Key Concepts and Formulas

• Conditional Probability: For any two events $E_1$ and $E_2$ where $P(E_2) > 0$, the probability of $E_1$ occurring given that $E_2$ has occurred is given by: $P(E_1 \mid E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)}$ In this problem, $E_1 = B$ and $E_2 = (A \cup \overline{B})$.
• Probability of Union (Addition Rule): For any two events $E_1$ and $E_2$: $P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2)$
• Complement Rule: For any event $E$: $P(\overline{E}) = 1 - P(E)$
• Set Difference/Intersection with Complement: The probability of event $A$ occurring but not event $B$ is: $P(A \cap \overline{B}) = P(A) - P(A \cap B)$
• Distributive Law for Sets: For any three sets $E_1, E_2, E_3$: $E_1 \cap (E_2 \cup E_3) = (E_1 \cap E_2) \cup (E_1 \cap E_3)$
• Intersection of a Set and its Complement: A set intersected with its complement is the empty set ($\emptyset$), hence its probability is 0: $P(E \cap \overline{E}) = 0$

Step-by-Step Solution

We are given:

• $P(A) = 0.7$
• $P(B) = 0.4$
• $P(A \cap \overline{B}) = 0.5$

Our goal is to calculate $P\left(B \mid (A \cup \overline{B})\right)$.

Step 1: Calculate $P(A \cap B)$

• What we are doing: We use the given $P(A)$ and $P(A \cap \overline{B})$ to find $P(A \cap B)$. This is a crucial intermediate step, as $P(A \cap B)$ is often a fundamental component in many probability calculations and will be needed for our numerator.
• Why we are doing it: The identity $P(A \cap \overline{B}) = P(A) - P(A \cap B)$ allows us to isolate $P(A \cap B)$.

Using the identity $P(A \cap \overline{B}) = P(A) - P(A \cap B)$: $0.5 = 0.7 - P(A \cap B)$ Solving for $P(A \cap B)$: $P(A \cap B) = 0.7 - 0.5$ $\implies P(A \cap B) = 0.2$

Step 2: Calculate the Denominator $P(A \cup \overline{B})$

• What we are doing: We need to find the probability of the event $(A \cup \overline{B})$, which will be the denominator of our conditional probability formula.
• Why we are doing it: This is a direct application of the Addition Rule for probabilities.

First, calculate $P(\overline{B})$ using the Complement Rule: $P(\overline{B}) = 1 - P(B)$ $P(\overline{B}) = 1 - 0.4$ $\implies P(\overline{B}) = 0.6$

Now, apply the Addition Rule for $P(A \cup \overline{B})$: $P(A \cup \overline{B}) = P(A) + P(\overline{B}) - P(A \cap \overline{B})$ Substitute the known values: $P(A \cup \overline{B}) = 0.7 + 0.6 - 0.5$ $P(A \cup \overline{B}) = 1.3 - 0.5$ $\implies P(A \cup \overline{B}) = 0.8$

Step 3: Calculate the Numerator $P(B \cap (A \cup \overline{B}))$

• What we are doing: We need to find the probability of the intersection of event $B$ with event $(A \cup \overline{B})$.
• Why we are doing it: The expression $B \cap (A \cup \overline{B})$ can be simplified using the Distributive Law for sets, which will allow us to relate it to probabilities we've already calculated.

Using the Distributive Law for sets, $E_1 \cap (E_2 \cup E_3) = (E_1 \cap E_2) \cup (E_1 \cap E_3)$: $B \cap (A \cup \overline{B}) = (B \cap A) \cup (B \cap \overline{B})$ Now, consider the term $B \cap \overline{B}$. This represents the event where $B$ occurs AND $B$ does not occur, which is an impossible event (the empty set $\emptyset$). Therefore, $P(B \cap \overline{B}) = 0$.

Substituting this back into the expression for the numerator: $P(B \cap (A \cup \overline{B})) = P((B \cap A) \cup \emptyset)$ The union of any set with the empty set is just the original set: $P((B \cap A) \cup \emptyset) = P(B \cap A)$ From Step 1, we know $P(B \cap A) = P(A \cap B) = 0.2$. $\implies P(B \cap (A \cup \overline{B})) = 0.2$

Step 4: Apply the Conditional Probability Formula

• What we are doing: We now have both the numerator and the denominator, so we can directly apply the conditional probability formula.
• Why we are doing it: This is the final step to answer the original question.

Using the conditional probability formula: $P\left(B \mid (A \cup \overline{B})\right) = \frac{P(B \cap (A \cup \overline{B}))}{P(A \cup \overline{B})}$ Substitute the values calculated in Step 3 and Step 2: $P\left(B \mid (A \cup \overline{B})\right) = \frac{0.2}{0.8}$ $P\left(B \mid (A \cup \overline{B})\right) = \frac{2}{8}$ $\implies P\left(B \mid (A \cup \overline{B})\right) = \frac{1}{4}$

Common Mistakes & Tips

• Misinterpreting Set Operations: Ensure you correctly interpret $\cap$ (intersection), $\cup$ (union), and $\overline{E}$ (complement). A common mistake is confusing $A \cap \overline{B}$ with $A \setminus B$, which are equivalent, but also with $P(A) - P(B)$ (which is generally incorrect).
• Incorrectly Applying Distributive Law: Be careful when distributing intersection over union. Remember that $E \cap \overline{E} = \emptyset$.
• Calculation Errors: Decimal arithmetic can be tricky. Convert to fractions if it helps prevent errors.

Summary

This problem required a systematic application of fundamental probability rules and set theory identities. We first calculated $P(A \cap B)$ from the given $P(A)$ and $P(A \cap \overline{B})$. Then, we determined the denominator $P(A \cup \overline{B})$ using the Addition Rule. The numerator $P(B \cap (A \cup \overline{B}))$ was simplified using the Distributive Law and the property of a set intersecting its complement, reducing it to $P(A \cap B)$. Finally, we divided the numerator by the denominator to find the conditional probability. The calculated probability is $\frac{1}{4}$.

Final Answer

The final answer is $\boxed{\frac{1}{4}}$, which corresponds to option (C).