Is there any quick way to do this problem?

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?

• 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). – Arthur Jan 12 '18 at 7:40
• @Arthur I didnt get you :(. – Daman deep Jan 12 '18 at 7:46
• 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? – Arthur Jan 12 '18 at 7:49
• Yes if we don't shift balls and i draw white ball that would give me 1/2. – Daman deep Jan 12 '18 at 7:52
• 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. – Arthur Jan 12 '18 at 8:03

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.