We place a coin in one of three identical boxes and put these three boxes into two accounts so that each account gets a box. What is the probability that if someone randomly selects an account and takes a box from it, they will find the coin,

a) if the box containing the coin is alone in an account?

b) if the box containing the coin is with another box in an account?

c) if the 3 boxes are placed randomly into the two accounts, without knowing which box contains the coin?
a) In this case, the drawing of the account already decides. The drawing of both accounts is equally probable, but only one of the drawings is favorable, and thus the sought probability is

$$
v_{1}=\frac{1}{2}
$$

b) The probability of drawing the account containing the coin is $\frac{1}{2}$, as in case a). If we have already drawn the favorable account, we still have to choose between the two boxes. The probability of selecting the coin-containing box is also $\frac{1}{2}$. Since the drawing of the account and the selection of the box are independent events, according to the multiplication rule, the sought probability is

$$
v_{2}=\frac{1}{2} \cdot \frac{1}{2}=\frac{1}{4}
$$

c) In this case, the coin has a $\frac{1}{3}$ probability of being in any of the boxes. There are two ways to find the coin:

1. We draw the account containing 1 box, which has a probability of $\frac{1}{2}$. The probability that the coin is in this box is $\frac{1}{3}$. Therefore, we find the coin with a probability of $\frac{1}{2} \cdot \frac{1}{3}=\frac{1}{6}$ in this way.
2. We draw the account containing 2 boxes, which has a probability of $\frac{1}{2}$. The probability that the coin is in this account is $\frac{2}{3}$. The probability of selecting the correct box is again $\frac{1}{2}$. Using the multiplication rule, the probability that we find the coin this way is $\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{1}{2}=\frac{1}{6}$.

Since the two cases above are mutually exclusive, we apply the addition rule (or-or) to find the sought probability:

$$
v_{3}=\frac{1}{6}+\frac{1}{6}=\frac{1}{3}
$$

We could also reason as follows: The three boxes now play completely equal roles, so the probability of finding the coin-containing box, regardless of the accounts, is the same as that of any other box, which is $\frac{1}{3}$.

However, the following line of thought is incorrect. We are either in case a) or case b). In the former case, we find the coin with a probability of $\frac{1}{2} \cdot \frac{1}{2}=\frac{1}{4}$, and in the latter case, with a probability of $\frac{1}{2} \cdot \frac{1}{4}=\frac{1}{8}$. Therefore, the sought probability is $\frac{1}{4} \cdot \frac{1}{8}=\frac{3}{8}$. The error lies in the fact that cases a) and b) are not equally probable, because the probability of case a) is $\frac{1}{3}$, while that of case b) is $\frac{2}{3}$. Indeed, $\frac{1}{3} \cdot \frac{1}{2}+\frac{2}{3} \cdot \frac{1}{4}=\frac{1}{6}+\frac{1}{6}=\frac{1}{3}$.