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Probability P(A' and B') of disjoint events.

  1. Sep 3, 2013 #1
    If we have two disjoint events then it comes to reason that P(A or B) = P(A) + P(B) - 0

    However, if I rewrite it as 1 - P(A' and B') how is the numerical result going to be the same? (without re-writing it back as 1 - P ((A or B)')).

    I assume it is 1 - P(A'|B')P(B')

    But how is P(A'|B') calculated in there?

    For example:

    Let's assume P(A) = 0.12 and P(B) = 0.17, and we know that P(A and B) = 0 by being disjoint.

    Hence P(A or B) = 0.12 + 0.17 -0 = 0.29

    Now, P(A or B) = 1 - P((A or B)') =
    = 1 - P(A' and B')

    If I do

    = 1 - P(A')P(B') = error, the result is 0.2696. It's as if the events were not disjoint obviously.

    if I do

    = 1 - P(A'|B')P(B')

    And I know that P(B') = 1 - P(B) = 0.83

    then it comes to reason that for the result to be correct

    P(A'|B') must be 0.8554

    Now how is that found on its own?

    I currently assume that I'm not supposed to find it directly at all but go with other methods around it.
     
  2. jcsd
  3. Sep 3, 2013 #2
    Oh wait, I think I found the secret: P(A'UB') in the calculation P(A' and B') = P(A') +P(B') - P(A'UB') is actually 1 since A and B is null hence it all results to 0.71 (1-0.12+1-0.13-1) hence 1-0.71 will net the 0.29.
     
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