In the illustrated toy, balls are placed at entrances $A, B$ or $C$ and always move downward, ending up in one of the boxes 1, 2, or 3. When they reach a point marked with $\mathbf{\Lambda}$, the balls have equal chances of going to either side.

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-07.jpg?height=420&width=534&top_left_y=658&top_left_x=675)

a) If a ball is placed at $C$, in which boxes can it end up? And if it is placed at $B$?

b) If a ball is placed at $A$, what is the probability that it will end up in box 2? And if it is placed at $B$, what is this probability?

c) If we place a ball at each entrance (one at a time), what is the probability that, in the end, there will be one ball in each box?
Solution

a) A ball placed at $C$ can only stop in boxes 2 or 3; if placed at $B$, it can stop in any of the boxes.

b) If it starts from $A$, to reach box 2 it must go right at both the first and second bifurcations. Since the ball has an equal chance of going right or left at each bifurcation, the probability of it reaching box 2 is $\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}$ or $25 \%$.

If the ball is placed at $B$, by the same reasoning, it can reach box 2 via two different paths: right, left or left, right; both occur with a probability of $\frac{1}{4}$. Since these events are disjoint, the probability of one of them occurring is the sum of the probabilities of each individual event. Therefore, the probability of the ball starting from $B$ and reaching box 2 is $\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$ or $50 \%$.

c) There are three possible scenarios for there to be a ball in each box at the end. We describe these scenarios in the table below, where (for example) the first row indicates the scenario in which a ball placed at $A$ falls into box 1, another placed at $B$ falls into box 2, and the last one, placed at $C$, falls into box 3.

|  | box 1 | box 2 | box 3 |
| :---: | :---: | :---: | :---: |
| $\mathbf{1}^{\mathbf{st}}$ scenario | $\mathrm{A}$ | $\mathrm{B}$ | $\mathrm{C}$ |
| $\mathbf{2}^{\mathbf{nd}}$ scenario | $\mathrm{A}$ | $\mathrm{C}$ | $\mathrm{B}$ |
| $\mathbf{3}^{\mathbf{rd}}$ scenario | $\mathrm{B}$ | $\mathrm{A}$ | $\mathrm{C}$ |

Observing that the events "ball placed at $X$ falls into box $Y$ " are independent and remembering that the probability of independent events occurring simultaneously is equal to the product of the probabilities of each event, the probability of each of these scenarios occurring is:

$$
\begin{aligned}
& 1^{\text{st}} \text { scenario: } \frac{3}{4} \times \frac{1}{2} \times \frac{3}{4}=\frac{9}{32} \\
& 2^{\text{nd}} \text { scenario: } \frac{1}{4} \times \frac{1}{4} \times \frac{3}{4}=\frac{3}{64} \\
& 3^{\text{rd}} \text { scenario: } \frac{3}{4} \times \frac{1}{4} \times \frac{1}{4}=\frac{3}{64}
\end{aligned}
$$

On the other hand, the occurrence of each of the configurations above is a disjoint event from the other two; the probability of at least one of them occurring is then equal to the sum of the probabilities of the individual events. Therefore, the probability that there is a ball in each box is

$$
\frac{9}{32}+\frac{3}{64}+\frac{3}{64}=\frac{24}{64}=\frac{3}{8}
$$

For reference, we list below the 12 possibilities for the distribution of three balls among the boxes and their respective probabilities.

| box 1 | box 2 | box 3 | probability |
| :---: | :---: | :---: | :---: |
| $\mathrm{A}$ | $\mathrm{B}$ | $\mathrm{C}$ | $18 / 64$ |
| $\mathrm{A}$ | $\mathrm{C}$ | $\mathrm{B}$ | $3 / 64$ |
| $\mathrm{A}$ | $\mathrm{BC}$ | empty | $6 / 64$ |
| $\mathrm{A}$ | empty | $\mathrm{BC}$ | $9 / 64$ |
| $\mathrm{AB}$ | empty | $\mathrm{C}$ | $9 / 64$ |
| $\mathrm{AB}$ | $\mathrm{C}$ | empty | $3 / 64$ |
| $\mathrm{B}$ | $\mathrm{A}$ | $\mathrm{C}$ | $3 / 64$ |
| $\mathrm{B}$ | $\mathrm{AC}$ | empty | $1 / 64$ |
| empty | $\mathrm{AB}$ | $\mathrm{C}$ | $6 / 64$ |
| empty | $\mathrm{ABC}$ | empty | $2 / 64$ |
| empty | $\mathrm{AC}$ | $\mathrm{B}$ | $1 / 64$ |
| empty | $\mathrm{A}$ | $\mathrm{BC}$ | $3 / 64$ |