# Urn Conditional Probability Problem

• Rosalie
In summary, the conversation discusses the problem of finding the probability of drawing a yellow ball as the third ball, given that the first ball drawn is green. The total number of balls is 45, with 20 green balls, 12 red balls, and 13 yellow balls. The conversation mentions using Bayes' formula and constructing a probability tree diagram to solve the problem. The final answer is determined to be 13/99, with the explanation that the second and third balls have the same probability distribution.
Rosalie
The problem is stated as follows:

You have an urn with 12 red balls, 20 green balls, and 13 yellow balls. Suppose 3 balls are drawn without replacement. What is the probability the third ball is yellow given the first ball is green.

I am pretty sure I use Bayes formula, but I am not certain how to apply it in this case. I am really just baffled! Can anyone help me?

Last edited:
First step is to work out the probability of each combination of 3 balls. Might be easiest with the aid of a probability tree diagram.

Im not certain I understand how to set up a probability tree for a problem like this. I am going to try to work it through. Give me a moment, I'll let you double check my results.

Ok! I think I got it. This is what I concluded.
Total Balls=45
Green(G)=20
Red(R)=12
Yellow(Y)

P(G,R,Y)=(20/45) X (12/44) X (13/43)=(52/1419)
P(G,G,Y)=(20/45) X (19/44) X (13/43)=(247/4257)
P(G,Y,Y)=(20/45) X (13/44) X (12/43)=(52/1419)
(52/1419) + (247/4257) + (52/1419) = (13/99)

If this is incorrect please let me know! It is very important I get this question correct. Thank you for all your help.

Rosalie said:
P(G,R,Y)=(20/45) X (12/44) X (13/43)=(52/1419)
P(G,G,Y)=(20/45) X (19/44) X (13/43)=(247/4257)
P(G,Y,Y)=(20/45) X (13/44) X (12/43)=(52/1419)
(52/1419) + (247/4257) + (52/1419) = (13/99)

You're half-way there - what you've worked out is the probability that 3rd is yellow AND 1st is green. Next step uses Bayes formula.

Rosalie said:
Ok! I think I got it. This is what I concluded.
Total Balls=45
Green(G)=20
Red(R)=12
Yellow(Y)

P(G,R,Y)=(20/45) X (12/44) X (13/43)=(52/1419)
P(G,G,Y)=(20/45) X (19/44) X (13/43)=(247/4257)
P(G,Y,Y)=(20/45) X (13/44) X (12/43)=(52/1419)
(52/1419) + (247/4257) + (52/1419) = (13/99)

If this is incorrect please let me know! It is very important I get this question correct. Thank you for all your help.

I can't resist. Did you notice that the value for P(G,Y) is also 13/99?

This is no accident. As long as you have no information about the second ball, the third ball and the second ball have exactly the same probability distributions. An easy way to get this first step would be to observe this equivalence and then just calculate P(G,Y) = (20/45)*(13/44)

It all makes no difference to the next step, where you apply Bayes theorem to get the required conditional probability.

Cheers -- sylas

## 1. What is the "Urn Conditional Probability Problem"?

The Urn Conditional Probability Problem is a famous probability problem that involves randomly selecting marbles from a bag or urn, with or without replacement. It is used to demonstrate and test understanding of conditional probability, which is the likelihood of an event occurring given that another event has already occurred.

## 2. How does the problem work?

The problem usually involves a bag or urn containing a certain number of marbles of different colors. The question asks for the probability of selecting a certain color marble, given certain conditions, such as whether marbles are replaced or not after each selection. The problem can also involve multiple selections and different scenarios.

## 3. Why is the Urn Conditional Probability Problem important?

The problem is important because it helps to illustrate the concept of conditional probability, which is widely used in statistics and real-world applications. It also allows for the testing and development of critical thinking and problem-solving skills.

## 4. What are some strategies for solving the problem?

One strategy for solving the Urn Conditional Probability Problem is to use a tree diagram to visually represent the different scenarios and outcomes. Another strategy is to use the formula for conditional probability, which is P(A|B) = P(A ∩ B)/P(B), where A and B are events and P(A ∩ B) is the probability of both events occurring.

## 5. Are there any real-life examples of the Urn Conditional Probability Problem?

Yes, the Urn Conditional Probability Problem can be applied to many real-life situations, such as predicting the likelihood of a certain disease given a positive test result, or the likelihood of winning a game with different scenarios and conditions. It can also be used in market research, weather forecasting, and other fields that involve predicting outcomes based on given conditions.

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