Probability Urn Problem with Replacement Kinda

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In summary: Continuing this way, what is the probability the special red ball is not drawn on the 2n th draw?We can multiply these probabilities together to get the probability it is not drawn in any of the remaining draws.So the probability it is drawn last is 1 minus this probability.In summary, the probability that the last ball removed from urn 1 is red is 1 minus the probability that the designated red ball is not drawn in any of the remaining draws, which can be calculated by multiplying the probabilities that it is not drawn on each individual draw. This method does not require Bayes' Rule and can be used for any designated red ball.
  • #1
RaeganW
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This probabilily summer course I am in is so bizarre; I have no idea where the professor is going during the lecture. Luckily I have awesome notes from another Prob & Stats course, but this problem has me stumped.

Homework Statement



Consider two urns. The fi rst one has N red balls and the second one has N blue balls. Balls are removed randomly from urn 1 in the following manner: after each removal from urn 1, a ball is taken from urn 2 (if urn 2 still has balls) and placed in urn 1. The process continues until all balls have been removed. What is the probability that the last ball removed from urn 1 is red?

Hint: consider first one particular red ball and compute the probability that it is the last one to be removed.

Homework Equations



P(Bi|A) = [P(A|Bi)*P(Bi)] / Ʃ 1->n P(A|Bi)*P(Bi)

Maybe?

The Attempt at a Solution



I figured out there will be 2n trials. During the first n trials there are n balls in urn 1, in the n+1th trial there are n-1 balls in urn 1, n+2th trial there are n-2 balls in urn 1, etc so the 2nth trial will draw the last ball out of urn 1.

In my other course, we didn't really get into Bayesian probability. We were shown how to make a tree, but we've never gone over that in this class so I don't know if the professor will accept that kind of answer. But, for a specific red ball to be drawn last:

mat431-hw1-q4.png


Hope that makes sense... branches to the left are the probability the specific red ball is chosen, branches to the right is the probability the specific red ball is not chosen.

For the first n trials, the probability that the specific red ball is not chosen is ((n-1)/n)^n.
For the second n trials, the probability that the specific red ball is not chosen is ∏ 1-> (n-1) 1-(1/(n-i)) which I know isn't in the image but I just figured out that's a better way of writing it.
The 2nth trial the probability of choosing the specific red ball is 1, it's the last ball.

But that's only for a particular red ball, not any red ball. No idea how to scale this up to that, and I'm pretty sure there's a more succinct way of doing this... one that doesn't involve drawing so many pictures...
 
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  • #2
I think you are close to the answer, and I don't think you need Baye's Rule.

You have already figured out that the probability the designated red ball is not chosen in the first n draws is [itex]\left( \frac{n-1}{n} \right) ^n[/itex].

So assume the ball has not been drawn yet, after n draws. There are still n balls in urn 1.
What is the probability it is not drawn on the n+1 th draw?

Now there are n-1 balls left in urn 1.
What is the probability the special red ball is not drawn on the n+2 th draw?
 

Related to Probability Urn Problem with Replacement Kinda

1. What is the Probability Urn Problem with Replacement Kinda?

The Probability Urn Problem with Replacement Kinda is a mathematical problem that involves selecting objects from an urn with replacement, where each object has a certain probability of being selected. The "Kinda" refers to the twist that the problem adds, where the probabilities of the objects may change after each selection.

2. How does the Probability Urn Problem with Replacement Kinda work?

The problem starts with an urn containing a certain number of objects with different probabilities of being selected. An object is randomly chosen from the urn and then placed back in the urn, along with a new object with a different probability. This process is repeated multiple times, and the goal is to determine the probability of selecting a specific object after a certain number of selections.

3. What is the difference between the Probability Urn Problem with Replacement and the Probability Urn Problem with Replacement Kinda?

The main difference between the two problems is the addition of the "Kinda" aspect. In the regular Probability Urn Problem with Replacement, the probabilities of the objects remain constant throughout the process. In the Kinda version, the probabilities may change after each selection, making the problem more complex and challenging to solve.

4. What are some real-life applications of the Probability Urn Problem with Replacement Kinda?

The Probability Urn Problem with Replacement Kinda has various applications in different fields, such as genetics, finance, and gambling. In genetics, it can be used to determine the probability of inheriting certain traits from parents. In finance, it can be used to model stock prices and predict future values. In gambling, it can be used to calculate the odds of winning a game or lottery.

5. How can I solve the Probability Urn Problem with Replacement Kinda?

The Probability Urn Problem with Replacement Kinda can be solved using mathematical equations and formulas, such as the Law of Total Probability and Bayes' Theorem. It is also possible to solve it using simulations or computer programs. Understanding the problem and its underlying concepts is crucial in finding the correct solution.

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