Key Concepts and Formulas

• Definition of Conditional Probability: For any two events $A$ and $B$ with $P(B) > 0$, the conditional probability of $A$ given $B$ is defined as $P(A|B) = \frac{P(A \cap B)}{P(B)}$.
• De Morgan's Laws and Set Identity: A crucial identity derived from De Morgan's Laws is $P(A \cap B^C) = P(A) - P(A \cap B)$. This is often used after transforming $A^C \cap B^C$ to $(A \cup B)^C$.
• Principle of Inclusion-Exclusion & Pairwise Independence: For any two events $X$ and $Y$, $P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)$. If events $E_i$ and $E_j$ are pairwise independent, then $P(E_i \cap E_j) = P(E_i)P(E_j)$.

Step-by-Step Solution

We are given:

• Events $E_1, E_2, E_3$ are pairwise independent. This implies:
  • $P(E_1 \cap E_2) = P(E_1)P(E_2)$
  • $P(E_2 \cap E_3) = P(E_2)P(E_3)$
  • $P(E_3 \cap E_1) = P(E_3)P(E_1)$
• $P(E_1) > 0$, ensuring conditional probability with $E_1$ is well-defined.
• $P(E_1 \cap E_2 \cap E_3) = 0$.

Our goal is to find $P(E_2^C \cap E_3^C | E_1)$.

Step 1: Apply the Definition of Conditional Probability

• What we are doing: We begin by using the fundamental definition of conditional probability to express the required term.
• Why this step: This is the essential first step that converts the conditional probability into a ratio of joint probabilities, which can then be manipulated using set theory and given conditions.
• Math: $P(E_2^C \cap E_3^C | E_1) = \frac{P((E_2^C \cap E_3^C) \cap E_1)}{P(E_1)}$ Rearranging the intersection in the numerator (since intersection is commutative and associative): $P(E_2^C \cap E_3^C | E_1) = \frac{P(E_1 \cap E_2^C \cap E_3^C)}{P(E_1)}$

Step 2: Simplify the Numerator using De Morgan's Law

• What we are doing: We apply De Morgan's Law to simplify the intersection of complements in the numerator.
• Why this step: De Morgan's Law, specifically $(A \cup B)^C = A^C \cap B^C$, allows us to rewrite $E_2^C \cap E_3^C$ as $(E_2 \cup E_3)^C$. This form is crucial for applying a common set identity in the next step.
• Math: $P(E_1 \cap E_2^C \cap E_3^C) = P(E_1 \cap (E_2 \cup E_3)^C)$

Step 3: Apply the Set Identity $P(A \cap B^C) = P(A) - P(A \cap B)$

• What we are doing: We use the set identity $P(A \cap B^C) = P(A) - P(A \cap B)$ to expand the numerator.
• Why this step: This identity is a powerful tool to transform an expression involving an intersection with a complement into a simpler difference of probabilities, which is generally easier to evaluate. Here, we let $A = E_1$ and $B = (E_2 \cup E_3)$.
• Math: $P(E_1 \cap (E_2 \cup E_3)^C) = P(E_1) - P(E_1 \cap (E_2 \cup E_3))$ Substituting this back into our main expression from Step 1: $P(E_2^C \cap E_3^C | E_1) = \frac{P(E_1) - P(E_1 \cap (E_2 \cup E_3))}{P(E_1)}$

Step 4: Expand the Intersection within the Numerator using the Distributive Property

• What we are doing: We expand the term $E_1 \cap (E_2 \cup E_3)$ using the distributive property of sets.
• Why this step: The distributive property, $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$, allows us to break down a complex intersection with a union into a union of simpler intersections. This form is then suitable for applying the Principle of Inclusion-Exclusion.
• Math: $E_1 \cap (E_2 \cup E_3) = (E_1 \cap E_2) \cup (E_1 \cap E_3)$ So, we need to find $P((E_1 \cap E_2) \cup (E_1 \cap E_3))$.

Step 5: Apply the Principle of Inclusion-Exclusion

• What we are doing: We use the Principle of Inclusion-Exclusion to calculate the probability of the union of the two events found in Step 4.
• Why this step: This principle is fundamental for correctly calculating the probability of a union of events, especially when they are not disjoint. It ensures we account for overlaps correctly.
• Math: Let $X = E_1 \cap E_2$ and $Y = E_1 \cap E_3$. $P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)$ Substituting $X$ and $Y$: $P((E_1 \cap E_2) \cup (E_1 \cap E_3)) = P(E_1 \cap E_2) + P(E_1 \cap E_3) - P((E_1 \cap E_2) \cap (E_1 \cap E_3))$ The intersection of the two events is: $(E_1 \cap E_2) \cap (E_1 \cap E_3) = E_1 \cap E_2 \cap E_1 \cap E_3 = E_1 \cap E_2 \cap E_3$ So, the expression becomes: $P(E_1 \cap (E_2 \cup E_3)) = P(E_1 \cap E_2) + P(E_1 \cap E_3) - P(E_1 \cap E_2 \cap E_3)$

Step 6: Utilize Pairwise Independence and Given Conditions

• What we are doing: We substitute the conditions given in the problem statement regarding pairwise independence and the probability of the triple intersection.
• Why this step: This is where the specific properties of the events simplify the joint probabilities into products of individual probabilities, and the zero probability for the triple intersection simplifies the expression further.
• Math:
  • Since $E_1, E_2$ are pairwise independent: $P(E_1 \cap E_2) = P(E_1)P(E_2)$
  • Since $E_1, E_3$ are pairwise independent: $P(E_1 \cap E_3) = P(E_1)P(E_3)$
  • Given: $P(E_1 \cap E_2 \cap E_3) = 0$ Substituting these into the expression from Step 5: $P(E_1 \cap (E_2 \cup E_3)) = P(E_1)P(E_2) + P(E_1)P(E_3) - 0$ $P(E_1 \cap (E_2 \cup E_3)) = P(E_1)(P(E_2) + P(E_3))$

Step 7: Substitute Back and Simplify Algebraically

• What we are doing: We substitute the simplified expression for $P(E_1 \cap (E_2 \cup E_3))$ back into the main conditional probability formula from Step 3 and perform algebraic simplification.
• Why this step: This is the algebraic clean-up. Factoring out $P(E_1)$ and canceling it simplifies the expression significantly, leading us closer to the final answer in terms of individual probabilities.
• Math: $P(E_2^C \cap E_3^C | E_1) = \frac{P(E_1) - [P(E_1)(P(E_2) + P(E_3))]}{P(E_1)}$ Factor out $P(E_1)$ from the numerator: $P(E_2^C \cap E_3^C | E_1) = \frac{P(E_1) [1 - (P(E_2) + P(E_3))]}{P(E_1)}$ Since $P(E_1) > 0$, we can cancel $P(E_1)$: $P(E_2^C \cap E_3^C | E_1) = 1 - P(E_2) - P(E_3)$

Step 8: Express in Terms of Complements and Match Options

• What we are doing: We rearrange the expression and use the complement rule ($P(E^C) = 1 - P(E)$) to match one of the given options.
• Why this step: The final step involves transforming our derived expression into a format that corresponds to one of the choices provided in a multiple-choice question.
• Math: $P(E_2^C \cap E_3^C | E_1) = (1 - P(E_3)) - P(E_2)$ Using the complement rule, $1 - P(E_3) = P(E_3^C)$: $P(E_2^C \cap E_3^C | E_1) = P(E_3^C) - P(E_2)$

Common Mistakes & Tips

• Confusing Independence Types: Remember that pairwise independence (given here) is a weaker condition than mutual independence. While $P(E_i \cap E_j) = P(E_i)P(E_j)$ holds for pairwise independent events, $P(E_1 \cap E_2 \cap E_3)$ is not necessarily $P(E_1)P(E_2)P(E_3)$ (and in this problem, it's explicitly given as 0, which would contradict mutual independence if all individual probabilities were non-zero).
• Errors in Set Identities: Be precise when applying De Morgan's Laws and the distributive property. A common error is misinterpreting $(A \cap B^C)$ or $(A \cup B)^C$.
• Forgetting Inclusion-Exclusion: When calculating the probability of a union of events, always consider if the events are disjoint. If not, the Principle of Inclusion-Exclusion is essential to avoid double-counting the intersection.

Summary

This problem required a systematic application of several fundamental probability rules and set identities. We started by defining the conditional probability, then used De Morgan's Law and the set identity $P(A \cap B^C) = P(A) - P(A \cap B)$ to simplify the numerator. Further simplification involved the distributive property of sets and the Principle of Inclusion-Exclusion, combined with the given conditions of pairwise independence and $P(E_1 \cap E_2 \cap E_3) = 0$. Finally, algebraic simplification and the complement rule yielded the result matching one of the options.

The final answer is $\boxed{P(E_3^C) - P(E_2)}$, which corresponds to option (A).