49. There are 100 lottery tickets. It is known that 5 tickets have a prize of 20 rubles, 10 tickets have a prize of 15 rubles, 15 tickets have a prize of 10 rubles, 25 tickets have a prize of 2 rubles, and the rest have nothing. Find the probability that the purchased ticket will have a prize of at least 10 rubles.
Solution. Let $A, B$ and $C$ be the events that the purchased ticket wins 20, 15, and 10 rubles, respectively. Since events $A, B$ and $C$ are mutually exclusive, then

$$
P(A+B+C)=P(A)+P(B)+P(C)=\frac{.5}{100}+\frac{10}{100}+\frac{15}{100}=0.3
$$