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?
 A: Imagine temporarily that each ball had its own color.  We see then that there are $4^4$ equally likely outcomes for where the balls are positioned.  We get as you already did the probabilities $P(A_0)=\dfrac{4!}{4^4}$ and $P(A_3)=\dfrac{4}{4^4}$
Now., for $P(A_1)$, count the numerator using the following multiplication principle argument:


*

*Pick which box was empty: $4$ choices

*Pick which box received two balls: $3$ choices

*Pick which two colored balls go into the box designated to receive two balls: $\binom{4}{2}$ choices

*Pick which colored ball goes into the left-most box designated to receive one ball: $2$ choices

*The final ball goes into the final box designated to receive one ball


This gives a total of $4\cdot 3\cdot \binom{4}{2}\cdot 2$ options here for a probability then of
$$P(A_1)=\dfrac{144}{4^4}$$
We can similarly calculate for the other case, taking advantage of the fact that the boxes are arranged from left to right, giving a total of $\binom{4}{2}\binom{4}{2} = 36$ if there were two boxes, each with two balls in them and $4\cdot 3\cdot 4=48$ if there was one box with three and another box with only one, for a total of $84$ outcomes and a probability of:
$$P(A_2)=\dfrac{84}{4^4}$$
Checking, $4!+144+84+4 = 256$ and so we do get our probabilities adding to one as expected.
A: 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
$$
 P(A_2) = \frac{1}{4} \left(\frac{1}{4} \cdot\frac{3}{4} + \frac{3}{4} \cdot \frac{1}{2} \right) + \frac{3}{4} \cdot\frac{1}{2} \cdot\frac{1}{2} = \frac{3 + 6 + 12}{64} = \frac{21}{64}.
$$
A: $P(A_0)=\frac{1}{35}$ because the only possibility is the every has exactly one ball.
$P(A_3)=\frac{4}{35}$ because there are 4 possibility to choose the one who isn't empty.
$P(A_1)=\frac{\binom{4}{1}\cdot3}{35}$ because the first binomial is the way to choose the one without the ball, and the way to give 4 balls to 3 people so that every of them has at least 1 ball.
$P(A_2)=\frac{\binom{4}{2}\cdot\binom{3}{2}}{35}$ because the first binomial is the way to choose the two without the ball, and the second binomial is the way to give 4 balls to 2 people so that every of them has at least 1 ball.
