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Math Help - Conditional probability.

  1. #1
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    Conditional probability.

    Hey guys, I'm having trouble with figuring this out:

    What is P(V) given the following information?

    P(F|V) = 0.02, P(F^c|V) = 0.98 P(F|V^c) = 0.4, P(F) = 0.2, P(F^c) = 0.8

    Where I've written F^c and V^c I mean 'not F' and 'not V'.

    I tried figuring it out using the probability rules, i.e.
    P(V) = P(F and V) + P(not F and V)
    but it seems like I don't have enough information to figure it out. Am I missing something?

    Thanks.
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  2. #2
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    This is probably not the simplest way, but consider focusing on P(F):
    P(F) = p(F and Not V) + P(F and V)



    P(F)=P(V^c)P(F|V^c) + P(V)P(F|V)

    0.2=P(V^c)0.4 + P(V)0.02

    0.2=(1-P(V))0.4 + P(V)0.02
    ...
    P(V)=\frac{10}{19}
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  3. #3
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    Quote Originally Posted by staralfur View Post
    Hey guys, I'm having trouble with figuring this out:

    What is P(V) given the following information?

    P(F|V) = 0.02, P(F^c|V) = 0.98 P(F|V^c) = 0.4, P(F) = 0.2, P(F^c) = 0.8

    Where I've written F^c and V^c I mean 'not F' and 'not V'.

    I tried figuring it out using the probability rules, i.e.
    P(V) = P(F and V) + P(not F and V)
    but it seems like I don't have enough information to figure it out. Am I missing something?

    Thanks.
    Drawing a tree diagram (the first two brances are V and V') makes it triival:

    Let Pr(V) = x so that Pr(V') = 1 - x.

    Then (x)(0.02) + (1 - x)(0.4) = 0.2 => x = 10/19.
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  4. #4
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    ...
    Last edited by staralfur; April 1st 2011 at 05:23 PM. Reason: Unnecessary
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  5. #5
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    Wait, sorry! Your answers make perfect sense, never mind.
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