## Task A-4.5.

In a certain school, students can learn two classical languages: Latin and Greek. Out of 100 students, 50 study Latin, 40 study Greek, and 20 study both languages. If we randomly select two students, what is the probability that at least one of them studies Latin and at least one of them studies Greek?
## First Solution.

From the given information, we conclude that in the observed school, 30 students study only Latin, 20 students study only Greek, 20 students study both Latin and Greek, and 30 students do not study either of the mentioned languages.

Two randomly selected students satisfy the condition that at least one of them studies Latin and at least one of them studies Greek if one of the following is true:

a) at least one of the two selected students studies both Latin and Greek,

b) one student studies only Latin while the other student studies only Greek.

Since these are disjoint cases, the desired probability is obtained as the sum of the probabilities of the events defined under a) and b).

The probability under a) can be obtained by calculating the complement of that case, i.e., that among the two randomly selected students, none of them studies both Latin and Greek. Since there are 80 students in the school who do not study at least one of the two languages, the number of (unordered) pairs of students who do not study both languages is $\binom{80}{2}$. Since there are $\binom{100}{2}$ different pairs of students in total, we conclude that the probability of the event under a) is

$$
1-\frac{\binom{80}{2}}{\binom{100}{2}}
$$

For the case under b), the corresponding pairs of students must include one student who studies only Latin and one student who studies only Greek. There are $30 \cdot 20$ such pairs. Since there are $\binom{100}{2}$ pairs of students in total, the probability of the case under b) is

$$
\frac{30 \cdot 20}{\binom{100}{2}}
$$

We conclude that the probability that at least one student studies Latin and at least one student studies Greek when two students are randomly selected is

$$
1-\frac{\binom{80}{2}}{\binom{100}{2}}+\frac{30 \cdot 20}{\binom{100}{2}}=\frac{239}{495}
$$