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An experiment is to toss two balls into four boxes in such a way that each ball is equally likely to fall in any box. Let $X$ denote the number of balls in the first box. Find the probability when $X=1$.

I think it's just a stars and bar problem. We know however we throw the ball, there are in total $10$ ways. The number of ways that there is $1$ ball in first box is clearly $3$. So I think $P(X=1)=\frac{3}{10}$. However the answer says it is $\frac{6}{16}$. Where is my error?

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Since each outcome is equally likely, we can simplify this into a counting problem.

There are 4 ways to place the first ball, and also 4 ways to place the second. So there are 16 total outcomes.

Now we need to count the number of ways where $X=1$. Given that the first ball went into box 1, there are 3 ways to place ball 2. Likewise, given the second ball went into box 1, there are 3 ways to place ball 1.

So the total probability is $$ p = \frac{3 + 3}{4 \times 4} = \frac{6}{16} $$

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It is not a stars and bars problem because that method counts distinct arrangements when we cannot distinguish between "stars", but these are not each equally probable outcomes for the physical system described by "dropping balls into boxes".

In short, while the observer may not be able to identify distinquishing features among the balls, they are actually physically discrete objects.


Vis: Take two boxes and two balls.   The stars and bars outcomes are **|, *|*, and |**.

Now: Labelling the balls we find that the equally probable outcomes are 12|, 1|2, 2|1, |12.   Since there are $2^2$ ways to assign one from two boxes independently for each of two balls.

tl;dr never use the stars-and-bars approach in probability calculations involving balls dropping in boxes, or such.   That never works.


For our four box, two balls, problem: There are two ways to select a ball to go in the preferec box, and it will do so with probability $1/4$ while the unselected ball will go into a different box with probability $3/4$.   So the required probability is $$\dfrac{2\cdot 3\cdot 1}{\quad 4\cdot 4}~=~\dfrac 3 8$$

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There are sixteen possibilities, two choices for 4 boxes or 2*2*2*2. 3/4 of the time the first ball does not land in box 1, so (3/4)^2 of the time the ball does not land in A both times, rewritten as 9/16. 1/16 of the time both balls will go in box 1, which is also invalid. We have 10 invalid possibilities out of 16, so we get 6/16 of the time exactly 1 lands in the first box.

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