Question

Let \(A\) and \(B\) be two events such that \( P(A) = \frac{5}{11}, \; P(B) = \frac{2}{11} \) and \( P(A \cup B) = \frac{3}{11} \), then \( P(A'|B') = \)_____

• A. \( \frac{8}{9} \)
• B. \( \frac{3}{5} \)
• C. \( \frac{1}{2} \)
• D. \( \frac{2}{9} \)

Hint:

Use complement rule: \( P(A' \cap B') = 1 - P(A \cup B) \).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
By De Morgan's Law and the definition of conditional probability: $P(A'|B') = \frac{P(A' \cap B')}{P(B')}$ and $A' \cap B' = (A \cup B)'$.
Step 2: Formula Application:
$P(B') = 1 - P(B) = 1 - 2/11 = 9/11$. $P(A' \cap B') = 1 - P(A \cup B)$. Using typical consistent values (where $P(A \cup B) = 3/11$ is likely a typo for $P(A \cap B)$): If $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 5/11 + 2/11 - P(A \cap B)$. Assuming the intended $P(A \cup B)$ results in $P(A' \cap B') = 8/11$.
Step 3: Explanation:
$P(A'|B') = \frac{8/11}{9/11} = \frac{8}{9}$.
Step 4: Final Answer:
The result is $8/9$.