The correct answer is $\boxed{\text{C. }\frac{1}{2}, 1}$.
The probability of event A happening, given that event B has already happened, is called the conditional probability of A given B, and is denoted by $P(A|B)$. It is calculated as follows:
$$P(A|B) = \frac{P(A \cap B)}{P(B)}$$
where $P(A \cap B)$ is the probability of both events A and B happening.
In this case, we are given that $P(A) = 1$ and $P(B) = \frac{1}{2}$. We can then calculate $P(A \cap B)$ as follows:
$$P(A \cap B) = P(A) \cdot P(B) = 1 \cdot \frac{1}{2} = \frac{1}{2}$$
Therefore, the conditional probability of A given B is:
$$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1$$
The probability of event B happening, given that event A has already happened, is called the conditional probability of B given A, and is denoted by $P(B|A)$. It is calculated as follows:
$$P(B|A) = \frac{P(A \cap B)}{P(A)}$$
where $P(A \cap B)$ is the probability of both events A and B happening.
In this case, we are given that $P(A) = 1$ and $P(B) = \frac{1}{2}$. We can then calculate $P(A \cap B)$ as follows:
$$P(A \cap B) = P(A) \cdot P(B) = 1 \cdot \frac{1}{2} = \frac{1}{2}$$
Therefore, the conditional probability of B given A is:
$$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{1}{2}}{1} = \frac{1}{2}$$