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$12$ identical balls are to be put into $3$ identical boxes.Find the probability that one of the boxes contain exactly $3$ balls.

The question would have been easy had both balls and boxes were distinct.It would have been plain enough had the boxes been distinct and balls been identical.

But both boxes as well as balls are identical leads to over counting of cases and therein lies the difficulty.

Maverick
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  • JEEMAIN4/4/15..? – RE60K Apr 05 '15 at 03:57
  • Yes, with all wrong options. – Arpan Apr 05 '15 at 03:57
  • for identical into identical "there is no real formula to calculate , but for small numbers you may count it by hand." – RE60K Apr 05 '15 at 03:59
  • If there is a question then there must be a solution even though options given are incorrect – Maverick Apr 05 '15 at 04:28
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    We must be given more information. For example if the balls are thrown one at a time, with each of the boxes equally likely, and the results independent, then the "identical" conditions are irrelevant. This is the most plausible kind of model, but at the other extreme we could assume that all distinguishable distributions are equally likely, – André Nicolas Apr 05 '15 at 05:18
  • @AndréNicolas in probability the outcome mostly doesn't depends upon the method used to acheive that. – RE60K Apr 05 '15 at 05:36
  • @ADG: One physically interesting example is Bose-Einstein Statistics. You used an analogous interpretation. In the (I think more natural) model I described in the comment, the $19$ outcomes of your answer are not equally likely. – André Nicolas Apr 05 '15 at 05:47
  • This question is a duplicate. There is an accepted answer here: http://math.stackexchange.com/questions/1219692/12-identical-balls-can-be-placed-into-3-identical-boxes/1219885#1219885 – Christian Blatter Apr 05 '15 at 11:26

2 Answers2

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12 identical balls into 3 identical boxes: $$\begin{array}{r|l} (0,0,12),(0,1,11),(0,2,10),\color{red}{(0,3,9)},(0,4,8),(0,5,7),(0,6,6)&7&\color{red}1\\ (1,1,10),(1,2,9),\color{red}{(1,3,8)},(1,4,7)(1,5,6)&5&\color{red}1\\ (2,2,8),\color{red}{(2,3,7)},(2,4,6),(2,5,5)&4&\color{red}1\\ \color{red}{(3,3,6)},\color{red}{(3,4,5)}&2&\color{red}2\\ (4,4,4)&1&\color{red}0\\\hline \text{Total}&19&\color{red}5 \end{array}$$ What a tasty prime number! :D. Favourable are coloured in red. Thus probability is a weird number: $$P=\frac5{19}$$

Brian M. Scott
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RE60K
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  • Consider a pair of identical dice .Can you say the pair (5,6) is as equally likely to turn up as the pair (3,3). I doubt whether it is as simple as this.Is formation of (4,4,4) as likely as formation of (3,3,6). – Maverick Apr 05 '15 at 04:30
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    I think there's an issue with you method. Consider, for example the triplet $(1,2,10)$, which can occur in $3!=6$ different ways (by moving the numbers $1,2$ and $10$). On the other hand, the triplet $(4,4,4)$ can occur only one way. It follows that they do not have the same probability, therefore you cannot count favorable cases and total cases and divide since it is not equiprobable (the way you've written it down). – Reveillark Apr 05 '15 at 04:55
  • @Reveillark since all boxes are identical 3! doesn't matter. – RE60K Apr 05 '15 at 05:04
  • @PankajSinha yes because both balls and bins are identical. – RE60K Apr 05 '15 at 05:05
  • @ADG What I meant with the $3!$ is that different triplets can occur a different number of times. For the probability of $A$ to be ${|A| \over |\Omega|}$, all elements in the sample space must have the same probability. $(4,4,4)$ and $(1,2,10)$ do not have the same probability. – Reveillark Apr 05 '15 at 05:07
  • no the have same probability i.e. $(1/3)^4(1/3)^4(1/3)^4$ and $(1/3)^1(1/3)^2(1/3)^{10}$ – RE60K Apr 05 '15 at 05:22
  • @Reveillark did you read the above comment? – RE60K Apr 05 '15 at 05:27
  • @ADG I did. Did you read mine? $(1,2,10)$ can happen in $3!$ ways, you can have the first box with 1, the second with 2, and the third with 10, or the first one with 2, the second one with 1, and the third one with 10, and so on, hence it has probability ${3! \over {(9+2)! \over 9!2!}}$. However, $(4,4,4)$ can only occur in one way: each box must contain 4 balls. Hence it has probability ${ 1 \over {(9+2)! \over 9!2!}}$. Regarding your second comment, I do not understand what you are doing, but in any case the two numbers you've written down are not the same. – Reveillark Apr 05 '15 at 14:43
  • @Reveillark how would you identify which box is first and which second or which third all are identical so any permutation of an arrangement would according to me counted as a single. – RE60K Apr 05 '15 at 14:47
  • @ADG I understand you can count it as a single, but then they are not equiprobable anymore. If you throw two identical dices, it is more probable to get a 2 and a 3, or double sixes? – Reveillark Apr 05 '15 at 14:54
  • @Reveillark both are equally probable. – RE60K Apr 05 '15 at 14:57
  • They are not, because you can obtain a 2 and a 3 in two different ways, while double 6 in only one way. You can get, (2,3) or (3,2) but only (6,6). In this connection: https://books.google.com.ar/books?id=pOQy6-qnVx8C&pg=PA41&lpg=PA41&dq=dice+galileo+probability&source=bl&ots=GTuPSbds_o&sig=CxvXimslrovZlFvK8E9XiB-tq5Y&hl=es&sa=X&ei=9U0hVaD9N8GXgwTVpIHADQ&ved=0CC0Q6AEwAg#v=onepage&q=dice%20galileo%20probability&f=false – Reveillark Apr 05 '15 at 15:02
  • @Reveillark you might be thinking that coming of same number on two die is less probable but how can the outcome of one dice affect the other? it is $(1/6)^2$ for both – RE60K Apr 05 '15 at 15:02
  • @Reveillark ok you are right then – RE60K Apr 05 '15 at 15:03
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Another approach:

Define the events:

$$A_i=\text{the $i$-th box contains exactly three balls}$$

for $i=1,2,3$

The the desired event is $A_1\cup A_2\cup A_3$. By the inclusion-exclusion principle, and by reasons of symmetry:

$$P(A_1\cup A_2\cup A_3)=P(A_1)+P(A_2)+P(A_3)-P(A_1 \cap A_2)-P(A_1 \cap A_3) -P(A_2 \cup A_3)+P(A_1\cap A_2\cap A_3)=3P(A_1)-3P(A_1\cap A_2)$$

(the probability $P(A_1 \cap A_2 \cap A_3)$ is trivially zero).

Observe now that, fixing three balls in the first box and moving around the remaining nine among the other two boxes we get: $$P(A_1)={{(9+1)! \over 1!9!} \over {(9+2)! \over 9!2!}}={2 \over 11}$$

And $A_1 \cap A_2$ means that there are $6$ balls in the third box, hence:

$$P(A_1\cap A_2)={1 \over {(9+2)! \over 9!2!}}={1 \over 55}$$

Hence the desired probability is $$3{2 \over 11}-3{1 \over 55}={27 \over 55}$$

Reveillark
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