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Homework Help: How do you find the intersection of the complements of two neg. dep events?

  1. Feb 12, 2012 #1
    Probability ? How do you find the intersection of the complements of two negatively dependent events?

    I'm given P(a), P(b), and P( A intersect B), but I need to find the conditional probability of 'the complement of B given the complement of A'. I don't know how to find it. I thought I only had to multiply the complements, but that obviously didn't work. Then I said maybe it's equal to the complement of the union of A and B: 1- ( A+B- A^B).

    Is that the right way of doing it?

    Thanks for your help.
  2. jcsd
  3. Feb 12, 2012 #2


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    Do you mean the complement of A instead of "complement of A' "?
    1 - (A+B-A^B) doesn't make any sense at all. You are mixing up numbers and arithmetic operations with with sets and and set operations. You might start with the formula for conditional probability:$$P( B'| A') =\ ?$$
  4. Feb 12, 2012 #3

    Thanks for responding! Yes, I know the formula.

    P (complement of B given complement of A) =

    [ (complement of A) Intersection (complement of B)] / (complement of A).

    I'm asking how to find the top part.

    since the events are negatively dependent, I can't just say (complement of A) * (complement of B), right?

    Ive worked it out these two ways (for the numerator):

    (complement of A) intersect ( complement of B) = 1 - A union B

    or do I say

    (complement of A) intersect (complement of B) = (B - A intersect B) * (A- A intersect B).

    I think it's the first one because the area outside the circles is supposed to be (complement of a) ^ (complement of b) = 1- (A+B-A^B).... right or wrong? lol
    Last edited: Feb 12, 2012
  5. Feb 12, 2012 #4


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    You are still being sloppy with your notation. The left side is a probability, which is a number, and the right side is a set. They can't be equal even if I know what you mean.
    That's right, you can't.
    They can't possibly be equal with sets mixed up with numbers. You are still mixing up sets with their probabilities.
    I know you are thinking to get the probability in the numerator you want to take 1 minus the probability of something. You will never get it right until you clear up your notation. You can't add and subtract numbers with sets. You add and subtract numbers and probabilities and you union and complement and intersect sets. Don't mix them. Try again.
  6. Feb 12, 2012 #5

    Ray Vickson

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    Using an "overbar" to denote the complement of something, you say you want
    [tex] P(\bar{B} | \bar{A}). [/tex]
    Use the definition of conditional probability:
    [tex] P(\bar{B} | \bar{A}) = \frac{P(\bar{B} \cap \bar{A})}{P(\bar{B})}. [/tex]
    If you know [itex] P(A), P(B) \mbox{ and } P(A \cap B) [/itex], can you see how to get [itex] P(\bar{B})[/itex] and [itex] P(\bar{B} \cap \bar{A})[/itex]? In particular, do you know who to relate
    [tex] \bar{B} \cap \bar{A}[/tex] to [itex] A,\: B \mbox{ and } A \cup B \mbox{ or } A \cap B[/itex]?

  7. Feb 15, 2012 #6

    @Ray Vickson...

    Thanks for your replies! I'm reworking the problem again and I understand about my lack of notation.
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