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Probability problem involving 4 Unions (inclusive OR)

  1. Jan 27, 2013 #1
    1. The problem statement, all variables and given/known data
    Given P(AUBUCUD), expand

    3. The attempt at a solution
    I approached the solution by procedurally drawing Vann Diagrams, building from AUB to AUBUCUD (Not included); Please do check if my line of reasoning is sound.
    (let "n" be the intersection, AND)

    1) P(AUB): P(A) + P(B) - P(AnB) [subtract the extra section, this is repeated at the end of subsequent steps]
    2) P((AUB)UC): P(AUB) + P(C) - P((AUB)nC)
    3) P((AUBUC)UD): P((AUB)UC) + P(D) - P((AUBUC)nD)

    Substituting backwards, hence

    A(AUBUCUD)= P(AUB) + P(C) - P((AUB)nC) + P(D) - P((AUBUC)nD)
    A(AUBUCUD)= P(A) + P(B) - P(AnB) + P(C) - P((AUB)nC) + P(D) - P((AUBUC)nD) (Solved)

    Also, do inform me if there is any more efficient way to put this. Thank you for your time.
     
  2. jcsd
  3. Jan 27, 2013 #2

    Ray Vickson

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    You have the right idea (and yes, it IS tedious!). However, I believe the question wants you to expand further, so it wants you expand ##P((A \cup B)\cap C)## and ##P((A \cup B \cup C)\cap D)##. Can you see how to do that?
     
  4. Jan 27, 2013 #3
    Hello Vickson,

    Thank you for your reminder; Using P(AnB)= P(A)*P(B) I believe that they can be solved as below? (some expressions were substituted backwards)

    1. P(AnB)= P(A)P(B) [solved]

    2. P((AUB)nC) = [P(A)+P(B)-P(AnB)]*P(C)
    = [P(A) + P(B) - P(A)P(B)]*P(C)
    = P(A)P(C) + P(B)P(C) - P(A)P(B)P(C) [solved]

    3. P((AUBUB)nD)= [P(AUB)+P(C)-P((AUB)nC)]*P(D)
    = {[P(A) + P(B) - P(AnB)] + P(C) - [P(A)P(C) + P(B)P(C) - P(A)P(B)P(C)]}*P(D)
    = [P(A) + P(B) - P(A)P(B) + P(C) - P(A)P(C) - P(B)P(C) + P(A)P(B)P(C)]*P(D)
    = P(A)P(D) + P(B)P(D) - P(A)P(B)P(D) + P(C)P(D) - P(A)P(C)P(D) - P(B)P(C)P(D) + P(A)P(B)P(C)P(D) [solved]

    What a jargon. Thank you for your time.
     
    Last edited: Jan 27, 2013
  5. Jan 27, 2013 #4

    Ray Vickson

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    No: what you wrote in 1) is true only if the events A and B are independent. Were you told that they are? If not, then what you wrote originally (in your first post) is correct: ##P(A \cup B) = P(A) + P(B) - P(AB), ## where I have used the simpler notation ##AB## instead of ##A \cap B##.

    So, 2) is wrong as well, at least for non-independent events. Start with the set identity ##(A \cup B) C = AB \cup AC##. This has nothing to do with probability; it is just "set algebra". OK, so you now have a probability of the form ##P(\text{this} \cup \text{that}),## and you have already figured how to expand such things.

    After fixing up 2), you can go on to 3) in a similar way. If you think it was tedious before, just keep going!
     
  6. Jan 27, 2013 #5
    Hello Vickson,

    So it would change the scenario if the events were independent. Then if events A, B, C and D could occur without the other(s) happening the 2nd post would be correct?

    I think I will get back to this after I familiarize myself with set algebra.
    Thanks for your inspection of the equations.
     
  7. Jan 28, 2013 #6

    Ray Vickson

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    The scenario does not change at all if the events are independent; the only thing that changes is whether or not you have P(AB) = P(A)*P(B) for example; you still have P(AB), etc., present in the final answer.
     
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