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I'm quite new to the whole probability branch of math, have been doing some problems, and I've come across a challenging one (in my inexperienced opinion):

There are 50 balls in an urn, 20 red, 30 white. If they are randomly being drawn without being returned, what's:

  1. The probability to have a white ball drawn in the 3rd draw
  2. The probability to have a red ball drawn in the 7th draw
  3. The probability to have drawn a red ball in the 7th draw if it's known that a white ball was drawn in the 3rd draw, and vice versa.

So, I've had 2 problems I just couldn't wrap my head around, and those are:

  • How to calculate the probabilities for 1. and 2. I went brute force for the probability of a white ball being drawn on the 3rd draw, made a total probability tree (diagram), summed up all the good outcomes, and got $\frac35$, which would be the same as if it was being drawn in the first draw. Then I checked the same for the 2nd draw and the outcome was the same, so I could assume that the answer to question 2. would be $\frac25$, but I'm still unsure as to why, and would really appreciate an explanation to this.
  • Even with those 2 solved, not quite sure how to answer question 3.

I'd be really thankful for all and any kind of explanations to these 2 problems.

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  • $\begingroup$ 1. Are any of the other balls+urn questions helpful? 2. What exact condition is meant by "vice versa" in part 3? $\endgroup$ – shoover Sep 8 '15 at 22:16
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This is a good question. Developing an understanding for why the probability is the same as for the first draw will help you find elegant solutions to probability problems that would otherwise require tedious calculations (like the ones you just performed).

To my mind, the most immediate reason that the probability of drawing a white ball must be the same on the third draw as on the first is that, since the balls are by definition identical (except for their colour, which doesn't influence their chances of being drawn), on any given draw all balls must have the same probability of being drawn. But that means that the probabilities of all events related to a single draw (including the event of drawing a white ball) must be the same for all draws.

Here are some more ways to think about this:

  • Draw three balls, then randomize their order, then look at the first ball. What's the probability of it being white?
  • In which direction would you expect the prior drawing of the first two balls to change the probability of drawing a white ball? Up or down? Why?
  • If the probability of drawing a white ball were, say, higher than on the first draw for some draws, it would have to be lower for others, since the average over all draws is the probability on the first draw.
  • Consider the drawing of the balls as fixing a permutation of them. Is any permutation more likely than any other permutation? Is any spot in a permutation different from any other spot?

For your third question, the fact that a white ball was drawn in the third draw simply eliminates one of the white balls, so the probability that a red ball is drawn in any of the other draws (before or after the third draw) is $20/49$.

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    $\begingroup$ Wow, thanks, that really clarified a lot. :) But one more thing, for the 3rd question, the vice versa part (drawing a white ball 3rd if the 7th is red), would be 30/49, according to this logic? $\endgroup$ – Domu Sep 8 '15 at 22:54
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    $\begingroup$ @Doma Yes, that is so. $\endgroup$ – Graham Kemp Sep 8 '15 at 23:09

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