Question

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

• A. $\frac{1}{4}$
• B. $\frac{1}{12}$
• C. $\frac{3}{16}$
• D. $\frac{1}{8}$

Hint:

$P(A' \cap B')$ is the probability that "None" of the events happen. In a Venn diagram, this is the area outside both circles.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We need to find the probability that neither event A nor event B occurs. This is solved using De Morgan's Law $P(A' \cap B') = 1 - P(A \cup B)$. We first use the conditional probability formula to find the intersection, then the addition rule for the union.

Step 2: Key Formula or Approach:
1. $P(A \cap B) = P(A \mid B) \times P(B)$
2. $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
3. $P(A' \cap B') = 1 - P(A \cup B)$

Step 3: Detailed Explanation:
1. Find $P(A \cap B)$: \[ P(A \cap B) = \frac{1}{4} \times \frac{1}{3} = \frac{1}{12} \] 2. Find $P(A \cup B)$: \[ P(A \cup B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{12} \] Find a common denominator (12): \[ P(A \cup B) = \frac{6}{12} + \frac{4}{12} - \frac{1}{12} = \frac{9}{12} = \frac{3}{4} \] 3. Find $P(A' \cap B')$: \[ P(A' \cap B') = 1 - \frac{3}{4} = \frac{1}{4} \]

Step 4: Final Answer
The value of $P(A' \cap B')$ is $\frac{1}{4}$.