A ten-member society played every combination of seven numbers in the Pick-5 lottery, one per ticket. During the draw, three numbers were drawn from the seven.

a) In what percentage of all (different) draws does this occur?

b) How much profit does each member make if they receive 7000 forints for a three-number match and 300 forints for a two-number match, and a lottery ticket costs $60 \mathrm{Ft}$?
In the case of the 5-number lottery, every combination of 7 numbers can occur in $\binom{7}{5}=21$ ways. Therefore, the company had to buy 21 tickets.

In the draw, five numbers can be drawn from ninety in $\binom{90}{5}=43949268$ ways. In our case, three numbers were drawn from the seven, and the remaining two were drawn from the other 83 numbers, so the number of favorable cases is:

$$
\binom{7}{3} \cdot\binom{83}{2}=35 \cdot 3403=119105
$$

The answer to question a) is: 119105 is 0.281% of 43949268, which is the percentage of such draws.

b) Let's calculate how many 3-hit and 2-hit tickets there were. Three numbers were drawn from the 7 numbers that were played. On the 3-hit tickets, all 3 numbers had to be present, and the remaining 2 numbers were chosen from the other 4 that were not drawn. The number of 3-hits is thus: $\binom{3}{3}\binom{4}{2}=6$. This resulted in a total of 42000 Ft being won by the players. The number of 2-hit tickets is: $\binom{3}{2}\binom{4}{3}=3 \cdot 4=12$, and this amounts to 3600 Ft.

The tickets cost $60 \cdot 21=1260 \mathrm{Ft}$, so the total profit was 44340 Ft, and since ten people played, each person won 4434 Ft from the game.

Hortobágyi Ágnes (Dunaújváros, Széchenyi I. Gymn., III. o.t.)