This question was given to me as a review for an upcoming exam:
$12$ baseballs are dropped into $5$ empty bins. Let $X$ be the number of empty bins after dropping the balls.
Calculate $\langle X \rangle$ (Expected value of $X$) and $P(X=4)$ and $P(X=3)$
My work thus far:
$$\langle X \rangle = \langle X_5 \rangle + \langle X_4 \rangle + \langle X_3 \rangle + \langle X_2 \rangle + \langle X_1 \rangle \\ \langle X \rangle = \left(\frac45 \right)^{12} \cdot (5+4+3+2+1) \approx 1.0308 \\ P(X=4)=5 \left(\frac15 \right)^{12} \approx 2.048 \cdot 10^{-8} \\ P(X=3)=\binom{5}{2} \left(\frac25 \right)^{12} \approx 0.000167 $$
Did I do anything incorrectly?