2
$\begingroup$

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 ?

$\endgroup$
  • 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
2
$\begingroup$

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 $r$.

$\endgroup$
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.