Question

For two events $A$ and $B$ such that $P(A) \neq 0$ and $P(B) \neq 1$, the conditional probability $P(A'/B')$ is equal to:

• 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 that $P(A'/B') \neq 1 - P(A/B)$. A complement inside conditional probabilities only follows the rule $P(A'/B) = 1 - P(A/B)$ when the condition event $B$ remains identical!

Answer 1

The Correct Option is D

Concept: The definition of conditional probability states that for any two events $X$ and $Y$: \[ P(X/Y) = \frac{P(X \cap Y)}{P(Y)} \] Additionally, De Morgan's Laws for sets state that the intersection of complements is the complement of the union: \[ A' \cap B' = (A \cup B)' \]

Step 1: Express the conditional probability formula.
Applying the standard conditional probability formula to $P(A'/B')$ gives: \[ P(A'/B') = \frac{P(A' \cap B')}{P(B')} \]

Step 2: Apply De Morgan's Law to the numerator.
According to De Morgan's Law, the simultaneous non-occurrence of $A$ and $B$ is equivalent to the complement of their union: \[ P(A' \cap B') = P((A \cup B)') \] Since the probability of any complement event is $1$ minus the probability of the event itself: \[ P((A \cup B)') = 1 - P(A \cup B) \]

Step 3: Substitute back into the main denominator equation.
Replacing the numerator in our step 1 formula with this expression gives: \[ P(A'/B') = \frac{1 - P(A \cup B)}{P(B')} \] This directly corresponds to option (D).