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Homework Help: Probability, Set Theory, Venn Diagrams

  1. Aug 30, 2015 #1
    1. The problem statement, all variables and given/known data
    Let A and B be two events such that

    P(A) = 0.4, P(B) = 0.7, P(A∪B) = 0.9

    Find P((A^c) - B)

    2. Relevant equations

    I can't think of any relevant equations except maybe the Inclusion Exclusion property.
    P(A∪B) = P(A) + P(B) - P(A∩B)
    This leads us to another thing
    P(A∩B^c) = P(A-B) = P(A) - P(A∩B)
    And P(A^c) = 1 - P(A)

    3. The attempt at a solution

    The primary problem I'm having is exchanging the equation for one I can easily understand.
    I know that P(A∩B) = 0.2 from the Inclusion Exclusion Property.
    However, I guess I'm having trouble comprehending the principles behind this math. I can't simply plug things in and expect it to work with the venn diagrams right?
    ex) I can't simply say A' = A^c = 1 - 0.4 = 0.6,
    and then have it in the form P(A'-B) = P(A') - P(A'∩B),
    Because then I have an equation with two unknowns...
    Using the In-Ex Prop, I'd have P(A'-B) = P(A') - P(A') + P(B) - P(A'∪B)
    Which basically leaves me in the same mess of too many variables.

    The logive behind Unions and Intersections really confuses me mathematically. I can visualize the venn diagrams for the most part, but translating that into a math and a function leaves me lost.
  2. jcsd
  3. Aug 30, 2015 #2

    Ray Vickson

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    You can't say ##A' = A^c = 1 - 0.4 = 0.6## because ##A'## is an event, not a probability. However, you can say ##P(A') = P(A^c) = 1 - 0.4 = 0.6##. Then, your equation ##P(A' - B) = P(A') - P(A' \cap B)## is true. You just need to find ##P(A' \cap B)##. Look at a Venn diagram, to see what this "region" ##A' \cap B## looks like, then use the information given in the problem (plus some side calculations) to see how you would reckon its probability. It might be helpful to work out the individual probabilities of the four relevant regions: R1 = in A but not B, R2 = in B but not A, R3 = in both A and B, and R4 = outside both A and B. You have all the information you need to do these four computations, and you have already presented evidence that you know the basic way of approaching them.
  4. Aug 30, 2015 #3
    I've been redoing this problem since last night, here's what I came up with...

    Using a Venn Diagram, I've begun with plotting P(A'). This would be everything outside of A, so B and the region outside of the overlapped circles. I've allocated the quantity 0.1 to this region because P(A∪B)' should be 1-0.9 which would be 0.1.

    Then I subtracted the P(B) Portion, which would leave the 0.1 and give me P(A' - B).

    Another way I came to this conclusion, and your reply helps cement my conviction, was by coming across this equality: P(A’ – B) = P(A’ ∩ B’) which would be all things that are neither in A or B.
  5. Aug 30, 2015 #4

    Ray Vickson

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    Exactly. You can also reason it out in words: A' = stuff not in A and so A' - B = stuff not in A but also outside B = the complement of A ∪ B. Note that (A ∪ B)' = A' ∩ B' (another "law" you should get to know by heart).
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