An urn contains 2 white and 2 black balls. Balls are drawn successively at random without replacement. An urn contains 2 white and 2 black balls. Balls are drawn successively  at random without replacement. What is the probability that black ball appears for the second time in the 4th draw?
I am trying in this way :
There can be 3 possibilities - 
a) B W W B
b) W B W B
c) W W B B
for a) $$(\frac{2}{4}) * (\frac{2}{3}) * (\frac{1}{2}) * 1 $$
for b) and c) $$(\frac{2}{4}) * (\frac{2}{3}) * (\frac{1}{2}) * 1 $$
as well.
So , it comes down to $$(\frac{1}{6}) * 3 = 0.5$$
Am I correct ?

First, let's fix up your solution (although the answer itself is correct):
There are $6$ combinations:


*

*BBWW

*BWBW

*BWWB

*WBBW

*WBWB

*WWBB


Out of which, in $3$ of them, a black ball appears for the 2nd time in the 4th draw:


*

*BWWB

*WBWB

*WWBB


Hence the probability that a black ball appears for the 2nd time in the 4th draw is $\frac36$.

Second, let's observe a much more simple way to answer this question:
Since there are $2$ black balls and $4$ balls altogether, the question can be rephrased as:

What is the probability that the last ball to be drawn is black?

And since the number of black balls and white balls is equal, the answer is simply $\frac12$.
