The first urn contains 2 white balls and 2 black balls and the second urn contains 2 white balls and 2 black balls. the third urn contains 2 and 2 respectively. We take at random 1 ball from the first urn 1 ball from the second urn and put them into the third urn. then a ball is randomly drawn from the third urn. what is the probability of it being white?
closed as off-topic by Michael Hoppe, C. Falcon, tired, Shailesh, user223391 Dec 19 '16 at 4:20
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25% of the time both balls drawn from the two urns will both be white, 25% of the time both will be black, and 50% of the time there will be one black and one white. After those balls have been added to the third urn, 25% of the time the third urn will contain 4 white and 2 black balls so the probability of drawing a white ball is 4/6= 2/3, 50% of the time it will contain 3 white and 3 black balls so the probability of drawing a white ball is 3/6= 1/2, and 25% of the time it will contain 2 white and 4 black balls so the probability of drawing a whit ball is 2/6= 1/3. The overall probability of drawing a white ball from the third urn is (.25)(2/3)+ (.50)(1/2)+ (.25)(1/3)= 1/6+ 1/4+ 1/12= (2+ 3+ 1)/12= 6/12= 1/2.
Actually, it should have been clear from the start that, since each urn contains the same number of white as black balls, the probabilities of drawing a white or black ball from the third urn are the same so 1/2.
Let $BW$ denote the event that a black ball is picked out of the first urn and a white ball out of the second. Denote $BB, WW, WB$ be the obvious counterparts. Let $W$ denote the probability that the ball picked is white. I claim the following:
Where $P(A|B)$ denotes the probability that event $A$ occurs given that probability $B$ occurs.
The rest, you should be able to do on your own. You'll also need to prove that equation (*) is true.