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Urn A contains 5 red balls and 5 black balls, urn B contains 4 red balls and 8 black balls, and Urn C contains 3 red balls and 6 black balls. A ball is drawn from A, color unknown, and put into B. Then a ball is drawn from B, color unknown, and put into C. what is the probability that a ball now drawn from C will be red?

I tried drawing a tree to model the different possible outcomes of the event, but it didn't seem to work.

Thanks a bunch

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  • $\begingroup$ Can you find the probability that at first hand urn C is enriched with a red ball? $\endgroup$ – drhab May 13 '17 at 12:51
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The Law of Total Probability says:$$\def\P{\operatorname{\mathsf P}} \P(C=r) ~=~ { {\quad\P(C=r\mid B=r)\P(B=r\mid A=r)\P(A=r)} \\ + {\P(C=r\mid B=r)\P(B=r\mid A=b)\P(A=b)} \\ + {\P(C=r\mid B=b)\P(B=b\mid A=r)\P(A=r)} \\ + {\P(C=r\mid B=b)\P(B=b\mid A=b)\P(A=b)} }$$

Where, $A,B,C$ represent the colour of the ball removed for each urn in the process, and $r,b$ represent the values.

Now $\P(C=r\mid B=r)$ is the probability for drawing a red ball from urn C given that the ball put into it (drawn from urn B) was red, which is $4/10$ (since there would be 4 red and 6 black balls in the urn when you make the draw).

And so on...

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