Eric has two identical number cubes. The faces of each cube are numbered 1–6. Eric will roll both of the number cubes at the same time. What is the probability that both of the numbers showing face-up will be multiples of 3?
A. two-ninths.
B. one-third.
C. start fraction 1 over 6 end fraction.
D. one-ninth.
A multiple of 3 is either 3 or 6. Out of the 6 possible outcomes of rolling a number cube, 2 of them are multiples of 3. Therefore, the probability of rolling a multiple of 3 on one cube is $\frac{2}{6}=\frac{1}{3}$. Since the two cubes are identical, this probability is the same for rolling a multiple of 3 on the other cube.
To find the probability of both events happening, we multiply the probabilities:
$\frac{1}{3} \cdot \frac{1}{3} = \frac{1}{9}$
So the answer is $\boxed{\textbf{(D) }one-ninth}.$
To find the probability that both numbers showing face-up on the number cubes are multiples of 3, we need to determine the total number of favorable outcomes and the total number of possible outcomes.
There are 6 possible outcomes for each roll of the number cube, as the faces are numbered 1-6.
Since both number cubes are identical, we can consider them as one combined experiment. So, the total number of possible outcomes is 6 * 6 = 36.
Now, let's determine the favorable outcomes. A number is a multiple of 3 if it can be divided evenly by 3. From the numbers 1-6, the multiples of 3 are 3 and 6.
Therefore, the favorable outcomes are (3, 3), (3, 6), (6, 3), and (6, 6). This gives us a total of 4 favorable outcomes.
Now, we can calculate the probability by dividing the favorable outcomes by the total number of outcomes:
Probability = Favorable outcomes / Total outcomes
= 4 / 36
= 1 / 9
Thus, the probability that both numbers showing face-up will be multiples of 3 is one-ninth.
Therefore, the answer is D. one-ninth.