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? asked in Science & MathematicsMathematics · 1 decade ago

How to work out this probability proof?

If Π(A) denotes the odds on A and Π(A|B) the odds on A given B, deduce that

Π(A|B) / Π(A) = P(B|A) / P(B |Ā )

I really am struggling with this this question, even though it seems pretty straightforward. Some help will be hugely appreciated.

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  • 1 decade ago
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    Hey! I thought this WAS a mathematics forum!

    If you'll excuse me, I'm going to write Ā as ~A, because I don't have an equivalent character for ~B and I'm going to need one.

    Now, let's simplify the whole business by considering all the combined cases possible with respect to the occurrence of events A and B: let

    P(A & B) = q/(q+r+s+t)

    P(A & ~B) = r/(q+r+s+t)

    P(~A & B) = s/(q+r+s+t)

    P(~A & ~B) = t/(q+r+s+t)

    We've just reduced the model to a typical Venn diagram consisting of two partially overlapping circles inside a larger circle, with the four spaces thus created marked q, r, s, and t. We can now express odds and probabilities using these four variables.

    Π(A|B) = q/s

    Π(A) = (q+r)/(s+t)

    P(B|A) = q/(q+r)

    P(B|~A) = s/(s+t)

    The proof is now just a matter of algebraic manipulation:

    Π(A|B) / Π(A)

    = (q/s) / [(q+r)/(s+t)]

    = (q/s) * [(s+t)/(q+r)]

    = [q(s+t)]/[s(q+r)]

    = [q/q+r)]/[s/(s+t)]

    = P(B|A) / P(B|~A)

    It is pretty straightforward, but only after finding a way to express odds and probabilities, including conditional ones, in terms of a common set of variables.

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  • 1 decade ago

    For future reference, I suggest you seek out a mathematics forums. Some good ones I recommend are :

    http://www.mymathforum.com/

    http://www.mathhelpforum.com/math-help/

    Both should have sections for Probability/Statistics.

    I hope that was useful!

    Source(s): I'm 1337 with my interwebs
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