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The following is a question I've made myself, but I need help in solving it:

Suppose there are 100 balls in a box. 20 balls are blue, 30 balls are green and 50 balls are yellow. Now we randomly pick out 10 balls out of the box (one ball after the other) and we don't put the balls back in the box.

What's the probability of the 10th ball being picked having color blue?

I tried thinking of all the possibilities of the the 10th ball being blue, divided by all of the possible combinations. I tried for several hours and couldn't figure out neither of them. Help is appreciated!

Blue
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    There's nothing special about the tenth ball. like all the others, the probability that it is blue is $\frac {20}{20+30+50}$. – lulu May 07 '16 at 15:12
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    @lulu I don't think so. Since he is not putting the balls back in the box, the probability will be different. – SS_C4 May 07 '16 at 15:25
  • Well, my thinking is somewhat along the lines of the answerer. – SS_C4 May 07 '16 at 15:31
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    The answer, as lulu pointed out, is $20/100$. But one can give more complicated-looking answers. After some computation, they will simplify to $20/100$. – André Nicolas May 07 '16 at 15:32
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    Without any knowledge of the color of the first 9 balls picked, lulu is correct. Also, whether you put them back or not is irrelevent if you do not know what color they were. – dansalmo May 07 '16 at 17:54
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    This became clear to me when I imagined the balls randomly arranged in a row. This question is equivalent to asking about the color of the 20th ball. – MackTuesday May 07 '16 at 20:12
  • Lulu is right. It's no different from the odds of the first. 1 in 5. – JTP - Apologise to Monica May 07 '16 at 23:45
  • Nice question. And many nice solutions. So does this mean that order does not matter in selection without replacement? Are there cases when it does matter? And when does replacement matter? – Hypergeometricx May 08 '16 at 07:42
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    @hypergeometric: If you were to, say, only put balls of some color back into the box, then the order would obviously matter. The order would also matter if you asked a question conditioned on some observation about the previous balls being true, like "If none of the first 9 balls is blue, what is the probability that the 10th ball will be blue?" – Ilmari Karonen May 08 '16 at 12:02
  • @IlmariKaronen - yes that's right! looks like probability problems of choosing coloured balls with/without replacement is no trivial matter! some of these rules should be formalised. – Hypergeometricx May 08 '16 at 17:50

9 Answers9

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Suppose that instead of picking ten balls, you pick all 100 balls and put them into a row in the order you picked them. Every one of the $100!$ possible orders is equally likely, and $20\cdot99!$ have a blue ball in the 10th position. Therefore the probability is exactly $\frac{1}{5}$.

f''
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    @jordan178 Note also that this argument can be applied in the same way to the ball in the first position, which demonstrates (again) what "lulu" says in the top comment... – Benjamin Dickman May 08 '16 at 05:04
  • Nice solution (+1). The probability of getting a blue ball as the 100th ball (if 100 balls are drawn) is also $0.2$. – Hypergeometricx May 08 '16 at 07:39
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Interesting question.

WLOG, assume that

(1) the balls are either blue (B) or not blue (N), and

(2) there are only two steps:

  • (i) Step One: the first 9 balls are chosen in one go ;

  • (ii) Step Two: the 10th ball is chosen.

Step One:

The probability of having $i \;(0\le i\le 9)$ blue balls out of $9$ chosen balls is $$\binom {20}i\binom {80}{9-i}\bigg/\binom {100}9$$. This leaves $20-i$ blue balls left, and $91$ balls left in total.

Step Two:

The probability of choosing a blue ball as the 10th ball is $\frac{20-i}{91}$.

In combination, the probability of choosing a blue ball as the 10th ball is $$\sum_{i=0}^9 \binom {20}i\binom {80}{9-i}\frac{20-i}{91}\bigg/\binom{100}9=0.2\qquad\blacksquare$$

EDIT: Just changed the lower limit of the summation from $1$ to $0$ and the result is $0.2$ (!).

Hypergeometricx
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Since you are (supposedly) treating all (remaining) balls equivalently at each draw, each ball has the same probability of being drawn at the 10th draw. Since the sum of these probabilities over all balls is $100\%$ (you are certain to draw exactly one of the balls as 10th) the probability for each individual ball must be exactly $100\%\div100=1\%$. Since there are $20$ blue balls, the probability that the ball drawn as 10th is one of those is $20\times1\%=20\%$.

This same symmetry argument applies regardless of the complications of the selection procedure used, as long as no information about its results are provided (if we were told the first 9 balls drawn were all blue that would certainly change the odds), and as long as the procedure is fair: it does not discriminate the balls in any way (which would for instance not be the case is say we put back a drawn ball if and only if it is blue).

Marc van Leeuwen
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Number the balls and ask: what is the probability to have ball number k as the 10th ball drawn?

This probability must be the same for every ball because none of them is favored, also probabilities must add up to 1; so it is $\frac{1}{100}$. There are 20 blue balls, each of them having $\frac{1}{100}$ chance to be the 10th.

Hence $\frac{20}{100}=\frac{1}{5}$

Kamil Maciorowski
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Let $B_i$ the event that we pick $i$ blue balls in the 9 first pick and $A$ the event that 10th picked ball is blue then using the total probability we have

$$P(A)=\sum_{i=0}^9P(A|B_i)P(B_i)$$

and we have

$$P(A|B_i)=\frac{20-i}{91}$$ and $$P(B_i)=\frac{{9\choose i}A_{20}^i\times A_{80}^{9-i}}{A_{100}^9}$$

user296113
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WLOG and to make it easy for us we can look at the $11$th ball instead.

On average you have picked $5$ yellow, $3$ green and $2$ blue balls on your first $10$ picks. Hence, when you are about to pick the $11$th ball there are $90$ balls in the box: $45$ yellow, $27$ green and $18$ blue balls.

The probability you will pick a blue ball on your next pick is

$$\frac{18}{18+27+45} = 0.2$$

JKnecht
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  • Yes, i explicitly wrote that we had picked 10 balls and only 90 remained in the box because you doubted the logic in lulu's comment. – JKnecht May 07 '16 at 16:22
  • @jordan178 You can think of it like this. You wrote "I tried thinking of all the possibilities of the the 10th ball being blue" This boils down to the average. – JKnecht May 07 '16 at 16:28
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If you are using JKnecht's way you will get the 10th ball probability on your next pick ranging from $$\frac{11}{11+30+50}\approx 0.1209$$ (if you always got the blue balls before) and $$\frac{20}{20+30+41}\approx 0.2198$$ if you always got the yellow balls. $0.2$ is not the answer then. The probability decreases as the blue balls are get picked out. The probability formula is $\frac{20-\mathrm{picked\,blue\,balls}}{91}$.

vps11
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I think the difficulty of this problem lies in the tricky concept of exchangeability. Because we don't put balls 1-9 back, the draws are not i.i.d. (the next draw is dependent upon the previous ones), but they are exchangeable and thus the probability $p(x_i = blue)$ for the $i^{th}$ ball should be the same regardless of which index is $i$.

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Suppose that you are blind folded and the balls are arranged in a row. Without seeing, you count the first 9 balls and as you touch the tenth ball open your eyes. What is the probability you will see a blue ball? This is same as the probability that the tenth ball is blue. It is now obvious that the answer is $\frac{1}{5}$.