GATE CS 2003 Q3 — Conditional Probability and Bayes Theorem

To find the conditional probabilities $P(A \mid B)$ and $P(B \mid A)$, we first need to determine the probability of their intersection, $P(A \cap B)$.

Using the probability of the union of two events:

Since any probability is at most $1$, we have $P(A \cup B) \le 1$. Given $P(A) = 1$ and $P(B) = \frac{1}{2}$:

Since $A \cap B \subseteq B$, it must also be true that $P(A \cap B) \le P(B) = \frac{1}{2}$. Therefore, we have:

Now, we can compute the required conditional probabilities:

1. $P(A \mid B)$: $$P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{1/2}{1/2} = 1$$
2. $P(B \mid A)$: $$P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{1/2}{1} = \frac{1}{2}$$

Thus, the values of $P(A \mid B)$ and $P(B \mid A)$ are $1$ and $\frac{1}{2}$, respectively.