William has a pair of identical, never cubes. The faces of each cube are numbered one through six. William will rule the cubed one time. What is the probability that the numbers showing Face-up after the role will have a sum of nine.
A. 1/18
B. 1/9
C. 3/4
D. 8/9
We can use a table to list all the possible outcomes of rolling the two identical cubes:
\[\begin{array}{ccc} \textbf{Cube 1} & & \textbf{Cube 2} \\ 1 & & 1 \\ 1 & & 2 \\ 1 & & 3 \\ 1 & & 4 \\ 1 & & 5 \\ 1 & & 6 \\ 2 & & 1 \\ 2 & & 2 \\ 2 & & 3 \\ 2 & & 4 \\ 2 & & 5 \\ 2 & & 6 \\ 3 & & 1 \\ 3 & & 2 \\ 3 & & 3 \\ 3 & & 4 \\ 3 & & 5 \\ 3 & & 6 \\ 4 & & 1 \\ 4 & & 2 \\ 4 & & 3 \\ 4 & & 4 \\ 4 & & 5 \\ 4 & & 6 \\ 5 & & 1 \\ 5 & & 2 \\ 5 & & 3 \\ 5 & & 4 \\ 5 & & 5 \\ 5 & & 6 \\ 6 & & 1 \\ 6 & & 2 \\ 6 & & 3 \\ 6 & & 4 \\ 6 & & 5 \\ 6 & & 6 \end{array}\]
There are 36 possible outcomes in total. To find the ones that have a sum of nine, we can simply look for the pairs of numbers that add up to nine:
\[\begin{array}{ccc} \textbf{Cube 1} & & \textbf{Cube 2} \\ 3 & & 6 \\ 4 & & 5 \\ 5 & & 4 \\ 6 & & 3 \end{array}\]
There are four pairs in total, so the probability of rolling a sum of nine is 4/36, which simplifies to 1/9. Therefore, the answer is $\boxed{\textbf{(B)}\ 1/9}$.