<s>[INST] Four balls are randomly dropped into four boxes Four balls are randomly dropped into four boxes, where any ball is equally likely to fall into
each box.
For a fixed $k = 0, 1, 2, 3$, let $A_k$ denote the event that exactly $k$ boxes are empty. Compute
$P(A_k)$ for each $k = 0, 1, 2, 3$.
I have computed $P(A_0)$ and $P(A_3)$ which are $\frac{4!}{4^4}=\frac{3}{32}$ and $\frac{4}{4^4}=\frac{1}{64}$ recpectively. But I don't know how to find $P(A_1)$ and $P(A_2)$. Could anybody help me?
 [/INST] I'll write out the computation "by hand" for $A_2$. There are slicker answers, but this doesn't require any clever observations.
First we drop the first ball, and it doesn't matter where it goes. Then we drop the second ball. There is a $1/4$ chance it lands in the same place as the first, and a $3/4$ chance otherwise. So
$$
P(A_2) = \frac{1}{4} P(A_2 | \text{same}) + \frac{3}{4} P(A_2 | \text{different}).
$$
If they landed in different places, then we don't care where balls 3 and 4 land as long as they land in the boxes already occupied. The probability of this is $1/2$ for each, and the events are independent, so
$$
   P(A_2) = \frac{1}{4} P(A_2 | \text{same}) + \frac{3}{4} \cdot\frac{1}{2} \cdot\frac{1}{2}.
$$ 
If the first two balls landed in the same place, then drop the third ball. Either it lands in the first place again, with probability $1/4$, or lands in a second place. If it lands in a second place, then we just need the fourth ball to land in one of the two already occupied.
$$
   P(A_2) = \frac{1}{4} \left(\frac{1}{4} P(A_2| 3 \text{same}) + \frac{3}{4} \cdot \frac{1}{2} \right) + \frac{3}{4} \cdot\frac{1}{2} \cdot\frac{1}{2}.
$$ 
If the first three balls all landed in the same place, then we need the fourth ball to land somewhere else, with probability $3/4$. So we have
$$
 P(A_2) = \frac{1}{4} \left(\frac{1}{4} \cdot\frac{3}{4} + \frac{3}{4} \cdot \frac{1}{2} \right) + \frac{3}{4} \cdot\frac{1}{2} \cdot\frac{1}{2} = \frac{3 + 6 + 12}{64} = \frac{21}{64}.
$$
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