To solve the given problems, we need to follow these steps:
a) Complete the sample space by listing all possible outcomes of rolling two 6-sided dice.
b) Calculate the probability of getting a sum of 1 when rolling two dice.
c) Calculate the probability of getting a sum of 5 when rolling two dice.
d) Calculate the probability of getting a sum of 12 when rolling two dice.
- Generate all possible outcomes of rolling two 6-sided dice.
- Count the number of outcomes that result in the desired sum.
- Calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes.
The sample space for rolling two 6-sided dice consists of all possible ordered pairs \((i, j)\) where \(i\) and \(j\) can take values from 1 to 6. The complete sample space is given by:
\[
\text{sample\_space} = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\}
\]
This results in a total of \(36\) possible outcomes.
To find the probability of obtaining a sum of \(1\), we note that there are no outcomes in the sample space that yield this sum. Therefore, the number of favorable outcomes is \(0\):
\[
\text{sum\_1} = \{\} \quad \Rightarrow \quad P(\text{sum} = 1) = \frac{0}{36} = 0.0
\]
Next, we calculate the probability of obtaining a sum of \(5\). The favorable outcomes that yield this sum are:
\[
\text{sum\_5} = \{(1, 4), (2, 3), (3, 2), (4, 1)\}
\]
There are \(4\) favorable outcomes, so the probability is:
\[
P(\text{sum} = 5) = \frac{4}{36} = \frac{1}{9} \approx 0.1111
\]
Finally, we find the probability of obtaining a sum of \(12\). The only favorable outcome for this sum is:
\[
\text{sum\_12} = \{(6, 6)\}
\]
There is \(1\) favorable outcome, thus the probability is:
\[
P(\text{sum} = 12) = \frac{1}{36} \approx 0.0278
\]
The probabilities for the respective sums are:
- Probability of getting a sum of \(1\): \(0.0000\)
- Probability of getting a sum of \(5\): \(0.1111\)
- Probability of getting a sum of \(12\): \(0.0278\)
Thus, the final answers are:
\[
\boxed{P(\text{sum} = 1) = 0.0000, \quad P(\text{sum} = 5) = 0.1111, \quad P(\text{sum} = 12) = 0.0278}
\]