William has a pair of identical number cubes. The faces of each cube are numbered 1 through 6. William will roll the cubes one time. What is the probability that the numbers showing face-up after the roll will have a sum of 9?
A. Start Fraction 1 over 18 End Fraction
B. start fraction 1 over 9 end fraction
C. three-fourths
D. start fraction 8 over 9 end fraction.
There are a total of $6\cdot6=36$ possible outcomes when rolling the two cubes. We want to count the number of outcomes that have a sum of 9. We can organize these outcomes in a table:
\begin{tabular}{|c|c|c|c|c|c|}
\hline
1+8 & 2+7 & 3+6 & 4+5 & 5+4 & 6+3 \\
\hline
7+2 & 8+1 & & & & \\
\hline
\end{tabular}
We have listed all possible outcomes with a sum of 9. We can count a total of 8 outcomes, out of 36 possible outcomes, that have a sum of 9. Therefore, the probability of rolling a sum of 9 is $\boxed{\textbf{(B) } \frac19}$.
To find the probability that the numbers showing face-up after the roll will have a sum of 9, we need to determine the number of favorable outcomes (when the sum is 9) and the total number of possible outcomes.
First, let's list all the possible outcomes when rolling two number cubes:
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
Out of these, the favorable outcomes (when the sum is 9) are:
3, 6
4, 5
5, 4
6, 3
Therefore, there are 4 favorable outcomes.
Since there are 36 total possible outcomes (6 options for the first cube multiplied by 6 options for the second cube), the probability is 4/36.
Simplifying this fraction, we get 1/9.
Hence, the correct answer is B. start fraction 1 over 9 end fraction.