Probability of getting white ball If there are x black balls, y white balls and z red balls in a bag , r balls are drawn without replacement and now after this one more ball is drawn . What is the probability of this ball being white ??
Given : r <(x+y)
How to solve this? What approach do you recommend for these kind of questions ?
 A: Imagine $r=1$.  The probability of drawing a white after drawing a single ball is
$$P(W|B) P(B) + P(W|W) P(W) + P(W|R) P(R)$$
which is
$$\frac{y}{x+y+z-1} \frac{x}{x+y+z} + \frac{y-1}{x+y+z-1} \frac{y}{x+y+z} + \frac{y}{x+y+z-1} \frac{z}{x+y+z}$$
which simplifies to $y/(x+y+z)$.  This experiment may be applied recursively, say $r$ times, so that the probability sought remains at  $y/(x+y+z)$ until there is a possibility of no more of a certain color of ball, i.e. $r \le \min\{x,y,x\}$.
A: As no one else is posting an answer...
Original setting: you draw balls one-by-one, without replacement, until you've drawn $r$ balls, and then one more ball.
Instead imagine the following modified setting: as you draw balls one-by-one, you place them in a row, in order (i.e., placing the latest drawn one at the end of the current row). Even after you've drawn $r+1$ balls, continue to draw balls one-by-one till all $x+y+z$ balls are exhausted.
We can see the following:


*

*The probability that the last ball drawn in the original setting is white, is the same as the probability that the ball at position $r+1$ in the row in the modified setting is white (in fact it's the same ball).

*The final row formed at the end, in the modified setting, is a uniformly random permutation of the $(x+y+z)$ balls.
Therefore, the answer, the probability that the ball at position $r+1$ is white, is $$\dfrac{y}{x+y+z}$$ as all balls are equally likely at that position, and there are $y$ white balls among the $x+y+z$ balls.
Note that this does not depend on $r$.
