1. Key Concepts and Formulas

To solve this problem, we will use the fundamental definition of conditional probability and several essential set theory identities and probability rules:

• Conditional Probability Formula: The probability of event $E$ occurring given that event $F$ has already occurred is defined as: $P(E|F) = \frac{P(E \cap F)}{P(F)}$ This formula is valid provided $P(F) > 0$.

• Set Theory Identities:

  • Distributive Law: For any sets $X, Y, Z$, $X \cap (Y \cup Z) = (X \cap Y) \cup (X \cap Z)$.
  • Complement Property: The intersection of an event and its complement is the empty set: $A \cap A' = \emptyset$.
  • Union with Empty Set: The union of any set with the empty set is the set itself: $X \cup \emptyset = X$.
  • De Morgan's Law: The complement of a union is the intersection of the complements, and the complement of an intersection is the union of the complements. Specifically, $(A \cap B)' = A' \cup B'$.

• Probability Rules:

  • Complement Rule: The probability of the complement of an event $E$ is $P(E') = 1 - P(E)$.
  • Probability of "A and not B": The probability of event $A$ occurring and event $B$ not occurring is $P(A \cap B') = P(A) - P(A \cap B)$. This is because event $A$ can be partitioned into two disjoint events: $(A \cap B)$ and $(A \cap B')$, so $P(A) = P(A \cap B) + P(A \cap B')$.

2. Step-by-Step Solution

We are asked to find $P(A | (A' \cup B'))$. Let $E = A$ and $F = (A' \cup B')$.

Step 1: Apply the Conditional Probability Formula First, we set up the expression using the definition of conditional probability: $P(A | (A' \cup B')) = \frac{P(A \cap (A' \cup B'))}{P(A' \cup B')}$ Our goal is to simplify the numerator and the denominator separately using set identities and probability rules, and then substitute the given values.

Step 2: Simplify the Numerator: $P(A \cap (A' \cup B'))$

• Step 2a: Simplify the set expression $A \cap (A' \cup B')$. We start by simplifying the set operation inside the probability function. We use the Distributive Law for sets, which states $X \cap (Y \cup Z) = (X \cap Y) \cup (X \cap Z)$. Applying this: $A \cap (A' \cup B') = (A \cap A') \cup (A \cap B')$ Next, we use the Complement Property $A \cap A' = \emptyset$ (the intersection of an event and its complement is the empty set, as an event cannot both occur and not occur simultaneously). So, the expression becomes: $\emptyset \cup (A \cap B')$ Finally, using the Union with Empty Set property $X \cup \emptyset = X$: $\emptyset \cup (A \cap B') = A \cap B'$ Therefore, the numerator simplifies to $P(A \cap B')$.

• Step 2b: Express $P(A \cap B')$ in terms of given probabilities. The event $A \cap B'$ means "event $A$ occurs AND event $B$ does NOT occur". This is often visualized as the part of $A$ that does not overlap with $B$ in a Venn diagram. We know that event $A$ can be divided into two mutually exclusive (disjoint) parts:

  1. $A \cap B$ (A occurs and B occurs)
  2. $A \cap B'$ (A occurs and B does not occur) Since these two events cover all possibilities for $A$ and are disjoint, their probabilities sum to $P(A)$: $P(A) = P(A \cap B) + P(A \cap B')$ Rearranging this to find $P(A \cap B')$: $P(A \cap B') = P(A) - P(A \cap B)$ This is a very useful identity for calculating the probability of "A only".

Step 3: Simplify the Denominator: $P(A' \cup B')$

• Step 3a: Simplify the set expression $A' \cup B'$. We use De Morgan's Law, which states that the union of complements is the complement of the intersection: $(A \cap B)' = A' \cup B'$. Applying this directly: $A' \cup B' = (A \cap B)'$ Therefore, the denominator simplifies to $P((A \cap B)')$.

• Step 3b: Express $P((A \cap B)')$ in terms of given probabilities. We use the Complement Rule, which states $P(X') = 1 - P(X)$. Applying this rule to the event $(A \cap B)$: $P((A \cap B)') = 1 - P(A \cap B)$

Step 4: Substitute and Calculate

Now we substitute the simplified expressions for the numerator and denominator back into the conditional probability formula: $P(A | (A' \cup B')) = \frac{P(A \cap B')}{P((A \cap B)')} = \frac{P(A) - P(A \cap B)}{1 - P(A \cap B)}$

We are given the following probabilities:

• $P(A) = \frac{2}{5}$
• $P(A \cap B) = \frac{3}{20}$

Substitute these values into the derived formula: $P(A | (A' \cup B')) = \frac{\frac{2}{5} - \frac{3}{20}}{1 - \frac{3}{20}}$

Perform the arithmetic:

• Calculate the numerator: To subtract the fractions, find a common denominator (20): $\frac{2}{5} - \frac{3}{20} = \frac{2 \times 4}{5 \times 4} - \frac{3}{20} = \frac{8}{20} - \frac{3}{20} = \frac{5}{20}$

• Calculate the denominator: To subtract the fraction from 1, write 1 as $\frac{20}{20}$: $1 - \frac{3}{20} = \frac{20}{20} - \frac{3}{20} = \frac{17}{20}$

• Divide the numerator by the denominator: $P(A | (A' \cup B')) = \frac{\frac{5}{20}}{\frac{17}{20}}$ Dividing by a fraction is equivalent to multiplying by its reciprocal: $P(A | (A' \cup B')) = \frac{5}{20} \times \frac{20}{17} = \frac{5}{17}$

3. Common Mistakes & Tips

• Misapplication of Set Identities: A common mistake is to confuse or misapply De Morgan's Laws or the distributive property. Always visualize with Venn diagrams if unsure. For instance, $A \cap (A' \cup B')$ is not simply $A \cap B'$ directly; the intermediate steps involving $A \cap A'$ are crucial.
• Incorrectly Calculating $P(A \cap B')$: Remember that $P(A \cap B')$ (A occurs, B does not) is $P(A) - P(A \cap B)$, not $P(A) - P(B)$.
• Arithmetic Errors: Fractions can be tricky. Ensure you find common denominators correctly for addition/subtraction and remember how to divide fractions.
• Simplify Set Expressions First: It's generally easier and less error-prone to simplify the set expressions (e.g., $A \cap (A' \cup B')$ and $A' \cup B'$) using set theory before applying probability rules.

4. Summary

We were asked to find the conditional probability $P(A | (A' \cup B'))$. We began by applying the definition of conditional probability. Then, we systematically simplified the numerator $P(A \cap (A' \cup B'))$ to $P(A) - P(A \cap B)$ using the distributive law and complement properties. Simultaneously, we simplified the denominator $P(A' \cup B')$ to $1 - P(A \cap B)$ using De Morgan's Law and the complement rule. Finally, we substituted the given probability values into the simplified expression and performed the arithmetic to arrive at the result $\frac{5}{17}$.

5. Final Answer

The conditional probability $P(A | (A' \cup B'))$ is $\frac{5}{17}$.

The final answer is $\boxed{\frac{5}{17}}$, which corresponds to option (B).