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Three balls are randomly placed in three empty boxes $\{B_1,B_2,B_3\}$. Let $N$ denote the total number boxes which are occupied and let $X_i$ denote the number of balls in the box $B_i$ where $i\in \{1,2,3\}$

Find the joint p.m.f. of $(N,X_1)$

Should the answer depend on whether the balls are alike or different ? What should be the answer. ?

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    $\begingroup$ Doesn't matter if the balls are alike, you are only asked about their numbers. As to the problem, just go case by case. If, say, $N=3$, then what can you say about $X_1$? $\endgroup$ – lulu Mar 30 '16 at 13:29
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The problem is not well-defined, since it says "randomly" without specifying a distribution. One might assume that when placing balls in boxes, each ball is equally likely to be placed in any given box. In this case it doesn't matter whether the balls are all alike, since this doesn't affect how they are placed. By contrast, if the balls are randomly placed such that each occupancy pattern of the boxes is equally likely, then it does matter whether occupancy patterns distinguish between different balls.

In case of the more likely intended meaning, where each ball is equally likely to be placed in any given box, there are $3^3=27$ possible placements, and for each pair of $N$ and $X_1$ you can count how many of them yield those values to arrive at the following joint probability mass function (multiplied by $27$ to avoid fractions):

\begin{array}{c|cc} X_1\setminus N&1&2&3\\\hline 0&2&6&0\\ 1&0&6&6\\ 2&0&6&0\\ 3&1&0&0 \end{array}

For example, for $N=2$ and $X_1=2$, there are $2$ choices for the box that contains the third ball and $3$ choices for which ball that is, for a total of $3\cdot2=6$ possibilities.

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  • $\begingroup$ If we assume the balls to be same , there are 10 possible cases for arrangement of balls into the boxes . These 10 cases should be equally probable or not ? $\endgroup$ – sniperykc Mar 31 '16 at 18:24
  • $\begingroup$ @sniperykc: You tell me. As I said, the problem isn't well-defined because you didn't specify a distribution. Usually, though, one thinks of putting balls into boxes in a way such that each ball goes into each box with the same probability. In that case, the answer is no, the $10$ cases you're thinking of are not equiprobable. You can see this immediately from the fact that in this case there are $27$ equiprobable cases and $\frac1{10}$ isn't a multiple of $\frac1{27}$. $\endgroup$ – joriki Mar 31 '16 at 18:49

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