Tossing n balls into n bins and proving independence

Question

Suppose we toss $n$ balls into $n$ bins with all outcomes equally likely. For clarity, assume the balls each have a unique number written on them, from $1$ to $n$, and also assume the bins are numbered from $1$ to $n$. Let $D$ be the event that all of the n balls land in different bins, $E_1$ be the event that the ball number $1$ lands in the first bin, and $E_2$ be the event that ball number $2$ lands in the second bin.

I have already proved that $D, E_1, E_2$ are all pairwise independent, but how would I go about proving that $D, E_1, E_2$ are not mutually independent so $ P(D \cap E1 \cap E2) \neq P(D)P(E1)P(E2) $ and that $E_1$ and $E_2$ are not conditionally independent given $D$.

score 2 · Accepted Answer · answered Mar 09 '17 at 16:36

2

You can just do the computation for $n=2$. Note that in that case if $E1$ and $E2$ are both true then $D$ is true automatically. For independence you should have $P(D \cap E1 \cap E2) \neq P(D)P(E1)P(E2)$, but the left side is $\frac 14$ and the right is $\frac 18$

answered Mar 09 '17 at 16:36

Ross Millikan

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That makes sense, but then how would I show they are not conditionally independent given D? – user1775500 Mar 09 '17 at 16:45
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The same way. In the $n=2$ case, if $D$ and $E1$ are both true, $E2$ is automatically true. – Ross Millikan Mar 09 '17 at 16:46

Tossing n balls into n bins and proving independence

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