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 ?

  • 5
    $\begingroup$ It's $y/(x+y+z)$. Just imagine arranging the $(x+y+z)$ balls in a row in a uniformly random order, and asking for the colour of the ball at position $(r+1)$. This is just an outline; someone may post a clearer answer. $\endgroup$ – ShreevatsaR Jul 8 '13 at 9:35
  • $\begingroup$ so it is independent of r ?? $\endgroup$ – Sayan Paul Jul 8 '13 at 10:06
  • $\begingroup$ $0$ if $ r \geq y $. $\endgroup$ – hjpotter92 Jul 8 '13 at 10:52
  • $\begingroup$ @hjpotter92 How?? if r =z and r>=y and only red balls are taken out then it is not zero ??? $\endgroup$ – Sayan Paul Jul 8 '13 at 11:02
  • $\begingroup$ What I meant was, you can find a lower limit and an upper limit of the probability. Exact probability can not be computed without knowing what $r$ holds. $\endgroup$ – hjpotter92 Jul 8 '13 at 11:04

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 $r$.


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. $r \le \min\{x,y,x\}$.

  • $\begingroup$ so what will be the answer if r > min{x,y,z} ? $\endgroup$ – Sayan Paul Jul 8 '13 at 11:26
  • $\begingroup$ I gave you the answer when $r \gt \min\{x,y,z\}$. When $r \lt \min\{x,y,z\}$, however, it gets messy. By then, you are $r$ levels down in a recursive tree and must sum across it; some of the branches of this tree have probabilities of zero which remain zero as more balls are drawn. $\endgroup$ – Ron Gordon Jul 8 '13 at 11:36
  • $\begingroup$ ok thanks for the answer .. so for r < min{x,y,z} , can you tell me more about the recursive procedure and how to come about a proper solution perhaps . $\endgroup$ – Sayan Paul Jul 8 '13 at 11:46
  • $\begingroup$ You don't need $r \le \min {x, y, z}$ for the answer to hold. :-) $\endgroup$ – ShreevatsaR Jul 9 '13 at 3:24

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