When rolling two dice, the number of possible outcomes is $V_{6}^{i, 2}=6^{2}=36$.

a) Rolls with a value greater than 9: $6+6,6+5,5+6,6+4,4+6,5+5$, which means the number of favorable outcomes is 6, and thus the sought probability is

$$
v_{1}=\frac{6}{36}=\frac{1}{6}(\approx 0.167)
$$

b) There are 10 ways to roll less than 6: $1+1,1+2,2+1,1+3,3+1,2+2,1+4,4+1,2+3$, $3+2$. Since the number of favorable outcomes is 10, the probability of rolling less than 6 in one roll is: $\frac{10}{36}=\frac{5}{18}$. Since the 3 rolls are independent, according to the multiplication rule, the sought probability is

$$
v_{2}=\left(\frac{5}{18}\right)^{3}=\frac{125}{5832}(\approx 0.021)
$$

Beke Éva and Mária (Bp. XIII., Mechanical Engineering Technology II. class)
What is the probability with two dice

a) when rolled once, to roll more than 9?

b) when rolled three times in a row, to roll less than 6 each time?

(For the example, see the article titled 》Elements of Probability Theory《 in this issue.)