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Homework Help: Combination of Mutually Exclusive and Independent events

  1. Sep 30, 2012 #1
    1. The problem statement, all variables and given/known data
    This isn't actually a homework question, I'm reviewing for a midterm I have coming up this week and came across this question in some of the practice exercises that were provided.

    Let A,B,C be three events. Suppose that A and B are independent, B and C are mutually exclusive, and that A and C are also independent. Given that P(A)=0.3, P(B)=0.4, P(C)=0.3 find the probability that at least one of the three events occur.

    2. Relevant equations
    For independent events / non-mutually exclusive P(A union B) = P(A) + P(B) - P(A intersect B)
    P(A intersect B) = P(A)P(B)
    for mutually exclusive events P(A union B) = P(A) + P(B)

    3. The attempt at a solution

    I know the correct answer is 0.79 (solutions provided) but I'm having trouble understanding how to approach the combination of independent / mutually exclusive events.

    Is it easier to count the complement? ie:the event that none of the 3 occurs and subtract it from 1?

    If I sum the probabilities of A and B and the probability of A AND C I get 0.79.
    ie: 0.3 + 0.4 + 0.3(0.4), but is this the correct method?

    Thank you in advance if you can help me understand this!
  2. jcsd
  3. Sep 30, 2012 #2

    Ray Vickson

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    Homework Helper

    Have you not seen the principle of inclusion/exclusion? For A and B it says P(A or B) = P(A) + P(B) - P(AB), where I am using AB to stand for "A and B". Do you see why it has to hold? Draw a Venn diagram to convince yourself, but the basic idea is that P(AB) is part of both P(A) and P(B), so when we add these two we are counting P(AB) twice. Therefore, we need to subtract it once in order to not double-count. The principle generalizes to any number of events. In particular, P(A or B or C) = P(A) + P(B) + P(C) - P(AB) - P(AC) - P(BC) + P(ABC).

  4. Sep 30, 2012 #3
    Hi RGV,

    Thanks for the help!
    For some reason I had myself convinced that because B and C are mutually exclusive this property wouldn't hold. I see now that it just means P(BC)=0 and P(ABC)=0 since B and C can never happen simultaneously.

    Again thank you, this helped a lot

    Luuk V.
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