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$12$ Identical balls can be placed into $3$ identical boxes, Then find probability that one of

the boxes contain exactly $3$ balls.

$\bf{My\; Try::}$ First we select $1$ bag out of $3$ and then put

$3$ balls into that bag and then put remaining balls into $2$ bag.

Which can be done by $\displaystyle \binom {3}{1}\times 1 \times \binom{1}{1}\times 1 = 3$

bcz here balls are identical . So we can arrange by only one ways.

But my answer is wrong.

plz help me , How can I get correct answer.

Thanks

Not a Salmon Fish
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juantheron
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    Jee Mains 2015? In my opinion the options were wrong. – Arpan Apr 04 '15 at 10:06
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    Your question is vague. What is the probability distribution of the placement of the balls? In other words, what mechanism was used to put the balls in the boxes? The most obvious way is to take the first ball and choose (at random with uniform probability) which box to put it in, then do that with the second ball, and so on. But there are other ways to place the balls in the boxes. Also, is "success" where at least one box has exactly three balls, or exactly one box has exactly three balls? – Rory Daulton Apr 04 '15 at 11:53

5 Answers5

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Take the simplest model: One after the other of the balls is thrown at random into one of the boxes $a$, $b$, $c$. There are $3^{12}$ different possible histories for that, all of them equiprobable. The number of histories leading to a particular final content $(r,s,t)$ of the three boxes is the coefficient of the term $a^r b^s c^t$ in the expansion of $(a+b+c)^{12}$, i.e., is given by ${12!\over r!\>s!\>t!}$.

There is the question whether "one of the boxes" means "at least one of the boxes", or "exactly one of the boxes". Since one sentence later they talk about "exactly three balls" my working hypothesis is that "at least one of the boxes" is meant.

For the probability in question we have to consider the contents $$(3,9,0), (3,8,1), (3,7,2), (3,5,4)$$ each of them in six orders, and $(6,3,3)$ in three orders. The total number $N$ of "admissible" histories is therefore given by $$N=6{12!\over3!}\left({1\over9!}+{1\over8!}+{1\over 7!\>2!}+{1\over 5!\>4!}\right)+3{12!\over 6!\>3!\>3!}=282\,480\ ,$$ and the required probability $P$ is $$P={N\over 3^{12}}\doteq0.531536\ .$$

Christian Blatter
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  • What do you mean by histories? – user34304 Apr 08 '15 at 13:02
  • @user34304: "One after the other of the balls is thrown at random into one of the boxes $a$, $b$, $c$." – Christian Blatter Apr 08 '15 at 14:08
  • @Christian Blatter Have you taken both boxes and balls to be distinct ? – ViX28 Apr 04 '16 at 06:29
  • @ViX28: For the counting of cases the balls as well as the boxes are distinct, but in the end I don't care about the numbers on the balls nor about which box(es) contain(s) exactly three balls. – Christian Blatter Apr 04 '16 at 07:55
  • @ChristianBlatter I am not getting why $3^{12}$ since balls are identical. I used to do problem when boxes were different and balls were identical then I used star and bars. Why $3^{12}$ here? – mathophile Dec 04 '22 at 07:02
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Suppose the experiment consists of randomly assigning $12$ balls to $3$ boxes. There are $3^{12}$ ways to do so. We want to find the probability that (at least) one of the boxes contains exactly $3$ balls. Let us number the boxes $1$, $2$ and $3$, and define $E_i$ as the event that box # $i$ contains exactly $3$ balls. We want to find the probability of the event $E_1 \cup E_2 \cup E_3$.

$\Pr(E_1 \cup E_2 \cup E_3) $ $= \Pr(E_1) + \Pr(E_2) + \Pr(E_3) - \Pr(E_1\cap E_2) - \Pr(E_1\cap E_3) - \Pr(E_2\cap E_3) + \Pr(E_1\cap E_2 \cap E_3)$ $= 3\Pr(E_1) - 3\Pr(E_1\cap E_2) + \Pr(E_1\cap E_2 \cap E_3) \ \ \ \ \color{blue}{\text{(By symmetry)}}$

We will now find $\Pr(E_1)$, $\Pr(E_1\cap E_2)$, and $\Pr(E_1\cap E_2 \cap E_3)$.

$\displaystyle \Pr(E_1) = {12\choose 3}\left(\frac{2^{9}}{3^{12}}\right)$

$\displaystyle \Pr(E_1\cap E_2) = {12\choose 3} {9\choose 3}\left(\frac{1^{6}}{3^{12}}\right)$

$\displaystyle \Pr(E_1\cap E_2 \cap E_3) = 0$

Therefore, $\Pr(E_1 \cup E_2 \cup E_3) $ $= 3\Pr(E_1) - 3\Pr(E_1\cap E_2) + \Pr(E_1\cap E_2 \cap E_3) $ $= \displaystyle 3\times {12\choose 3}\left(\frac{2^{9}}{3^{12}}\right) - 3 \times {12\choose 3} {9\choose 3}\left(\frac{1^{6}}{3^{12}}\right)$ $\approx 0.5315$

Here is the code to find the above value. You will also find the computation of the required probability using simulation method in the code: http://www.codeskulptor.org/#user42_g9uaFBlARwEdnL6_1.py

Amit
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4

Keep in mind that the bags are identical. So if you choose to put $3$ into one bag, then you have to put $9$ into the rest:

3 0 9
3 1 8
3 2 7
3 3 6
3 4 5

If you continued to 3 5 4 then you would have a duplicate, because the bags are identical. So there are $5$ ways to have at least $1$ bag with exactly $3$ balls and $4$ ways to have exactly one bag with exactly $3$ balls.

Since 12 is such a small number, it is easier to just enumerate the placements than to try some counting trick (although they do exist) :

0 0 12   1 1 10   2 2 8    3 3 6    4 4 4
0 1 11   1 2 9    2 3 7    3 4 5
0 2 10   1 3 8    2 4 6
0 3 9    1 4 7    2 5 5
0 4 8    1 5 6
0 5 7    
0 6 6

So there are $7 + 5 + 4 + 2 + 1 = 19$ total placement options.

DanielV
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    You’re assuming that the final results are equally likely. This is one interpretation of the problem, and I’ve upvoted it on that basis, but I think that Christian’s is a more likely interpretation. – Brian M. Scott Apr 04 '15 at 14:41
  • @BrianM.Scott I see, thanks for pointing that out. – DanielV Apr 04 '15 at 14:43
  • @BrianM.Scott What do you mean by 'assuming final results are equally likely'? – user34304 Apr 08 '15 at 13:01
  • @user34304: Assuming that, for instance, the 3 4 5 outcome is exactly as probable as the 0 0 12 outcome. This is not the case if, for example, we simply throw $12$ bags in such a way that each has probability $\frac13$ of ending up in each box. – Brian M. Scott Apr 08 '15 at 16:34
  • @BrianM.Scott why Christian Blatter's probablity does not match with DanielV's probability ($5/19$ )? I think Christian has taken both boxes and balls to be distinct instead of identical. am I right? – ViX28 Apr 04 '16 at 06:33
  • @ViX28: Because Christian and Daniel made different assumptions about the underlying distribution of the outcomes. Christian used the model that he described, of tossing $12$ balls in turn into three boxes, and then removing any identifying features from both. Daniel, on the other hand, assumed that each final distribution is equally likely; this is a possible assumption, but I can’t offhand think of a straightforward model experiment to which it would apply, so I take Christian’s answer to be almost certainly the intended one. \ No, Christian has assumed that the boxes and the balls ... – Brian M. Scott Apr 04 '16 at 06:43
  • ... are both indistinguishable. His calculation first works with distinguishable boxes and balls, but then he divides by the appropriate factors to get the result for indistinguishable boxes and balls – Brian M. Scott Apr 04 '16 at 06:43
  • ohh thanks for clarifying :) @BrianM.Scott but still that particular thing bugs me..it may irritate you. but can you again clarify that "equally likely distribution term" .. whats wrong in assuming $(3,4,5)$ is equiprobable as $(0,0,12)$ ..what I think, those are $19$ cases which can happen with equal probability and among which our needs satisfy 5 cases.. so $5/19$ .. how christian has avoided that equally likely thing? – ViX28 Apr 04 '16 at 07:13
  • @ViX28: I can’t think of any natural experimental design that would result in ${!!{3,4,5}!!}$ and ${!!{0,0,12}!!}$ occurring with equal probability. – Brian M. Scott Apr 04 '16 at 22:35
  • @BrianM.Scott: How about an experiment in which you instruct a child to place in all possible ways , balls into three non-descending heaps including those with no balls ? To the child, ${3,4,5}$ and ${0,0,12}$ are each one of the many such heaps, thus equi-probable ? – true blue anil Sep 19 '24 at 14:40
  • @trueblueanil: On consideration it seems to me that this in effect labels the boxes first, second, third. – Brian M. Scott Sep 21 '24 at 06:07
  • @BrianM.Scott: [1] My stipulation of non-descending heaps was to avoid duplication, so that ${3,4,5}$ e.g. would not be counted multiple times. Would you still consider the boxes as getting labeled ? Say three "buildings" made of heights $3,4,5$ with the blocks labeled $1-12$ and they are just $3$ heaps. [2] Why I am asking is that it seems paradoxical that the probability of putting distinct balls into identical boxes should be the same as putting distinct balls into distinct boxes, once we assume that balls are being placed uniformly at random in the boxes, and not by design – true blue anil Sep 21 '24 at 07:07
  • @trueblueanil: I understood that the stipulation was to avoid duplication, but I do see it as implicitly giving the boxes labels; I don’t understand your buildings sentence, I’m afraid, or where [2] is coming from. – Brian M. Scott Sep 21 '24 at 19:06
  • @BrianM.Scott: [1] Sorry for introducing confusing "buidngs" etc In a nutshell, for this particular question, if I have been instructed to make as many distinct heaps of $3$ from $12$ balls as possible, why can't we treat this as a possible real life experiment, and take each such heap as equiprobable instead of assumng that each ball has been placed uniformly at random in the boxes. [2] Thanks for sparing your valuable time – true blue anil Sep 23 '24 at 03:54
  • @trueblueanil: After much consideration I think that it will depend on one’s intuition. My feeling is that the process of making the three heaps in effect labels them, in the sense that interchanging two of them of different sizes produces a different arrangement, and I don’t see any way to word the instructions to avoid that short of writing eachof the satisfactory multisets on a slip of paper and drawing one of the slips from a bag. It seems that you’ve a different intuition here, and it may be that this is something on which general agreement is simply lacking. – Brian M. Scott Sep 28 '24 at 06:50
  • @BrianM.Scott: Thanks a lot, I go by your considered judgment, although as a novice in this field, it is puzzling why there is only one model for putting balls into boxes, which results in the apparent paradox that the probability remains the same regardless of balls/boxes being distinct or identical – true blue anil Sep 28 '24 at 07:54
  • @BrianM.Scott: By happenstance, recently encountered a problem in probability that goes to the heart of the matter. Why can't balls be put into equiprobable heaps by some given instruction/design instead of always being put in uniformly at random ? – true blue anil Oct 05 '24 at 08:03
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Probability does not depend on whether the boxes are identical or not because $(0,0,12)$ has more chances than the event $(4,4,4)$. So we will assume the boxes are different. Now let $A$ be the event that the first ball has exactly 3 balls. Similarly define $B$ and $C$. The required probability is $Pr(A\cup B\cup C) = 3Pr(A)-3P(A\cap B)$,(due to symmetry). Note that $P(ABC)=0$. Twelve identical balls can be distributed in 3 distinct boxes in $\binom{14}{2}= 91$ ways. This can be proved using dots and sticks. If $A$ contains 3 balls then remaining 9 balls can be placed in other two boxes in $\binom{10}{1}=10$ ways. Hence $Pr(A) = \frac{10}{91}.$ Similarly, $Pr(A\cap B) = \frac{1}{91}.$ Hence our required probability is $3\cdot \frac{10}{91}-\frac{3}{91}= \frac{27}{91}.$

Samir Ch. Mandal
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Required probability {(3c1*12c3*2 to the power 9) - (3c2*12c3*9c3)} ÷ (3 to the power 12) this is the right answer

John B
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CHANDRA SEN
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