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Problem:

If $n$ balls are distributed at random into $r$ boxes (where $r \geq 3$), what is the probability that box $1$ at exactly $j$ balls for $0 \leq j \leq n$ and box $2$ contains exactly $k$ balls for $0 \leq k \leq n$ ?

Answer:
Let $p$ be the probability that we seek. First we consider a special case. If $j + k > n$ then $p = 0$. Let $p_1$ be the probability that a ball is placed in box $1$. Let $p_2$ be the probability that a ball is placed in box $2$. Let $p_3$ be the probability that a ball is placed in box other than box $1$ and box $2$. \begin{align*} p_1 &= \frac{1}{r} \\ p_2 &= \frac{1}{r} \\ p_3 &= \frac{r-2}{r} \end{align*} Now we have a multinomial distribution. \begin{align*} P &= \left( \frac{n!}{j!k!(n-j-k)!} \right) \left( \frac{1}{r}\right) ^j \left( \frac{1}{r}\right) ^k \left( \frac{r-2}{r} \right)^{n - j - k} \\ P &= \left( \frac{n!}{j!k!(n-j-k)!} \right) \left( \frac{1}{r}\right) ^{j+k} \left( \frac{r-2}{r} \right)^{n - j - k} \\ P &= \left( \frac{n!}{j!k!(n-j-k)!} \right) \frac{ (r-2)^{n-j-k} } {r^n} \end{align*} So for example, if we have: $j = 2$, $k = 2$, $n = 8$ and $r = 8$ then \begin{align*} P &= \left( \frac{8!}{2!2!(8-2-2)!} \right) \frac{ (8-2)^{8-2-2} } {8^8} = \left( \frac{8!}{4(8-2-2)!} \right) \frac{ (6)^{4} } {8^8} \\ P &= \left( \frac{8!}{4(4)!} \right) \frac{ (6)^{4} } {8^8} = \left( \frac{8(7)(6)(5)}{4} \right) \frac{ 2^4(3^4) } {8^8} \\ P &= \frac{ 8(7)(6)(5)( 2^4)(3^4) } { 4(8^8) } = \frac{ 2(7)(6)(5)( 2^4)(3^4) } { 4(8^7) } \\ P &= \frac{ 7(6)(5)( 2^4)(3^4) } { 2(8^7) } = \frac{ 7(6)(5)( 2^3 )(3^4) } { (8^7) }\\ P &= \frac{ 7(6)(5)(3^4) } { 8^6 } = \frac{ 17010 } { 262144 } \\ P &= \frac{8505 } { 131072 } \\ P &\doteq 0.064888 \end{align*} I think I have it right now. Do I?

Noa Even
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Bob
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    Did you mean to write $0 \leq k \leq r$ for box $2$? – N. F. Taussig Jan 05 '20 at 21:58
  • @N.F.Taussig Yes I did. I will fix it. – Bob Jan 05 '20 at 22:12
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    There are some other typos. For instance, you meant $p_1 = \frac{1}{r}$ and the $m$ in the exponent in your second expression for $P$ should be an $n$. I assume $p_3$ is supposed to represent the probability that a ball is placed in a box other than boxes $1$ and $2$. You are missing a factor of $(n - j - k)!$ in the denominator of the multinomial coefficient. In your example, you are missing a factor of $(8 - 2 - 2)! = 4!$ in the denominator of your multinomial coefficient. – N. F. Taussig Jan 05 '20 at 22:23
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    The problem statement has n balls and r boxes, but the question has a limit of r balls per box, and your analysis has $0$ probability for $j+k\gt r$. You need to clarify. – herb steinberg Jan 05 '20 at 22:27
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    Following up on herb steinberg's comment, did you mean $0 \leq j \leq n$ and $0 \leq k \leq n$ since you have a total of $n$ balls? – N. F. Taussig Jan 05 '20 at 23:20
  • @N.F.Taussig That is what I meant. Herb's comments was right on. – Bob Jan 05 '20 at 23:57
  • @N.F.Taussig I believe that I have fixed all the mistakes you have found. Is my solution right now? – Bob Jan 05 '20 at 23:58
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    I believe it is. – N. F. Taussig Jan 06 '20 at 00:17
  • @N.F.Taussig I am thinking that you should answer my question, with the answer: I believe it is. Once you do that, I will accept your answer and you will get points. – Bob Jan 06 '20 at 14:06

1 Answers1

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You have corrected the mistakes that were your initial solution. Your work is now correct.

N. F. Taussig
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