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We have an urn with $17$ balls, $3$ are green, $4$ are blue, $5$ are red and $5$ are brown. We take balls out of the urn until there are none left. What is the probability that the sixth ball picked will be blue?

I tried to do this by case work, because in order for the sixth pick to be blue we can't have the first five picks use up all the blue marbles. So we can split it into three cases for the first five picks.

  1. No blue ball was picked
  2. One blue ball was picked
  3. Two blue balls were picked
  4. Three blue balls were picked

I quickly realized that there is too much adding and calculating doing it like this. Since there are $5$ subcases for $2$, and $10$ subcases each for $3$ and $4$. Is there a better way to do this problem? Thank you!

Not a Salmon Fish
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StevenLang
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  • On your question, can you do this for the first marble being blue and then independently for the second marble being blue? Compare the answers. – Math Lover Jun 01 '21 at 18:34
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    It is clearly $4/17$. Just like if it were the fisrt ball or any other specific one. – ajotatxe Jun 01 '21 at 18:50
  • @ajotatxe it is not independent – Not a Salmon Fish Jun 01 '21 at 18:55
  • @Bulbasaur the answer is indeed $4/17$. If you let $X$ denote the number of blue balls selected among the first five balls, you can use total law of probability to compute $P(E)$ where $E$ is the event that the sixth ball is blue. – Matthew H. Jun 01 '21 at 19:05
  • @ajotatxe I was just going through your deleted answer. Not sure why you deleted it. Just because someone downvoted, the answer does not become wrong. In my comment earlier, that is why I hinted to OP to do this for first and second and conclude for themselves. – Math Lover Jun 01 '21 at 19:13
  • The punchline at the end of this and related questions is that despite the fact that the outcomes of each draw is dependent on one another, the draws themselves are still identically distributed. The outcome of the tenth ball is distributed in exactly the same way as the outcome of the first ball. – JMoravitz Jun 01 '21 at 19:34
  • You can see this in a variety of ways... recursion, tedious calculation by breaking into cases, and such are highly inefficient. Much better are arguments appealing to symmetry of the situation, or appealing to counting arguments or bijective arguments. There is a clear bijection between outcomes where the tenth ball is blue and where the first ball is blue. Thanks to such a bijection, we know the sizes of the events are the same, and as such the probabilities the same as well. – JMoravitz Jun 01 '21 at 19:36
  • @JMoravitz if the balls is replaced , the answer would be the same $4/17$ , is it a coincide ? – Not a Salmon Fish Jun 01 '21 at 19:47
  • @Bulbasaur No coincidence at all. Whether the balls were replaced or not, it should be clear that each of the $17$ balls if they were labeled are equally likely to have been chosen as the tenth pull, nothing favoring one as being drawn before any other. The total probability being $1$, each ball corresponding to the same proportion of it, each individual ball being the tenth $\frac{1}{17}$ of the time, four of them being blue, hence $\frac{4}{17}$ – JMoravitz Jun 01 '21 at 19:49
  • @JMoravitz thanks for nice clarification ! – Not a Salmon Fish Jun 02 '21 at 08:55

2 Answers2

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@ajotatxe has the right answer. I am not sure if I am entirely convinced with his reasoning but I did end up solving the problem in another way. I will post this solution just for the sake of wrapping this question.

We make distinction of blue balls and not blue balls. 13 not blue balls, 4 blue balls. If we fix the blue ball at the sixth position and permute the rest we get C(16, 3) ways to have a blue ball at position sixth. We know we can have a total of C(17, 4) ways to permute all of the balls (we permute the 4 blue balls and just randomly fill in the rest with not-blue balls). If we divide C(16, 3) over C(17, 4) we get 4/17 in the end.

StevenLang
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Hint: a way to choose a random ball is picking 5 random balls and choosing the sixth.

Another hint: rearrange randomly the $17$ balls. What is the probability that the first is blue? And the sencond? And the last? And the sixth?

ajotatxe
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  • Hi, thank you for your answer! I have written up something based on your hint. Please tell me if I am thinking about this correctly.

    We have 1312111094 ways to pick balls such that the sixth is blue and no balls before that is blue. We have 1312111043 ways such that the sixth is blue and one ball before that is blue. We do the same thing for 2 balls before sixth pick is blue and 3 before 6th is blue cases. We add them up and then divide over 1716151413*12 (which is presumably the total amount of cases) for our desired probability.

     – StevenLang Jun 01 '21 at 18:23
  • You don't have to consider the picks before the sixth. The sixth ball is as random as the first (and the others). – ajotatxe Jun 01 '21 at 18:29
  • What if the first five picks picked 4 blue balls? I would have no blue balls for the sixth pick then right? – StevenLang Jun 01 '21 at 18:32
  • IF the first four balls are blue then the probability that the sixth is blue is $0$, of course. But if the four last balls are blue, it is $0$ too. And if the fourth is blue, then it is $1$. But we needn't consider every possible "if".

    See the problem this way: close your eyes and take out five balls. Then pick another one and see its color. This sixth ball could be any of the balls. Exactly like the first one.

     – ajotatxe Jun 01 '21 at 18:35
  • What I say is that the probability is $4/17$, exactly as instead of the sixth ball we considered the first, the last, the 11th, or every other one. – ajotatxe Jun 01 '21 at 18:41
  • Hmmm yes I see what you mean. But I am afraid I don't see how I can use this way of thinking to solve the problem, since I want the sixth pick to be exactly blue. – StevenLang Jun 01 '21 at 18:44
  • Oh sorry I missed your response. We don't replace balls so if we picked a blue ball out in say the first pick we only have 3 blue balls left, and the total would become 16. Then the probability of picking out a blue ball second pick would be 3/16. – StevenLang Jun 01 '21 at 18:45