1532. Find the probability that a randomly taken two-digit number will be divisible by either 2, or 5, or both at the same time.
Solution. Let $A$ be the event that a randomly chosen two-digit number is divisible by 2, and $B$ be the event that it is divisible by 5. We need to find $P(A+B)$. Since $A$ and $B$ are compatible events, we have

$$
P(A+B)=P(A)+P(B)-P(A B)
$$

Two-digit numbers are $10,11, \ldots, 98,99$. There are 90 of them in total. Clearly, 45 of them are divisible by 2 (favoring the occurrence of $A$), 18 are divisible by 5 (favoring the occurrence of $B$), and finally, 9 are divisible by both 2 and 5 simultaneously (favoring the occurrence of $A B$). Therefore,

$$
P(A)=\frac{45}{90}=0.5 ; P(B)=\frac{18}{90}=0.2 ; P(A B)=\frac{9}{90}=0.1
$$

and, consequently,

$$
P(A+B)=0.5+0.2-0.1=0.6
$$