Question

If A and B are the two events such that $P(A) = \frac{1}{2}$, $P(B) = \frac{1}{3}$ and $P(A|B) = \frac{1}{4}$, then $P(A' \cap B')$ is

• A. $\frac{1}{4}$
• B. $\frac{1}{12}$
• C. $\frac{3}{16}$
• D. $\frac{3}{4}$

Answer 1

Answer: (A)

$\frac{1}{4}$

Answer 2

To find the probability of the intersection of the complements of events A and B, we can use the complement rule:
$P(A' \cap B') = 1 - P(A \cup B)$
We know that $P(A|B) = \frac{P(A \cap B)}{P(B)}$
Rearranging the equation, we have:
$P(A \cap B) = P(A|B) \cdot P(B)$
Given that $P(A) = \frac{1}{2}, \quad P(B) = \frac{1}{3}$, and $P(A|B) = \frac{1}{4}$
we can substitute these values into the equation:
$P(A \cap B) = \left(\frac{1}{4}\right) \times \left(\frac{1}{3}\right) = \frac{1}{12}$
Now, we can find the probability of the union of events A and B:
$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{12} = \frac{6}{12} + \frac{4}{12} - \frac{1}{12} = \frac{9}{12} = \frac{3}{4}$
Finally, we can find the probability of the intersection of the complements:
$P(A' \cap B') = 1 - P(A \cup B) = 1 - \left(\frac{3}{4}\right) = \frac{1}{4}$
Therefore, the correct option is (A) $\frac{1}{4}$