$\frac{15}{36}$
The question asks for the probability of obtaining a prime number sum when rolling an unbiased six-sided dice (faces 1 to 6) twice in succession.
An unbiased six-sided dice has 6 possible outcomes for each roll (1, 2, 3, 4, 5, 6).
When rolling the dice twice, the total number of unique outcomes is calculated by multiplying the number of outcomes for each roll:
Total Possible Outcomes = $6 \times 6 = 36$
The minimum sum achievable is $1 + 1 = 2$.
The maximum sum achievable is $6 + 6 = 12$.
The prime numbers between 2 and 12 (inclusive) are: 2, 3, 5, 7, and 11.
We need to find all pairs of outcomes from the two rolls whose sum is one of these prime numbers.
| Prime Sum | Favorable Pairs (Roll 1, Roll 2) | Count |
| 2 | (1, 1) | 1 |
| 3 | (1, 2), (2, 1) | 2 |
| 5 | (1, 4), (4, 1), (2, 3), (3, 2) | 4 |
| 7 | (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3) | 6 |
| 11 | (5, 6), (6, 5) | 2 |
Summing the counts for each prime sum gives the total number of favorable outcomes:
Total Favorable Outcomes = $1 + 2 + 4 + 6 + 2 = 15$
The probability is the ratio of favorable outcomes to the total possible outcomes.
Probability (Sum is Prime) = $\frac{\text{Total Favorable Outcomes}}{\text{Total Possible Outcomes}}$
Probability = $\frac{15}{36}$
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