# $12$ Identical balls can be placed into $3$ identical boxes,

$12$ 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.

plz help me , How can I get correct answer.

Thanks

• Jee Mains 2015? In my opinion the options were wrong. – Arpan Apr 4 '15 at 10:06
• 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 4 '15 at 11:53

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 $$P={N\over 3^{12}}\doteq0.531536\ .$$

• What do you mean by histories? – user34304 Apr 8 '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 8 '15 at 14:08
• @Christian Blatter Have you taken both boxes and balls to be distinct ? – ViX28 Apr 4 '16 at 6:29
• @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 4 '16 at 7:55

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.

• 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 4 '15 at 14:41
• @BrianM.Scott I see, thanks for pointing that out. – DanielV Apr 4 '15 at 14:43
• @BrianM.Scott What do you mean by 'assuming final results are equally likely'? – user34304 Apr 8 '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 8 '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 4 '16 at 6:33

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

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