26. A fair cube die is rolled twice, and the sum of the numbers on the top faces is divided by 5, with the remainder denoted as $i$. The probability of this remainder being $i$ is denoted as $P_{i}$. Then the relationship between $P_{0}$ and $P_{4}$ is ( ).
(A) $P_{0}>P_{4}$
(B) $P_{0}<P_{4}$
(C) $P_{0}=P_{4}$
(D) Cannot be determined
Answer: (C)
Solution: The number of ordered pairs of real numbers formed by tossing a die twice is $6 \times 6=36$. The sum of the numbers on the faces that are up has 36 possibilities. Listing them, we get
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline 3 & 4 & 5 & 6 & 7 & 9 & 10 \\
\hline 4 & 5 & 6 & 7 & 8 & 10 & 11 \\
\hline 5 & 6 & 7 & 8 & 10 & 12 \\
\hline 6 & 7 & 8 & 9 & 10 & 7 \\
\hline
\end{tabular}

Among these, there are 7 cases where the remainder when divided by 5 is 0 and 7 cases where the remainder is 4,

Therefore,
$$
P_{0}=P_{4} \text { . }
$$