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I got a problem with probability.

Two vases are not distinguishable (V1 and V2). V1 contains 3 red, 5 white and 7 blue balls. V2 contains 7 red, 5 white and 3 blue balls.

Someone takes a ball from V1 and puts it in V2 unseen. One draws randomly a ball from V2. This ball is red. What is the probability that the ball, unseen from V1 in V2 is laid, is white?

This is what I thought: Probability = P(White ball AND V1) / P(V1) = (5/30) / (15/30).

I feel I need to do something with the red ball that is taken. But I don't know what.

Thanks,

Bob

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Note that containers for balls in probability problems are conventionally called "urns" in English, not "vases". –  Henning Makholm Jan 30 '12 at 23:34
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I think it would be nice if they were called vases, though. Vases make me think of flowers, urns make me think of cremation. –  Michael Lugo Jan 30 '12 at 23:38
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If the vases are not distinguishable then how do they know they are taking the ball from V1? –  opt Jan 30 '12 at 23:46
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1 Answer 1

This is a Bayes' Theorem question. Let $A$ be the event that the ball transferred from $V_1$ to $V_2$ was white, and $B$ the event that the ball removed from $V_2$ was red. You want to find the conditional probability $P(A|B)$. You need to find $P(A)$, $P(B|A)$ and $P(B|A^c)$ and put them into the Bayes formula.

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