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Consider $3$ urns. Urn $A$ contains $2$ white balls and $4$ red balls; Urn $B$ contains $8$ white balls and $4$ red balls; Urn $C$ contains $1$ white ball and $3$ red balls. We draw $1$ ball from each Urn. What is the probability to draw a white ball from Urn $A$ if we know that we drew exactly a total of $2$ white balls total?

I am asked to proceed with Bayes theorem but i am a little confused on how to do it. Any hint would help me alot.

Thank you.

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  • $\begingroup$ I prefer to use the definition of conditional probability, since in my experience students are more likely to get the right answer that way than by manipulating complements. $\endgroup$ – André Nicolas Oct 12 '15 at 0:39
  • $\begingroup$ How would you proceed with the definition of conditional probability? $\endgroup$ – spexel Oct 12 '15 at 0:50
  • $\begingroup$ I have written out most of the details of an answer. There is only some arithmetic left to do. $\endgroup$ – André Nicolas Oct 12 '15 at 1:06
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Let $A$ be the event we picked a white from Urn A, and let $X$ be the event we picked exactly two white, We want $\Pr(A\mid X)$. By the definition of conditional probability, we have $$\Pr(A\mid X)=\frac{\Pr(A\cap X)}{\Pr(X)}.\tag{1}$$ We need to find the two probabilities on the right of (1).

First find $\Pr(X)$. This can happen in three ways: (i) white from A, white from B, red from C; (ii) white from A, red from B, white from C: (iii) red from A, white from B, white from C.

The probability of (i) is $(2/6)(8/12)(3/4)$. Find similar expressions for (ii) and (iii), and add up to find $\Pr(X)$.

Next we find $\Pr(A\cap X)$. This is the sum of terms (i) and (ii) above.

Now we know both the numerator and denominator of the right side of (1), so we are finished.

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  • $\begingroup$ Thanks! finally could get my head around it $\endgroup$ – spexel Oct 12 '15 at 1:43
  • $\begingroup$ You are welcome. The gneral approach above can be used as a template for conditional probability problems. $\endgroup$ – André Nicolas Oct 12 '15 at 2:51

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