What is the probability that there will be at least one empty box?

Question

Consider four boxes, numbered $1$ to $4$. We throw four balls in the boxes in such a way that each ball ends up in any particular box with probability $1/4$, and in such a way that different balls are thrown independently. What is the probability that there will be at least one empty box?

My method produced the wrong answer and I would like to know why and how this method is wrong. The correct answer is given by $\frac{29}{32}$.

What I did: Consider the event that precisely $i$ boxes are empty, $A_i$. The probability of the event that at least $1$ is empty is given by $P(A_1 \cup A_2 \cup A_3) = P(A_1) + P(A_2) + P(A_3) - P(A_1 \cap A_2) - P(A_2 \cap A_3) - P(A_3 \cap A_1)$ + $P(A_1 \cap A_2 \cap A_3)$ clearly the intersections correspond to the empty set because there are no outcomes where there are exactly $2$ empty boxes and exactly $1$ empty box for example. Hence,

$$ P(A_1 \cup A_2 \cup A_3) = P(A_1) + P(A_2) + P(A_3) $$

Where, For instance we define, $$ P(A_1) = \binom{4}{1} (\frac{3}{4})^{1}(\frac{1}{4})^{3} $$

Because we can think of it as performing $4$ different independent coin tosses with probability $\frac{3}{4}$ of "succeeding" and $\frac{1}{4}$ of not "succeeding" with $\binom{4}{1}$ ways of choosing a single empty box. Subsequently,

$$ P(A_2) = \binom{4}{2}(\frac{3}{4})^2(\frac{1}{4})^2 $$

and,

$$ P(A_3) = \binom{4}{3}(\frac{3}{4})^{3}(\frac{1}{3})^{1} $$

Their sum doesn't correspond to the correct answer. Any explanation on why the correct answer happens to be $\frac{29}{32}$ would also be appreciated.

Answer 1

The correct answer can be more easily computed using complementary counting. That is, we compute the probability that there are no empty boxes and subtract it from 1.

Since there are 4 balls and 4 boxes, there are no empty boxes iff each box contains exactly one ball. The first ball can go in any box, the second ball has a $\frac{3}{4}$ chance of going in a box not already chosen, the third ball has a $\frac{2}{4}$ chance of doing so, and the last ball has a $\frac{1}{4}$ chance. Hence, the probability that all boxes have exactly one ball is $\frac{3}{4}\cdot\frac{2}{4}\cdot\frac{1}{4}=\frac{3}{32}$. The answer is just the complement, or $\frac{29}{32}$.

As for your method, just because the balls are thrown independently doesn't mean the boxes being empty are independent. For example, if we consider the case for 2 boxes and 2 independently thrown balls, telling you that the first box is empty guarantees the second one is not; but if I don't tell you anything, then the second one could or could not be empty; their probability differs. The emptiness of the boxes depends on each other because there is only a certain total number of balls that can be thrown.

Generally, for probability questions on a discrete variable asking for "at least $1$," I would suggest looking at the "exactly $0$" case as a complement to simplify the calculations.

Answer 2

There are $4^4 = 256$ ways to put balls in boxes.

There are $4! = 24$ ways to put 1 ball in each box.

That means that there $256 - 24 = 232$ ways that at least one box is empty.

$\frac {232}{256} = \frac {29}{32}$

Another way to do this is inclusion-exclusion.

$3^4{4\choose 1} - 2^4{4\choose 2} + 4$

Or you can be exhaustive.

The number of balls in each box can be
4,0,0,0 -- $4$ ways to do this
3,1,0,0 -- ${4\choose 2}(2){4\choose 3} = 48$ ways
2,2,0,0 -- ${4\choose 2}{4\choose 2} = 36$ ways
2,1,1,0 -- ${4\choose 2}(2){4\choose 2}(2) = 144$ ways
1,1,1,1 -- $4! = 24$ ways

Answer 3

Your computation of $P(A_1)$, for example, seems bizarre to me. I do not understand what do you mean by

We can think of it as performing 4 different independent coin tosses with probability $3/4$ of 'succeeding' and 1/4 of not 'succeeding'.

What does "succeeding" correspond to here?

What I will do to compute $P(A_1)$ is to count directly: There are $4^4$ different ways for 4 balls to be in each box. First assume only box 1 is empty. Then exactly one of box 2,3,4 should contain two balls, while the rest should contain one, with the order distinguished, which gives us $$\binom{4}{2} \cdot2 \cdot3=36$$ differen possibilities. But since there are four ways for exactly one box to be empty, we simply have $$P(A_1)=\frac{4\cdot 36}{4^4}=\frac{9}{16}.$$

But as already mentioned in other comments, it is the easiest to solve this problem is by complementary counting.