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 ?

• Yes, it is correct (more accurately, there are $3$ out of $6$ possible combinations). But since there are $2$ black balls and $4$ draws, you may as well calculate the probability that the last ball to be drawn is black. And since the number of black balls is equal to the number of white balls, this probability is simply $\frac12$. – barak manos Oct 22 '15 at 15:08

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 $\frac12$.

You ask for the color of the remaining ball, after all other three have been taken out. That is equivalent to just taking out one. (Iff the last ball is black, it will be the second time a black one appears, as there were only two of them.)

i.e. $p = 0.5$