Given two events *A* and *B* with the following probabilities:

- $P(A \cup B) = \frac{3}{4}$
- $P(A \cap B) = \frac{1}{4}$
- $P(A') = \frac{2}{3}$

Calculate $P(A' \cap B)$ and determine whether *A* and *B* are independent, mutually exclusive, or dependent events.

Question

Given two events *A* and *B* with the following probabilities:

- $P(A \cup B) = \frac{3}{4}$
- $P(A \cap B) = \frac{1}{4}$
- $P(A') = \frac{2}{3}$

Calculate $P(A' \cap B)$ and determine whether *A* and *B* are independent, mutually exclusive, or dependent events.

Accepted Answer

**Solution:**

1. Given:
   - $P(A \cup B) = \frac{3}{4}$
   - $P(A \cap B) = \frac{1}{4}$
   - $P(A') = \frac{2}{3}$

2. Find $P(A)$ using complement rule:
   $$P(A) = 1 - P(A') = 1 - \frac{2}{3} = \frac{1}{3}$$

3. Calculate $P(A' \cap B)$:
   Using the formula:
   $$P(A' \cap B) = P(B) - P(A \cap B)$$

But first, find $P(B)$ using the union formula:
   $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
   Substitute known values:
   $$\frac{3}{4} = \frac{1}{3} + P(B) - \frac{1}{4}$$

Rearranging:
   $$P(B) = \frac{3}{4} - \frac{1}{3} + \frac{1}{4} = \frac{3}{4} + \frac{1}{4} - \frac{1}{3} = 1 - \frac{1}{3} = \frac{2}{3}$$

4. Now calculate $P(A' \cap B)$:
   $$P(A' \cap B) = P(B) - P(A \cap B) = \frac{2}{3} - \frac{1}{4} = \frac{8}{12} - \frac{3}{12} = \frac{5}{12}$$

5. Check if *A* and *B* are independent:
   For independence:
   $$P(A \cap B) = P(A) 	imes P(B)$$

Calculate RHS:
   $$P(A) 	imes P(B) = \frac{1}{3} 	imes \frac{2}{3} = \frac{2}{9} \approx 0.222$$

Compare with LHS:
   $$P(A \cap B) = \frac{1}{4} = 0.25$$

Since $0.25 
eq 0.222$, *A* and *B* are **not independent**.

However, the question's original options mark "Independent" as correct, which suggests a re-check.

6. Re-examining the problem:
   - Given $P(A \cap B) = \frac{1}{4}$
   - $P(A) = \frac{1}{3}$
   - $P(B) = \frac{2}{3}$

Calculate $P(A) 	imes P(B) = \frac{1}{3} 	imes \frac{2}{3} = \frac{2}{9} \approx 0.222$

Since $P(A \cap B) = 0.25$ is close but not equal, strictly speaking, *A* and *B* are **dependent**.

7. Check if *A* and *B* are mutually exclusive:
   Since $P(A \cap B) 
eq 0$, they are **not mutually exclusive**.

**Answer:**
- $P(A' \cap B) = \frac{5}{12}$
- *A* and *B* are **dependent** (not independent or mutually exclusive).

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