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    Events A and B are independent with P(A) = 1/4 and P(AUB) = 2/3
    Find:
    b) P(B)
    Independent so P(A)P(B) = P(ANB) therefore = 1/4P(B)
    So P(A) + P(B) - P(ANB) = P(AUB)
    Therefore 1/4 + 3/4P(B) = 2/3
    Therefore 2/3 - 1/4 = 5/12
    P(B) = 5/12 over 3/4 = 5/9 right??
    c) P(A'|B) = ??
    d) P(B'|A) = ??
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    Wait it's 5/12 and 1/9 respectively right??
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    (Original post by AlphaArgonian)
    Events A and B are independent with P(A) = 1/4 and P(AUB) = 2/3
    Find:
    b) P(B)
    Independent so P(A)P(B) = P(ANB) therefore = 1/4P(B)
    So P(A) + P(B) - P(ANB) = P(AUB)
    Therefore 1/4 + 3/4P(B) = 2/3
    Therefore 2/3 - 1/4 = 5/12
    P(B) = 5/12 over 3/4 = 5/9 right??
    If you have 2/3 = 1/4 * P(B) - then how can you find P(B)?
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    (Original post by Zacken)
    If you have 2/3 = 1/4 * P(B) - then how can you find P(B)?
    Divide both sides by 1/4
    Wait why?
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    (Original post by AlphaArgonian)
    Divide both sides by 1/4
    Wait why?
    And that's your answer, isn't it?
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    Oh, and: moved to maths.
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    (Original post by Zacken)
    And that's your answer, isn't it?
    Uhhh I get 8/3 from that
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    (Original post by AlphaArgonian)
    Divide both sides by 1/4
    Wait why?
    If ax = b then: \displaystyle \frac{ax}{a} = \frac{b}{a} \iff x = \frac{b}{a}, in this case, a=\frac{1}{4}.
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    (Original post by AlphaArgonian)
    Uhhh I gett 8/3 from that
    Argh!! Ignore me, I read your union and intersection signs wrong!!
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    (Original post by Zacken)
    Argh!! Ignore me, I read your union and intersection signs wrong!!
    Haha, that's fine I'll probably do that in the real exam this year
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    (Original post by AlphaArgonian)
    Events A and B are independent with P(A) = 1/4 and P(AUB) = 2/3
    Find:
    b) P(B)
    Independent so P(A)P(B) = P(ANB) therefore = 1/4P(B)
    So P(A) + P(B) - P(ANB) = P(AUB)
    Therefore 1/4 + 3/4P(B) = 2/3
    Therefore 2/3 - 1/4 = 5/12
    P(B) = 5/12 over 3/4 = 5/9 right??
    c) P(A'|B) = ??
    d) P(B'|A) = ??

    Yeah, you get 5/9. That's correct.

    For the next part, try looking at the conditional probability formula and recall how you can write P(A' if you know P(A).
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    (Original post by Zacken)
    Yeah, you get 5/9. That's correct.

    For the next part, try looking at the conditional probability formula and recall how you can write P(A' if you know P(A).
    Ah thanks
    1-P(A)?
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    (Original post by AlphaArgonian)
    Ah thanks
    1-P(A)?
    Yep.
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    (Original post by Zacken)
    Yep.
    Cheers
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    (Original post by Zacken)
    Yeah, you get 5/9. That's correct.

    For the next part, try looking at the conditional probability formula and recall how you can write P(A' if you know P(A).
    (Original post by AlphaArgonian)
    Cheers
    .
    Also, if A and B are independent, A' and B are also independent, so P(A'|B) = P(A')
 
 
 
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