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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?

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    $\begingroup$ Any thoughts on the problem? $\endgroup$ – saz Dec 18 '16 at 20:31
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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.

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  • $\begingroup$ thank you so much for your clear answer sir,,,, i really appreciate ur help on time $\endgroup$ – Diwakar Ragupathi Dec 18 '16 at 21:16
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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:

(*) $P(W)=P(BW)P(W|BW)+P(WB)P(W|WB)+P(WW)P(W|WW)+P(BB)P(W|BB)$

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.

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  • $\begingroup$ sSorry to say this i am new to probability. and i m learning distance education. couldnt understand what u say. can u please explain me in numbers as per the question' $\endgroup$ – Diwakar Ragupathi Dec 18 '16 at 20:47
  • $\begingroup$ @DiwakarRagupathi $P(E)$ is the probability that event $E$ occurs. $P(BW)$ would be the probability that a black ball is picked from the first urn and a white ball from the second. $\endgroup$ – Kevin Long Dec 18 '16 at 20:50
  • $\begingroup$ thanks for your explanation sir $\endgroup$ – Diwakar Ragupathi Dec 18 '16 at 20:52
  • $\begingroup$ so P(BW)=P(WB)=P(WW)=P(BB)=(1/2*1/2)... is this correct $\endgroup$ – Diwakar Ragupathi Dec 18 '16 at 20:58
  • $\begingroup$ and what is P(W|BW) $\endgroup$ – Diwakar Ragupathi Dec 18 '16 at 20:58

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