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

To find \(P(A' \cap B')\), we will use the given information about events \(A\) and \(B\) and the properties of probability.

1. We know:
   • \(P(A) = \dfrac{1}{2}\)
   • \(P(B) = \dfrac{1}{3}\)
   • \(P(A \mid B) = \dfrac{1}{4}\)
2. The conditional probability formula is: \(P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}\)
3. Substituting the given values into the formula: \(\dfrac{1}{4} = \dfrac{P(A \cap B)}{\dfrac{1}{3}}\)
4. Solving for \(P(A \cap B)\): \(P(A \cap B) = \dfrac{1}{4} \times \dfrac{1}{3} = \dfrac{1}{12}\)
5. We now use the formula for the probability of the complement of union of two events: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
6. Substitute the known probabilities: \(P(A \cup B) = \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{12}\)
7. Common denominator for calculation is 12: \(P(A \cup B) = \dfrac{6}{12} + \dfrac{4}{12} - \dfrac{1}{12} = \dfrac{9}{12} = \dfrac{3}{4}\)
8. The probability of the complement of \(A \cup B\) is: \(P(A' \cap B') = 1 - P(A \cup B)\)
9. Substitute \(P(A \cup B)\): \(P(A' \cap B') = 1 - \dfrac{3}{4} = \dfrac{1}{4}\)

Therefore, the probability \(P(A' \cap B')\) is \(\dfrac{1}{4}\).