# conditional probability

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Hi everyone,

I have a math question on conditional probability that I am completely stuck on. Would you mind explaining the step by step process (and why I should do each step) to me?

The question is as follows:

Bag A contains 6 blue counters and 4 green counters. Bag B contains 9 blue counters and 5 green counters. A counter is drawn at random from bag A and two counters are drawn at random from bag B. Tbe counters are not replaced.

Given that there are two blue counters and one green counter, what is the probability that the green counter was drawn from bag B?

I have the answer, which is supposed to be 0.628, but I have no idea how to get to that point.

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The hard part of this problem is hypergeometric distributions, not actually the conditional probability. Although you could just count the events, without necessarily using the distributions per se, but familiarity with them will help in this problem. I actually am fairly confident in my answer, which is 0.652 (15/23). I actually ran a simulation and despite only few thousands samples, the probability still hover around 0.65 and not 0.63

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29 minutes ago, kw0573 said:

The hard part of this problem is hypergeometric distributions, not actually the conditional probability. Although you could just count the events, without necessarily using the distributions per se, but familiarity with them will help in this problem. I actually am fairly confident in my answer, which is 0.652 (15/23). I actually ran a simulation and despite only few thousands samples, the probability still hover around 0.65 and not 0.63

Would you mind explaining how you solved the question (ie. how you got your answer)?

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If it's ok, I'll explain the concepts and hope you can fill out the blanks (the actual calculations just to see if you understand).

1. http://www.probabilityformula.org/hypergeometric-distribution-examples.html Take a look all 4 sample problems; see if you are familiar with hypergeometric distributions. This is the hard part.

2. The actions of picking from Bag A and Bag B are independent events (multiplication).

3. There are 2 mutually exclusive cases (addition). i) blue from bag A; blue and green from bag B. We call this the favorable outcome. ii) green from bag A; blue and blue from bag B. This is essentially the complement. In terms of the conditional probability formula ,  i) is ; both cases combined is . The quickest approach is to not find P(A) but directly the intersection because A and B are neither independent nor mutually exclusive so it gets complicated. Notation is that upright A and B mean the bag, not the formulae above. Color is counter color.

4. Case i) blue in Bag A . Find probability of . Multiply together (see step 2) to get case i) probability.

5. Case ii) green in Bag A is . Find probability of . Multiply to find the complement probability.

6. Apply conditional probability formula (Step 3) to get 15/23.

7. In the future, you actually don't need to calculate the denominator in this of type problems because it cancels out. So you really need to count favorable events and total events only.

Feel free to ask for clarification at any step.

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@kw0573 Thank you so much for your help. I haven't seen hypergeometric distributions before. I understand everything up to the last step when the final value is calculated (the substitution of variables and their logic is clear to me. But taking the first problem, how do you get from P (X =2) = ((10 --2) ( 10 --3)) / (20--5)  = 0.087 ? Because if I use combinations I get a different result of around 0.34.

These are the steps I did:

10 nCr 2 = 45

10 nCr 3 = 120

20 nCr 5 = 15504

45*120 /15504 = 0.348

Would you please tell me where I went wrong? Or did I completely misunderstand how to do hypergeometric distributions?

Info: I did not know how to write the 10 above the 2, and so on for the binomials on this forum, so (10 --2) is supposed to stand for 10 nCr 2.

The rest seems to be clear to me once I understand how to do these hypergeometric distributions.

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I get 0.34 too, I think it was just a big typo on the website's part.

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Thank you so much!

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