Suppose $N$ blue balls, $N$ red balls and $N$ yellow balls are placed in $N$ boxes such that each box has $3$ balls. What is the expected number of boxes with a blue ball?
I know it's going to involve an indicator function, but ...
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I interpret "with a blue ball" as meaning at least one blue. Let $X_i=1$ if Box $i$ has at least one blue ball, and let $X_i=0$ otherwise. If $Y$ is the number of boxes with a blue ball, then $Y=X_1+\cdots+X_N$. So by the linearity of expectation, and symmetry, $E(Y)=NE(X_1)$.
We find the probability $p$ that Box $1$ has no blue ball. Think of the balls as distinct. There are $\binom{3N}{3}$ equally likely ways to choose the balls that will go into Box $1$. There are $\binom{2N}{3}$ ways to select non-blues. It follows that $$p=\frac{\binom{2N}{3}}{\binom{3N}{3}}.$$ Now $\Pr(X_i=1)=1-p$, and our required expectation is $N(1-p)$. One can "simplify."
Basically the same strategy can be used if we interpret "with a blue ball" as meaning exactly one blue.
André Nicolas
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