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Looks like a simple probability calculation but am I getting it right?

  1. Aug 14, 2009 #1
    All of a sudden, I seem to be confused about the union of probabilities. I wanted to see how this would work out but ended up confusing myself even more. Here's a scenario I'm considering:

    I have three friends. I know that if a friend posts me a letter upon being asked, it reaches me with a probability of p. I can contact my first friend A, directly but I would have to contact C through B. I wanted to calculate the probability of getting a letter from one of A or C. So, in the end, I have to get one letter atmost. To do this, I ended up saying the following:

    P(getting at most one letter)
    = P(I receive a letter from A or I receive a letter from C)
    = P(I receive a letter from A) + P(I receive a letter from C)
    = P(choosing A)*P(letter from A reaching me) + P(choosing C)*P(letter from C reaching me)
    = [tex]\frac{1}{2}[/tex]p + P(choosing C)*P(B receives the letter from C)*P(I receive the letter from B)
    = [tex]\frac{1}{2}[/tex]p + [tex]\frac{1}{2}[/tex]p*p
    = [tex]\frac{1}{2}[/tex]p + [tex]\frac{1}{2}[/tex]p2

    I felt something was wrong so I ended up subtracting the P(getting a letter from both A and C) to make the solution as [tex]\frac{1}{2}[/tex]p + [tex]\frac{1}{2}[/tex]p2 - [tex]\frac{1}{4}[/tex]p3 from the axiom of probability of unions.

    I've complicated the problem so much that I'm now confused... I am making a fundamental mistake in understanding the problem and hope someone can help me out in understanding it right...

    PS: I made up this problem myself to try out something interesting so I might have been wrong in framing the question itself... If thats the case, please advice...
    Last edited: Aug 14, 2009
  2. jcsd
  3. Aug 16, 2009 #2
    You are trying to find the probability that you receive a letter from exactly one person. I am assuming here that you are contacting both A and B (with the message to be delivered to C). If this is correct, then this is not a conditional probability question since you are not selecting who might reply (they are). Your assumption about the 50-50 split between A and B is off.

    The probability that you will receive a reply from A is p; the probability that you'll receive a reply from C (as I understand your question) is p2. Since receiving replies from A or C are independent (your question does not give any hint they shouldn't be) then we have:

    [tex]P(AC)=P(A) \cdot P(C)[/tex]

    (Here [itex] AC = A \cap C[/itex].)

    As always [itex] P(XY') = P(X) - P(XY)[/itex]

    You need

    [tex]P(AC' \cup A'C)= \left[ P(A)-P(AC) \right] + \left[ P(C) - P(AC) \right][/tex]


    [tex]P(AC' \cup A'C) = (p - p \cdot p^2) + (p^2 - p \cdot p^2)[/tex]

    [tex]= p + p^2 - 2p^3.[/tex]

  4. Aug 17, 2009 #3
    Thanks so much... Because the formulation involves union of probabilities, am I right in assuming that if the number of persons increase, then the calculation become more and more complex?

    To make it clear, if I add a fourth person with a probability p again, I would end up calculating:

    [tex]P(AB'C' \cup A'BC' \cup A'B'C) = 3*P(ABC) + P(A) + P(B) + P(C) - 2(P(AB) + P(BC) + P(CA))[/tex]

    in which case, the calculation turns to be crazy. Do you think I'm on the right track? (If I am, is there a general expansion that can be used to do this?)
    Last edited: Aug 17, 2009
  5. Aug 17, 2009 #4
    In the case where the probabilities of each A, B, C, etc are the same, then maybe. Otherwise, no. There are always n combinations that include exactly one success, but calculating the probabilities becomes an exercise in tedium using the inclusion/exclusion principle.


    EDIT: Sorry, forgot to mention your calculation is correct.
    Last edited: Aug 17, 2009
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