Probability remains same Why? 
A box has $m$ black and $n$ white balls. A ball is drawn at random and put back with $k$ additional balls of the same colour as that of the drawn ball. Now a ball is drawn again. Find the probability that it is a white ball.

I got the answer $\frac{n}{m+n}$.
I just want to understand why it's not dependent on $k$.
 A: Look at this probability from the other side. Consider two disjoint events. 
Event $B_1$ means that the second ball was in the urn from the very beginning. Event $B_2$ means that the second ball is taken out of those balls that were added to the urn. Event $A$ means that the second ball is white. Consider conditional probabilities  $\mathbb P(A\mid B_1)$ and $\mathbb P(A\mid B_2)$:
$$\mathbb P(A\mid B_1) = \frac{n}{n+m}$$
since no ball added is taken into account when $B_1$ happens. 
And also 
$$\mathbb P(A\mid B_2) = \frac{n}{n+m}.$$
This is the probability of the second ball to be white if it is selected from $k$ balls added. In other words, this is the probability that $k$ white balls are added into the urn. This is exactly the probability that the first ball was white.
Since conditional probabilities of $A$ under both $B_1$ and $B_2$ are the same, the probabilities of $B_1$ and $B_2$ are irrelevant and 
$$\mathbb P(A) = \mathbb P(B_1)\frac{n}{n+m}+\mathbb P(B_2)\frac{n}{n+m}=\frac{n}{n+m}.$$
Note that only $\mathbb P(B_1)$ and $\mathbb P(B_2)$ depend on $k$.
