An urn contains $a$ white and $b$ black balls, where $a$ and $b$ are positive integers. One ball at a time is randomly drawn until the first white ball is drawn. find the expected number of black balls that are drawn, by considering conditional probabilities.
I can solve this using recurrence relation and induction to get the answer, i.e. $\frac{b}{a+1}$. However, I cannot think of an approach that uses conditional probabilities. Below is my approach.
Let $X_{a,b}$ denote the number of black balls drawn if we have $a$ white and $b$ black balls in the beginning. By conditional expectation,
$$\mathbb{E}(X_{a,b}) = \frac{b}{a+b}(1+\mathbb{E}(X_{a,b-1})).$$Solving the recurrence relation using induction, I get $\frac{b}{a+1}$. However, I will be very happy if anyone could share his/her approach using conditional probabilities. I have seen people posted similar questions previously, but not using a conditional probability approach.