In a basket, there are 10 tokens, 5 of which are blank, while the others are labeled with the numbers 1, 2, 3, 4, 5. We draw a token three times in a row, returning the drawn token before the next draw. What is the probability that the sum of the numbers on the 3 drawn tokens is 10? What is the probability if the drawn token is not returned?
a) If we put the marbles back, the number of possible cases is $l=V_{10}^{i, 3}=10^{3}$. Favorable cases are the permutations of $0+5+5, 1+4+5$, $2+3+5, 2+4+4$, and $3+3+4$. However, each permutation of the first group can occur in 5 different ways because there are 5 unmarked marbles. Therefore, the number of favorable cases is $k=5 \cdot P_{3}^{2}+P_{3}+P_{3}+P_{3}^{2}+P_{3}^{2}=7 \cdot 3+2 \cdot 3!=21+12=33$. Thus, the sought probability is

$$
v_{a}=\frac{k}{l}=\frac{33}{1000}=0.033
$$

b) If we do not put the marbles back, the number of possible cases (since repetition cannot occur, and we distinguish the 5 unmarked marbles from each other) is $l=V_{10}^{3}=10 \cdot 9 \cdot 8=720$. The number of favorable cases is now only the permutations of $1+4+5$ and $2+3+5$, that is, $k=2 \cdot P_{3}=2 \cdot 3!=12$. Thus, the sought probability is

$$
v=\frac{k}{l}=\frac{12}{720}=\frac{1}{60} \approx 0.017
$$

In case b), we can work more simply with combinations:

$$
l=C_{10}^{3}=\binom{10}{3}=\frac{10 \cdot 9 \cdot 8}{1 \cdot 2 \cdot 3}=120, \quad k=2
$$

and thus the sought probability is

$$
v=\frac{k}{l}=\frac{2}{120}=\frac{1}{60}
$$

Mihály Makkai (Bp., V., Eötvös József g. I. o. t.)