1. A die is rolled twice, and the sum of the points that come up is calculated and recorded. This procedure is repeated three times (a total of six rolls). Find the probability that only one of the three recorded sums is divisible by three. (12 points)
Solution. The sum of points with two throws equals 3 in two variants: $(1 ; 2)$ and (2; 1). The sum of points with two throws equals 6 in five variants: $(1 ; 5),(5 ; 1),(2 ; 4),(4 ; 2)$ and $(3 ; 3)$. The sum of points with two throws equals 9 in four variants: $(4 ; 5),(5 ; 4)$, $(3 ; 6)$ and ( $6 ; 3$ ). The sum of points with two throws equals 12 in one variant: ( $6 ; 6$ ). Thus, the total number of favorable outcomes in one procedure is 12. The probability of "success" $p=\frac{12}{36}=\frac{1}{3}$, the probability of "failure" $q=\frac{2}{3}$. The probability of one "success" in three procedures $P=3 \cdot \frac{1}{3}\left(\frac{2}{3}\right)^{2}=\frac{4}{9}$. Answer: $\frac{4}{9}$.