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Basic Probability question

  1. Nov 24, 2009 #1
    I have never been very good at probability, and I am confused with this rather simple statement:

    Let E be any event, and F and G be events such that one and only one of the events F and G will occur. Then

    P(E) = P(F)*P(E given F) + P(G)*P(E given G)"

    Where P(E) = the probability of E occurring. And the same for the others.

    To be honest I don't even understand what it is asking me to do procedurally. What does it mean "E given F"? Is that the probability of E occurring if F occurs? Why is that pertinent? More than that I don't understand the reasoning nor do I have any intuitive inkling as to why this expression would yield the correct answer.
    Can someone give an example perhaps?
    (The book I was given just assumes the reader automatically understands this property).

    Any help would be greatly appreciated.
  2. jcsd
  3. Nov 24, 2009 #2


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    Science Advisor

    The best way to understand it is by using a Venn diagram. The descriptions of F and G are such that they don't overlap and together fill the entire event space. Place E on the diagram and you see that part of it may overlap F while the rest would overlap G.

    To add things up properly, P(E given F) means the area of the part of E overlapping F divided by the area of F. Similarly for P(E given G). To get the E area, multiply each piece by the area of F or G as needed.
  4. Nov 25, 2009 #3
    Okay. I still have a couple questions.
    If P(F) and P(G) fill the "entire event space" does that mean P(F) + P(G) always = 1?
    And is that just because one and only one of the events must occur?
    That seems to make sense.
    So E is dependent on F and G, and this description is finding the probability of E as it depends on the outcomes of F or G?
    I hope I'm not misunderstanding because it seems to fit now.

    On a side note, do you know if there is a name for this sort of thing so that I can look into it more, or should I just look into general probability basics?
    Thanks :)
  5. Nov 25, 2009 #4

    P(F v G)=1; P(F^G)= 0; P(F) = 1 - P(G); P(G)= 1 - P(F).

    These types of problems are about the most basic probability examples, such as coin tosses with fair or biased coins. I guess you could call them strict dichotomies.
    Last edited: Nov 25, 2009
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