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Probability of all elements question

  1. Aug 31, 2007 #1
    1. The problem statement, all variables and given/known data

    If P(A) =0.4, P(B)=0.5, and P(A∪B)=0.7, find P(A’∪B’)




    3. The attempt at a solution


    P(A’) = (Probability of all elements in S that are not in A) = 1 - P(A) = 0.6

    P(B’) = (Probability of all elements in S that are not in B) = 1 – P(B) = 0.5

    P(A’∪B’)= The union of A’ and B’ = 1 - 0.7 = 0.3

    P(A’∪B’) = P(A’) + P(B’) – P(A’∩ B’) = 0.6 + 0.5 – 0.3 = 0.8

    So, P(A’∪B’) = 0.8

    I'm not sure if i did this correctly.

    Thanks
     
  2. jcsd
  3. Aug 31, 2007 #2

    EnumaElish

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    How do you know P(A’∩ B’) = 0.3?

    The easiest method is to figure P(A ∩ B). (Can you tell me why?)
     
    Last edited: Aug 31, 2007
  4. Aug 31, 2007 #3
    I just assumed that you could do that.

    P(A ∩ B) = .4 + .5 - .7 = .2

    So would P(A' ∩ B') = 1 - .2 = .8 ???
     
  5. Aug 31, 2007 #4
    You're correct but you seem to be bent on making this too difficult =P. P(A intersect B) = P(A) * P(B) = 0.4 * 0.5 = 0.2

    I know you probably get this all the time, but drawing pictures for these things really does help. Or even just visualizing a Venn Diagram for these events in your head. You can see that the complement of (A' union B') is (A intersect B) and P(A intersect B) is easy to find.
     
  6. Sep 1, 2007 #5

    EnumaElish

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    Correct; and that's exactly how I would have calculated this.

    No. P(A ∩ B)' = 1 - .2 = .8 is correct. What is the relationship between (A ∩ B)' and A' U B' ?
     
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