5. A. A fair cube die has the numbers $1, 2, 3, 4, 5, 6$ on its six faces. When the die is rolled twice, let the remainders when the sum of the numbers on the top faces is divided by 4 be $0, 1, 2, 3$ with probabilities $P_{0}, P_{1}, P_{2}, P_{3}$, respectively. Then the largest among $P_{0}, P_{1}, P_{2}, P_{3}$ is $(\quad)$.
(A) $P_{0}$
(B) $P_{1}$
(C) $P_{2}$
(D) $P_{3}$
5. A. D.

Rolling a die twice, the ordered pairs formed by the numbers on the two faces up total 36, and the remainders when their sums are divided by 4 are $0, 1, 2, 3$, with 9, 8, 9, and 10 ordered pairs respectively. Thus,
$$
P_{0}=\frac{9}{36}, P_{1}=\frac{8}{36}, P_{2}=\frac{9}{36}, P_{3}=\frac{10}{36} .
$$

Therefore, $P_{3}$ is the largest.