If $A$ and $B$ are two events such that $P(A) = \frac{3}{4}$ and $P(B) = \frac{5}{8}$ then:
$P(A \cup B) \leq \frac{3}{4}$
$P(A^c \cap B) \leq \frac{1}{4}$
$\frac{3}{8} \leq P(A \cap B) \leq \frac{5}{8}$
$\frac{1}{8} \leq P(A \cap B^c) \leq \frac{3}{8}$
Step-by-Step Solution
Key Concept: Using the constraints P(A) = 3/4 and P(B) = 5/8, apply the fundamental relations: P(A ∪ B) = P(A) + P(B) - P(A ∩ B) ≤ 1, P(A ∩ B) ≤ min(P(A), P(B)), and P(A^c ∩ B) = P(B) - P(A ∩ B) to establish tight bounds on derived probabilities.
Given $A \subseteq A \cup B$, we have $P(A \cup B) \geq P(A)$, so $P(A \cup B) \geq \frac{3}{4}$. Using $P(A \cup B) = P(A) + P(B) - P(A \cap B) \geq \frac{3}{4} + \frac{5}{8} - 1 = \frac{3}{8}$. Since $(A \cap B) \subseteq B$, we have $P(A \cap B) \leq \frac{5}{8}$. For $P(A \cap B')$, we combine these constraints to show $\frac{1}{8} \leq P(A \cap B') \leq \frac{3}{8}$, and ultimately $0 \leq P(A'\cap B) \leq \frac{1}{4}$.
Correct Answer: 1,2,3,4