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

Updated On: Apr 20, 2024
  • \(\frac{1}{4}\)
  • \(\frac{1}{12}\)
  • \(\frac{3}{16}\)
  • \(\frac{3}{4}\)
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The Correct Option is A

Solution and Explanation

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) \(\frac{1}{4}\)

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