The problem asks us to compute the conditional probabilities $P(A|B)$ and $P(B|A)$ given the following probabilities:

1. $P(A) = \frac{1}{2}$,
2. $P(B) = \frac{3}{4}$,
3. $P(A \cup B) = \frac{11}{12}$.

We know that:

$P(A \cup B) = P(A) + P(B) - P(A \cap B).$

Step 1: Find $P(A \cap B)$

Substitute the given values into the formula:

$\frac{11}{12} = \frac{1}{2} + \frac{3}{4} - P(A \cap B).$

First, express all terms with a denominator of 12: $\frac{11}{12} = \frac{6}{12} + \frac{9}{12} - P(A \cap B).$

Simplify: $\frac{11}{12} = \frac{15}{12} - P(A \cap B).$

Solve for $P(A \cap B)$: $P(A \cap B) = \frac{15}{12} - \frac{11}{12} = \frac{4}{12} = \frac{1}{3}.$

Step 2: Calculate $P(A|B)$

The conditional probability formula is: $P(A|B) = \frac{P(A \cap B)}{P(B)}.$

Substitute $P(A \cap B) = \frac{1}{3}$ and $P(B) = \frac{3}{4}$: $P(A|B) = \frac{\frac{1}{3}}{\frac{3}{4}} = \frac{1}{3} \times \frac{4}{3} = \frac{4}{9}.$

Step 3: Calculate $P(B|A)$

The conditional probability formula is: $P(B|A) = \frac{P(A \cap B)}{P(A)}.$

Substitute $P(A \cap B) = \frac{1}{3}$ and $P(A) = \frac{1}{2}$: $P(B|A) = \frac{\frac{1}{3}}{\frac{1}{2}} = \frac{1}{3} \times 2 = \frac{2}{3}.$

Final Answers:

1. $P(A|B) = \frac{4}{9}$,
2. $P(B|A) = \frac{2}{3}$.

Would you like a more detailed explanation or help with a similar problem?

Here are 5 related questions to expand your understanding:

1. What is the significance of conditional probability in real-world scenarios?
2. How is $P(A|B)$ interpreted in terms of dependency between events $A$ and $B$?
3. If $A$ and $B$ are independent, how would $P(A|B)$ and $P(B|A)$ change?
4. How can we compute $P(A \cap B)$ if $P(A|B)$ and $P(B)$ are given?
5. What is the relationship between $P(A \cap B)$ and $P(B \cap A)$?

Tip: Always confirm whether the given events $A$ and $B$ are mutually exclusive, as it simplifies calculations.