How to solve this urn problem in without replacement model.

An urn contains $a$ white balls and $b$ black balls. Balls are drawn randomly from the urn without replacement. Find the probability that a white ball is drawn at the $k$th draw.

In this question I have tried to put binomial theorem at $k^{th}$ instant. As without replacement is given so at $k^{th}$ instant or before $k-1$ instant there is a chance of white ball appear. So how to proceed?

Solution 1:

I think this problem is probably easiest to solve by applying combinatorics to the order of ball selection. Even if we stop after the $k$th draw, pretend we draw out all the balls, and line them up in the order they are drawn. Now there are $\binom{a+b}{a}$ ways to arrange the white balls in these orders, and $\binom{a+b-1}{a-1}$ ways to arrange them such that a white ball is in the $k$th position. Then the probability of a white ball being drawn on the $k$th draw is:

$$P=\frac{\binom{a+b-1}{a-1}}{\binom{a+b}{a}}=\frac{\frac{(a+b-1)!}{(a-1)!b!}}{\frac{(a+b)!}{a!b!}}=\frac{a}{a+b}$$

As @lulu mentions, this answer makes sense because the $k$th draw should be no different from the first draw.