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

Step 1 (Formula / Governing Law & Given Data): [0.5 Mark]

By the definition of conditional probability, for any two events $E$ and $F$ where $P(F) \neq 0$:

$$P(E/F) = \frac{P(E \cap F)}{P(F)}$$

Applying this to the given expression where $E = A'$ and $F = B'$:

$$P(A'/B') = \frac{P(A' \cap B')}{P(B')}$$

Step 2 (Step-by-Step Mathematical Working): [0.25 Mark]

Using De Morgan's Law from Set Theory, we know that the intersection of the complements of two sets is the complement of their union:

$$A' \cap B' = (A \cup B)'$$

Therefore, the probability of the intersection of the complements is:

$$P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B)$$

Substituting this result back into the conditional probability formula:

$$P(A'/B') = \frac{1 - P(A \cup B)}{P(B')}$$

Step 3 (Final Concluded Answer): [0.25 Mark]

Comparing the derived expression with the given options, we conclude that the correct option is (D).

Final Answer: $\frac{1 - P(A \cup B)}{P(B')}$

💡 EL10 Concept Insight: Think of $P(A'/B')$ as looking only at the "outside" of $B$ (which is $B'$). Within that specific area, we want to know how much is also "outside" of $A$. Since the area outside both $A$ and $B$ is everything except the combined footprint of $A$ and $B$, we simply take the total probability (1) minus the union of $A$ and $B$, then divide by the size of the "outside of $B$" region.