Conditional probability with urns

In summary, the probability of selecting a white ball from urn A is 2/3 if exactly two white balls were drawn.
  • #1
CAF123
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Homework Statement


Consider 3 urns. Urn A contains 2 white and 4 red balls, Urn B contains 8 white and 4 red balls and Urn C contains 1 white and 3 red balls. If 1 ball is selected from each urn, what is the probability that the ball chosen from urn A was white given that exactly 2 white balls were selected?

The Attempt at a Solution


Let E be the event that a white ball was chosen from urn A
Let F be the event that exactly 2 white balls were selected.

Given that the same colour of balls are indistinguishable , we have |s| = (2 choose 1) x (2 choose 1) x (2 choose 1) = 8 possibilities
Therefore F = {(WRW),(WWR),(RWW)} and E = {(WWW),(WRW),(WWR),(WRR)}
So, [tex] P(E|F) = \frac{P(EF)}{P(F)} = \frac{\frac{2}{8}}{\frac{3}{8}} = 2/3 [/tex]
The given answer is 7/11, which is slightly below this - where is the error?
Many thanks.
 
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  • #2
Your basic concept is correct but you appear to have done the arithmetic incorrectly- at least your "P(EF)= 2/9" and "P(F)= 3/8" are wrong but you don't show how you got those.

Yes, P(F), the probability that exactly two white balls are drawn can be calculated by looking at "WRW", "WWR", and "RWW". Since the first urn contains two white balls and four red, the probability that the first ball drawn is white is 2/6= 1/3 and the probability that it is red is 4/6= 2/3. The second urn contains eight white and four red balls so that the probability the second ball drawn is white is 8/12= 2/3 and the probability it is red is 4/12= 1/3. The third urn contains one white and three red balls so the probability the third ball drawn is white is 1/4 and the probability it is red is 3/4.

So the probability of "WRW" is (1/3)(1/3)(1/4)= 1/36. The probability of "WWR" is (1/3)(2/3)(3/4)= 1/6. The probability of "RWW" is (2/3)(2/3)(1/4)= 1/9. The probability of "two white", in any order, is P(F)= 1/36+ 1/6+ 1/9= 1/36+ 6/36+ 4/36= 11/36, not "3/8".

The event "the first ball drawn is white and two white balls are drawn" just drops "RWW" from the possible cases so the probability of that is 1/36+ 1/6= 1/36+ 6/36= 7/36.
 
  • #3
I said EF = {(WRW),(WWR)} and the probability of attaining this out of the sample space was 2/8. Similarly, F = {(WRW),(WWR),(RWW)} which gives P(F) = 3/8. (3 elements out of sample space of 8 elements)
This method does not take into account the numbers of red and white balls in ech urn, but at the same time, I don't see anything wrong with what I have done.
 

Related to Conditional probability with urns

1. What is conditional probability with urns?

Conditional probability with urns is a mathematical concept used to calculate the likelihood of an event occurring, given that another event has already happened. In the context of urns, it refers to the probability of drawing a specific colored ball, given the information about the number and color of balls already in the urn.

2. How do you calculate conditional probability with urns?

To calculate conditional probability with urns, you use the formula P(A|B) = P(A and B) / P(B). This means that you take the probability of both events A and B occurring together and divide it by the probability of event B occurring. This will give you the likelihood of event A happening, given that event B has already occurred.

3. What is an example of conditional probability with urns?

One example of conditional probability with urns is the classic "Monty Hall" problem. In this scenario, there are three doors, one of which has a car behind it and the other two have goats. If you pick a door at random, the probability of choosing the car is 1/3. However, if the host opens one of the other doors to reveal a goat, the probability of your chosen door containing the car increases to 1/2. This is an example of conditional probability, as the likelihood of choosing the car is dependent on the information about the other doors.

4. What is the difference between marginal probability and conditional probability?

Marginal probability refers to the likelihood of an event occurring without taking into consideration any other information. On the other hand, conditional probability takes into account additional information or conditions that affect the likelihood of the event. In the case of urns, marginal probability would be the probability of drawing a specific colored ball without any prior knowledge, while conditional probability would involve taking into account the color and number of balls already in the urn.

5. Why is conditional probability important in scientific research?

Conditional probability is important in scientific research because it allows us to make more accurate predictions and decisions based on the available information. By taking into account the relationship between different events, we can better understand the underlying factors and make more informed conclusions. In fields such as genetics, epidemiology, and psychology, conditional probability is used to analyze data and make meaningful inferences about complex systems.

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