Question

For two events \( A \) and \( B \) such that \( P(A) \neq 0 \) and \( P(B) \neq 1 \), \( P(A' / B') = \)

• A. \( 1 - P(A / B) \)
• B. \( 1 - P(A' / B) \)
• C. \( \frac{1 - P(A \cap B)}{P(B')} \)
• D. \( \frac{1 - P(A \cup B)}{P(B')} \)

Hint:

Remember \( P(A/B) + P(A'/B) = 1 \). This helps eliminate incorrect options like (A) and (B).
De Morgan's Laws are essential for probability problems involving "neither A nor B".

Answer 1

The Correct Option is D

Concept:

• Conditional Probability: \( P(X / Y) = \frac{P(X \cap Y)}{P(Y)} \).
• De Morgan's Law: \( A' \cap B' = (A \cup B)' \).
• Complementary Probability: \( P(E') = 1 - P(E) \).

Step 1: Apply the conditional probability formula
By definition:
\[ P(A' / B') = \frac{P(A' \cap B')}{P(B')} \]

Step 2: Simplify the numerator using De Morgan's Law
From set theory, the intersection of complements is the complement of the union:
\[ A' \cap B' = (A \cup B)' \]
Therefore, the probability is:
\[ P(A' \cap B') = P((A \cup B)') \]

Step 3: Convert to standard probability form
Using the property \( P(E') = 1 - P(E) \):
\[ P((A \cup B)') = 1 - P(A \cup B) \]

Step 4: Substitute back into the expression
Replacing the numerator in the expression from
Step 1:
\[ P(A' / B') = \frac{1 - P(A \cup B)}{P(B')} \]
This matches option (D).