B hits a 10-ring target with probabilities of $\frac{1}{5}, \frac{1}{4}, \frac{1}{6}, \frac{1}{8}, \frac{1}{10}$ for 10, 9, 8, 7, 6 points respectively. The probability of other (from 5 to 1) hits is $\frac{1}{12}$. $A$ pays $B$ for every hit of at least 6 points the amount in forints equal to the score. For any other hit, $A$ pays 1.7 forints. How much should $B$ pay $A$ in case of a miss, so that the bet is fair?
The probability of a mistake according to the task is:

$$
1-\left(\frac{1}{5}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12}\right)=1-\frac{111}{120}=\frac{9}{120}=\frac{3}{40} .
$$

Suppose that in a fair bet, $B$ has to pay $x$ forints for each missed shot, then $B$'s expected profit is

$$
M=\frac{1}{5} \cdot 10+\frac{1}{4} \cdot 9+\frac{1}{6} \cdot 8+\frac{1}{8} \cdot 7-\frac{1}{10} \cdot 6+\frac{1}{12} \cdot 1.7-\frac{3}{40} x=0
$$

Multiplying both sides by 120

$$
240+270+160+105+72+17-9 x=0
$$

which means

$$
9 x=864
$$

from which

$$
x=96 \text{ forints. }
$$

Gáti Gyula (Debrecen, Chemical Engineering Technology II. year)