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There are two urns $U_1$ and $U_2$ , $U_1$ contains four white and four black balls, and $U_2$ is empty. Four balls are drawn at random from $U_1$ and transferred to $U_2$.Then a ball is drawn at random from $U_2$.The probability that the ball drawn from $U_2$ is white is?

(A) $\ \dfrac{1}{3}$

(B) $\ \dfrac{1}{2}$

(C)$\ \dfrac{2}{3}$

(D)$\ \dfrac{3}{4}$

I know one way of solving this problem is with The Law of total probability along with hyper geometric distribution but this question came in $2$ marks.Is there any quick version?

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    $\begingroup$ If you do calculate it the long way, I'm certain that a suspicion about the short way of doing it should appear to you just by looking at the answer. You would probably be right in that suspicion (as in, yes, there is a short way, and once you know the answer the obvious short and simple way of getting to that answer is correct). $\endgroup$
    – Arthur
    Jan 12, 2018 at 7:40
  • $\begingroup$ @Arthur I didnt get you :(. $\endgroup$
    – Daman
    Jan 12, 2018 at 7:46
  • $\begingroup$ Well, there is an answer below telling you what the answer is, so that cat is out of the bag. At any rate, from the numbers you've been given, is there an obvious, simple way to make $\frac12$ that seems related to the original problem of drawing a white ball? $\endgroup$
    – Arthur
    Jan 12, 2018 at 7:49
  • $\begingroup$ Yes if we don't shift balls and i draw white ball that would give me 1/2. $\endgroup$
    – Daman
    Jan 12, 2018 at 7:52
  • $\begingroup$ And you'd be right! The shifting does nothing, and each ball is just as likely to be picked as any other, no matter how many steps you throw in to try to confuse people. $\endgroup$
    – Arthur
    Jan 12, 2018 at 8:03

2 Answers 2

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1/2, it is symmetric. I think it is a quick version.

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  • $\begingroup$ Please elaborate to me. $\endgroup$
    – Daman
    Jan 12, 2018 at 7:46
  • $\begingroup$ You are drawing balls at random from equal amount of black and white balls, which means any combination of colors(say bbwb) have the same probability under the act of changing colors(wwbw). Therefore, drawing a white ball from any combination has a symmetric combination with the same probability of drawing a black ball. $\endgroup$
    – Atbey
    Jan 12, 2018 at 8:06
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Suppose you painted all the white balls black, and all the (original) black balls white. You'd have exactly the same problem, so the probability wouldn't change: $P_{\text{new}}(W)=P_{\text{orig}}(W)$. But since a white ball in the new setup was originally black, $P_{\text{new}}(W)=P_{\text{orig}}(B)$, and so the probabilities of white and black are equal.

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