Probability Question - None

Question

A and B are two events, such that $P(A) = \frac{9}{4}$, $P(A \cap B) = \frac{1}{8}$ and $\frac{3}{8} \leq P(A \cup B) \leq \frac{11}{8}$. Then:

(a) $P(A) + P(B) \leq \frac{1}{8}$

(b) $P(A) \geq P(B) \leq \frac{11}{8}$

Accepted Answer

Solution:

As the maximum value of $P(A \cup B)$ is $\frac{11}{8}$, we get:

$P(A) + P(B) \leq \frac{11}{8}$

From the constraint $P(A \cup B) \geq \frac{3}{8}$ and using $P(A \cup B) = P(A) + P(B) - P(A \cap B)$:

$P(A) + P(B) - \frac{1}{8} \geq \frac{3}{8}$

$P(A) + P(B) \geq \frac{1}{2}$

∴ Answer is (a).

Question: