[Combinatorics]
Problem: Two fair dice, each having six faces numbered 1 to 6 , are thrown. What is the probability that the product of the two numbers on the top faces is divisible by 5 ?

Solution: There are 36 possibilities for the pair of numbers on the faces when the dice are thrown. For the product of the two numbers, each of which is between 1 and 6 , to be divisible by 5 , one of the two numbers must be equal to 5 .

Therefore, the possible pairs for the faces are

$$
(1,5),(2,5),(3,5),(4,5),(5,5),(6,5),(5,1),(5,2),(5,3),(5,4),(5,6)
$$

ie. there are 11 possibilities.

Thus, the probability is $\frac{11}{36}$.
For the product of the two numbers, each of which is between 1 and 6 , to be divisible by 5 , one of the two numbers must be equal to 5 .

When the two dice are thrown, the probability that the first die has a 5 on the top face and any number appears on the second die has any number on the top face is $\frac{1}{6} \times 1=\frac{1}{6}$.

Also, the probability that any number appears on the first die and a 5 appears on the second die is $1 \times \frac{1}{6}=\frac{1}{6}$.



If we consider the sum of these probabilities, we have double-counted the possibility that a 5 occurs on both dice, which happens with probability $\frac{1}{6} \times \frac{1}{6}=\frac{1}{36}$.

Therefore, the required probability is $\frac{1}{6}+\frac{1}{6}-\frac{1}{36}=\frac{11}{36}$.

Final answer: $\frac{11}{36}$