ≈ 0.1351.