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Conditional Probability

  1. Feb 11, 2007 #1
    Given: P(A)= .4, P(B)=.3, P(A n B)=.11, P(C| not A)=.5

    If P(C U A) = .66, then find P[(C U A) | (C n A)].

    I have been trying to manipulate this thing for a while now with no luck.
    Could you try and show the work if not thats alright, I'll work it out.
    Thanks.
     
    Last edited: Feb 11, 2007
  2. jcsd
  3. Feb 11, 2007 #2

    cristo

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    Well, there are a few ways to do it. What is the equation for conditional probabilty P(F|G), say?

    Alternatively, what are the definitions of C u A and C n A? You want to find the probability that C u A occurs given that C n A has occured. Or, you want to find the probabilty that x is in the union of C and A given that it is in the intersection of C and A.

    (This seems a bit easy; are you sure you've written the question correctly?)
     
    Last edited: Feb 11, 2007
  4. Feb 11, 2007 #3
    That would be, P(F|G) = P(F n G)/P(G)

    This is the correct question. It is just throwing me off that it is the conditional probability of the union and intersection of two events instead of just the conditional probability of two events.

    I've gotten as far as: P[(C U A) n (C n A)] / P(C n A).

    Is that correct.
     
  5. Feb 11, 2007 #4

    cristo

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    Yes. Now, can you simplify (C u A)n(C n A)? If you can't straight away, try drawing a venn diagram.

    Remember, if x is in (C u A)n(C n A), then x is in C u A and C n A.
     
  6. Feb 11, 2007 #5
    P[(C U A) n (C n A)] = P(C n A)?
    (C n A) is a subset of (C U A)?

    So then the equation would be P(C n A) / P(C n A)?

    Is that why you said it seems too easy or is this wrong?
     
  7. Feb 11, 2007 #6

    cristo

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    Yes, that's correct, which is why I thought you made a mistake writing the question up.

    Intuitively, P[(C U A) | (C n A)] is the probability that C happens, or A happens, or they both happen, given that C and A have both happened, and so this must equal 1!
     
  8. Feb 11, 2007 #7
    Does it make a difference though that there are three events: A, B, and C. When drawing the Venn Diagram I also have to take event B into consideration also right, or that still does not make a difference.
     
  9. Feb 11, 2007 #8

    cristo

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    It doesn't make a difference. We are only considering A and C in this case. If you've drawn a Venn diagram with the 3 sets, then that's fine, but to tell whether x is in the intersection of A and C, we do not need to know whether it is in B or not.
     
  10. Feb 11, 2007 #9
    You are much appreciated. While I have you here and if your up for it, I doubt this is challenging for you but could you check this work.

    There are 6 males and 4 females awaiting to see a teller at a bank.
    It is the end of the day and there is only one teller, so only 4 of the people can be served one-at-a-time.

    1) How many ways can four of the people be picked and served one at a time, if they must include two(2) men and two(2) women?

    Solution: (6c2) * (4c2) = 90


    2) If indeed the four people are picked randomly, what is the probability that the four will include two (2) men and two (2) women?

    Solution: (6c2) * (4c2) / (10c4) = 3/7 = .428571
     
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