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Probability Question

  1. Jan 16, 2015 #1
    1. The problem statement, all variables and given/known data
    We have three events A, B and C. Find an expression for the probability that exactly one event occur. Note the answer must be written in terms of probabilities of the individual events, probabilities of simultaneous occurrence of pairs of events and the probability of the simultaneous occurrences of all three events.

    2. Relevant equations


    3. The attempt at a solution

    So
    I be honest, I'm actually very confused about this question. I don't understand how to approach it. I think I need to solve for the probability of event A occurring while not B or C. And same thing for B and C. But I'm not sure how to express this. And expressing the probabilities of simultaneous occurrence of pairs of events and the probability of the simultaneous occurrences of three events don't make much sense to me. Any assistance would be greatly appreciated.
     
  2. jcsd
  3. Jan 16, 2015 #2

    haruspex

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    Let me explain what the question is asking for: an expression for the probability of (A or B or C), in terms of the six probabilities P(A), P(B), P(C), P(A&B), P(B&C), P(C&A), P(A&B&C).
    Start with just A and B. How would you express P(A or B) in terms of P(A), P(B), P(A&B)?
    Drawing a Venn diagram might help.
     
  4. Jan 17, 2015 #3

    Ray Vickson

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    You are asked to find the probability that exactly one of the three evens occur; that is not "A or B or C", but rather "A only, or B only, or C only". As suggested, drawing a Venn diagram is a good first step. An even better first step would be to deal initially with the simpler case of two events A, B and finding the probability that exactly one of the two occur. Can you find P(A only) in terms of P(A), P(B) and P(AB)? (Here I am using the simpler notation AB instead of A##\cap##B.) Can you also find P(B only)? Can you then find P(A only or B only)?

    Doing it for three or more events uses a similar principle, but is more complicated. For three events A,B,C you need to express probabilities like P(A only) in terms of P(A), P(B), P(C), P(AB), P(AC), P(BC) and P(ABC).
     
  5. Jan 17, 2015 #4

    Bystander

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    The problem statement does not include a minimum constraint that at least one of the three occur.
     
  6. Jan 17, 2015 #5
    Ok that makes more sense. Basically, we are solving for P(A ∪ B ∪ C) then. Taking your suggestions I can start by computing these simple probabilities like P(A or B) = P(A υ B) = P(A) + P(B) - P(A ∩ B). Similarly, P(A or C) = P(A) + P(C) - P(A ∩ C) and P(B or C) = P(B) + P(C) - P(B ∩ C).

    And Ray Vickson do you mean find the probability of P(A and not B and not C) ?

    And also can't i just express P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A∩B) - P(A∩C) - P(B∩C) +P(A∩B∩C) which was a theorem we discussed in class.
     
  7. Jan 17, 2015 #6

    haruspex

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    Exactly right. This is an example of the principle of inclusion and exclusion.
     
  8. Jan 17, 2015 #7
    But then isn't that the answer since we are solving for P(A or B or C) right ?
     
  9. Jan 17, 2015 #8

    haruspex

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    Yes, it's the answer. Sorry, I thought that was clear.
     
  10. Jan 17, 2015 #9
    but I find that to be rather strange since our prof presented this theorem in class. And theorem is the answer.
     
  11. Jan 17, 2015 #10

    Ray Vickson

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    (i) Yes to your first question.
    (ii) Yes, you can express P(A ##\cup## B ##\cup##C) as you have done, but that is NOT P(A only, or B only, or C only). That is why I tried to lead you gently to a solution, by asking you if you can express P(A only) in a similar way to what you have done above. Then, do the same for P(B only) and P(C only). The answer you seek is
    P(exactly one of A,B,C) = P(A only) + P(B only) + P(C only),
    because the events on the right are now mutually exclusive. Looking at a Venn diagram should help you a lot.

    It turns out that there is a generalization of the inclusion-exclusion principle that allows for computation of things like
    [tex]P\{\text{exactly}\; r \; \text{of the events} \; A_1, A_2, \ldots, A_n \; \text{occur} \} [/tex]
    in terms of the sums
    [tex] \begin{array}{rcl}S_1 &=& \sum P(A_i), \\S_2 &=& \sum_{i < j} P(A_i A_j), \\S_3 &=& \sum_{i < j < k} P(A_i A_j A_k), \\ & \vdots &\\S_n &=& P(A_1 A_2 \cdots A_n)
    \end{array}[/tex]
    You can find it in the classic textbook W. Feller, "Introduction to Probability Theory and its Applications", Vol. 1, Wiley (1968), for example. Other textbooks, such as those of Sheldon Ross develop the same results in more "modern" (but IMHO more obscure and less enlightening) ways.

    Note added in edit: is your problem really exactly as you wrote it in post #1? There you basically said "exactly one of the events occur", which is very different from "at least one of the events occur". The latter would be "A or B or C", but the former would not be.
     
    Last edited: Jan 17, 2015
  12. Jan 17, 2015 #11
    The question states "...probability that exactly one event occur."

    And also is there an explicit way to expand the intersection of multiple events. i.e.. P(A and not B and not C)-without using venn diagram. Otherwise, by looking at the venn diagram I find P(A ∩ ~B ∩ not C)= P(A) - P(A ∩ B) - P(A ∩ C) + P(A ∩ B ∩ C). From here I can repeat for only B and only C and I'm done.

    Nonetheless I appreciate your help. The problem is much simpler now.
     
    Last edited: Jan 17, 2015
  13. Jan 17, 2015 #12

    haruspex

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    Glad you picked that up Ray. I must have misread it.
     
  14. Jan 30, 2015 #13

    Stephen Tashi

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    A brute force approach: If we have 3 subsets of a "universal set" [itex]S[/itex] then the best description we can make of an element [itex] x[/itex] in [itex] S [/itex] in terms of those 3 sets is to say whether [itex] x[/itex] does or does not belong to each set.

    For example we might have [itex] x \in A \cap B\cap C^c [/itex] where [itex] C^c [/itex] denotes the complement of [itex] C [/itex]. Or we might have [itex] x \in A\cap B^c\cap C [/itex] etc. So can write the set equation [itex] S = (A\cap B \cap C) \cup A\cap B\cap C^c) \cup (A \cap B^c\cap C) \cup (A\cap B^c \cap C^c) ... [/itex] etc. where the intersection terms go through all possible combinations of the sets and their complements.

    This expression is the union of mutually exclusive sets, so:
    [itex] P(S) = 1 [/itex]
    [itex] = P(A\cap B \cap C) + P( A\cap B\cap C^c) + P(A \cap B^c\cap C) + P (A\cap B^c \cap C^c) + ... [/itex]

    You can solve the numerical equation for the term on the right hand side that you want to express as a function of the others.
     
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