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Thread: Events A and B with complements

  1. #1
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    Events A and B with complements

    Hi all,

    I will show the quest first and my work after.

    Given: P(A) = 0.3, P(B) = 0.8, P(AUB) = 0.9

    A) Find P(A|B), P(A'nB), P(B'UA')
    B) Are A and B independant? Why?

    My Work so far...

    I know that P(A') = 0.7, P(B') = 0.2

    To find P(A|B) I need to use the formula P(A|B) = P(AnB)/P(B).
    I dont have P(AnB) so I must find it first using
    P(AUB) = P(A)+P(B)-P(AnB) becomes
    P(AnB) = P(A)+P(B)-P(AUB)
    P(AnB) = 0.3 + 0.8 - 0.9
    P(AnB) = 0.2

    Now I can use P(A|B) = P(AnB)/P(B) = 0.2/0.8 = 0.25

    Can I use the P(AnB) = P(A)P(B|A) to find P(A'nB)? If I just sub in the complements? Like this? P(A'nB) = P(A')P(B|A') If I can't do this, what should I do?

    Finding P(B'UA') is a similar problem. Unforunatly this problem is just using numbers so I cannot think of anything in real life to help me. I also thing drawing a venn diagram may help but I'm not sure where to start.

    B) I used the P(AUB) = P(A)+P(B)-P(AnB) formula to find P(AnB) = 0.2 however the events can only be independant if P(A)*P(B) = P(AnB).
    Actually doing the multiplication given the starting info 0.3*0.8 = .24 means is not indepentant.


  2. #2
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    Re: Events A and B with complements

    The events \{B, B^c\} actually forms a partition over the
    sample space \Omega.

    So P(A) = P(A\cap\Omega) = P(A\cap(B\cup B^c)) 
= P((A\cap B)\cup(A\cap B^c)) = P(A\cap B) + P(A\cap B^c)

    And also using De Morgan's Law, you can quickly simplify:
    P(A^c \cup B^c) = P((A \cap B)^c)

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