3-Color Urn: Expected Remaining Balls After Exhaustion
Hey everyone! Today, we're diving deep into a fascinating probability problem that builds upon the classic urn problem. We're going to explore the expected number of remaining balls after one color is exhausted in a 3-color urn process. This problem touches on several key concepts in probability, including expected value, conditional probability, and conditional expectation. So, buckle up and let's get started!
Understanding the Problem: A 3-Color Urn Scenario
Let's kick things off by clearly defining the problem we're tackling. Imagine we have a bag, or an urn, filled with balls of three different colors: red, blue, and green. To make things concrete, let's say we initially have r red balls, b blue balls, and g green balls. The total number of balls in the urn is then N = r + b + g.
Now, here's the process we're going to follow: We'll randomly draw balls from the urn one at a time, without replacement. This means that once a ball is drawn, it's not put back into the urn. We continue drawing balls until one of the colors is completely exhausted – that is, until we've drawn all the balls of that color.
The central question we want to answer is: What is the expected number of balls remaining in the urn when one color is exhausted? This isn't as straightforward as it might seem at first glance. We need to consider all the possible scenarios and their probabilities to arrive at the expected value.
To really grasp this, let’s think about the different ways one color can be exhausted first. It could be that we draw all the red balls first, leaving some blue and green balls. Or, we might exhaust the blue balls first, leaving red and green. Finally, we could run out of green balls first, leaving red and blue. Each of these scenarios has a certain probability of occurring, and the number of balls remaining will be different in each case. To calculate the expected value, we'll need to carefully weigh these possibilities.
This problem is a classic example of how probability can be used to model real-world situations. Imagine, for instance, that these colored balls represent different types of resources, and we're interested in how long we can use each resource before one runs out. The urn problem provides a framework for analyzing such scenarios and making predictions about the expected outcomes.
Setting Up the Framework: Key Variables and Probabilities
To solve this problem, we need to establish some clear notation and define the key variables we'll be working with. As we mentioned earlier, let r, b, and g represent the initial number of red, blue, and green balls, respectively. Let N = r + b + g be the total number of balls.
Let's define a random variable, R, to represent the number of balls remaining in the urn when one color is exhausted. This is the quantity we're ultimately interested in finding the expected value of, denoted as E[R].
Now, let's introduce some events to help us break down the problem:
- Let A be the event that the red balls are exhausted first.
- Let B be the event that the blue balls are exhausted first.
- Let C be the event that the green balls are exhausted first.
These events are mutually exclusive, meaning that only one of them can occur. They also cover all the possibilities, so the sum of their probabilities must be 1: P(A) + P(B) + P(C) = 1.
To calculate the expected value E[R], we can use the law of total expectation. This law states that the expected value of a random variable can be calculated as the weighted average of its conditional expected values, where the weights are the probabilities of the conditioning events. In our case, this means:
E[R] = E[R | A]P(A) + E[R | B]P(B) + E[R | C]P(C)
This formula is the key to solving our problem. It tells us that the expected number of balls remaining is the sum of three terms: the expected number of balls remaining if red is exhausted first, multiplied by the probability that red is exhausted first; plus the expected number of balls remaining if blue is exhausted first, multiplied by the probability that blue is exhausted first; and so on.
So, our next step is to figure out how to calculate the conditional expected values (E[R | A], E[R | B], and E[R | C]) and the probabilities (P(A), P(B), and P(C)). This is where things get a bit more interesting, and we'll need to employ some clever probabilistic reasoning.
Calculating Probabilities: When Does a Color Get Exhausted First?
Let's start by figuring out how to calculate the probabilities P(A), P(B), and P(C), which represent the chances that red, blue, or green balls are exhausted first, respectively. This might seem tricky, but there's a neat trick we can use that involves thinking about the order in which the balls are drawn.
Consider the event A, where red balls are exhausted first. This means that the last red ball must be drawn before the last blue ball and the last green ball. Imagine we line up all the balls in a random order. The probability that the last red ball appears before the last blue ball and the last green ball is the same as the probability that a randomly chosen ball from the urn is red. Why? Because the position of the last red ball in the line determines whether the red balls are exhausted first.
Therefore, the probability that the red balls are exhausted first is simply:
P(A) = r / (r + b + g) = r / N
Similarly, we can reason that the probability that the blue balls are exhausted first is:
P(B) = b / (r + b + g) = b / N
And the probability that the green balls are exhausted first is:
P(C) = g / (r + b + g) = g / N
These probabilities make intuitive sense. The more balls of a particular color there are initially, the higher the chance that that color will be exhausted first. This is a crucial piece of the puzzle, as it gives us the weights we need for the law of total expectation.
Now that we have the probabilities, let's move on to calculating the conditional expected values. This is where we'll need to think about how many balls of the other colors are likely to be left when a particular color is exhausted.
Conditional Expected Values: How Many Balls Remain?
Next up, we need to figure out the conditional expected values: E[R | A], E[R | B], and E[R | C]. These represent the expected number of balls remaining in the urn, given that red, blue, or green balls are exhausted first, respectively.
Let's focus on E[R | A], the expected number of balls remaining when the red balls are exhausted first. If all the red balls are gone, then the remaining balls must be blue and green. The number of remaining balls in this case is b + g minus the number of blue and green balls that have already been drawn.
To calculate this, we need to think about how many blue and green balls are likely to have been drawn when the last red ball is drawn. Imagine we arrange all the balls in a line again. When the last red ball is drawn, we've drawn r red balls, and some number of blue and green balls. Let's say we've drawn x blue balls and y green balls. Then, the total number of balls drawn is r + x + y.
Since the last ball drawn was red, we know that we haven't drawn all the blue balls or all the green balls yet. So, x < b and y < g. The number of blue balls remaining is b - x, and the number of green balls remaining is g - y. The total number of balls remaining is then: R = (b - x) + (g - y).
The expected number of blue balls drawn when the red balls are exhausted can be calculated using a symmetry argument. Think of it this way: out of the r + b red and blue balls, the last red ball is equally likely to be in any of the positions. Therefore, the expected number of blue balls drawn before the last red ball is: E[x] = b * (r / (r + b)).
Similarly, the expected number of green balls drawn when the red balls are exhausted is: E[y] = g * (r / (r + g)).
Therefore, the expected number of balls remaining when the red balls are exhausted first is:
E[R | A] = E[(b - x) + (g - y)] = b - E[x] + g - E[y]
Substituting the values we found for E[x] and E[y], we get:
E[R | A] = b - (b * (r / (r + b))) + g - (g * (r / (r + g)))
Simplifying this expression, we get:
E[R | A] = (b * b / (r + b)) + (g * g / (r + g))
We can use similar reasoning to find the conditional expected values for when the blue balls are exhausted first and when the green balls are exhausted first:
E[R | B] = (r * r / (b + r)) + (g * g / (b + g))
E[R | C] = (r * r / (g + r)) + (b * b / (g + b))
Now we have all the pieces we need to calculate the overall expected value!
Putting It All Together: Calculating the Expected Number of Remaining Balls
Okay, we've done the heavy lifting! We've calculated the probabilities of each color being exhausted first, and we've calculated the conditional expected values for the number of balls remaining in each of those scenarios. Now, it's time to plug everything into the law of total expectation and get our final answer.
Recall the formula for the expected number of remaining balls:
E[R] = E[R | A]P(A) + E[R | B]P(B) + E[R | C]P(C)
We know that:
- P(A) = r / N
- P(B) = b / N
- P(C) = g / N
- E[R | A] = (b * b / (r + b)) + (g * g / (r + g))
- E[R | B] = (r * r / (b + r)) + (g * g / (b + g))
- E[R | C] = (r * r / (g + r)) + (b * b / (g + b))
Substituting these values into the formula, we get:
E[R] = [(b * b / (r + b)) + (g * g / (r + g))] * (r / N) + [(r * r / (b + r)) + (g * g / (b + g))] * (b / N) + [(r * r / (g + r)) + (b * b / (g + b))] * (g / N)
This looks like a bit of a beast, but it's just a matter of carefully plugging in the numbers. This formula gives us the expected number of balls remaining in the urn after one color is exhausted, based on the initial number of balls of each color.
Conclusion: The Power of Probability in Action
Wow, we've really dug into this problem! We started with a seemingly simple scenario – an urn with colored balls – and used the tools of probability to calculate a non-trivial quantity: the expected number of balls remaining after one color is exhausted.
This problem highlights the power of probabilistic thinking. By breaking down the problem into smaller parts, defining key events and variables, and applying fundamental concepts like the law of total expectation, we were able to arrive at a solution.
Expected value, conditional probability, and conditional expectation are powerful tools in probability theory, and this 3-color urn process problem gives us a fantastic context to explore them. By understanding these concepts, we can tackle a wide range of problems in various fields, from finance to engineering to everyday decision-making.
So, the next time you encounter a situation with uncertainty, remember the urn problem and the power of probabilistic reasoning! You might be surprised at what you can figure out.
