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Step-by-Step Solution
Step 1: Identify the sample space
Each die has 6 faces with numbers {1, 2, 3, 5, 7, 11}. When two such dice are rolled, the total number of possible outcomes is
$6 \times 6 = 36.$
Each outcome is an ordered pair $(a, b)$ where $a$ and $b$ are numbers from {1, 2, 3, 5, 7, 11}.
Step 2: Determine the condition for favorable outcomes
We are interested in the outcomes where the sum of the top faces is less than or equal to 8, i.e.
$a + b \le 8.$
Step 3: List all possible favorable outcomes
We enumerate the pairs where $a + b \le 8$:
(1, 1) : sum = 2
(1, 2) : sum = 3
(1, 3) : sum = 4
(1, 5) : sum = 6
(1, 7) : sum = 8
(2, 1) : sum = 3
(2, 2) : sum = 4
(2, 3) : sum = 5
(2, 5) : sum = 7
(3, 1) : sum = 4
(3, 2) : sum = 5
(3, 3) : sum = 6
(3, 5) : sum = 8
(5, 1) : sum = 6
(5, 2) : sum = 7
(5, 3) : sum = 8
(7, 1) : sum = 8
There are 17 such pairs in total.
Step 4: Calculate the probability
The probability is given by the number of favorable outcomes divided by the total number of outcomes:
$ \displaystyle \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{17}{36}. $
Step 5: State the final answer
Hence, the probability that the sum of the numbers on the top faces is less than or equal to 8 is
$ \displaystyle \frac{17}{36}.$
