Key Concepts and Formulas

To solve this problem, we'll utilize fundamental concepts and formulas from probability theory.

1. Addition Rule for Probabilities: This rule helps determine the probability that at least one of two events, $A$ or $B$, occurs. $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ This formula can be rearranged to find the probability of the intersection: $P(A \cap B) = P(A) + P(B) - P(A \cup B)$ Explanation: When we sum $P(A)$ and $P(B)$, the probability of their overlap, $P(A \cap B)$, is counted twice. Subtracting it once corrects for this double-counting.

2. Probability of a Complementary Event: The complement of an event $E$, denoted as $E'$, is the event that $E$ does not occur. $P(E') = 1 - P(E)$ Explanation: An event either occurs or does not occur, so the sum of its probability and the probability of its complement must be 1.

3. Probability of a Difference Event: This refers to the probability that event $A$ occurs, but event $B$ does not, denoted as $P(A \cap B')$. $P(A \cap B') = P(A) - P(A \cap B)$ Similarly, for event $B$ occurring and event $A$ not occurring: $P(B \cap A') = P(B) - P(A \cap B)$ Explanation: In a Venn diagram, $A \cap B'$ is the part of event $A$ that does not overlap with event $B$. We get this by taking the entire probability of $A$ and subtracting the part that is shared with $B$.

4. Conditional Probability: This measures the probability of an event $A$ occurring, given that another event $B$ has already occurred. The sample space is effectively reduced to event $B$. $P(A \mid B) = \frac{P(A \cap B)}{P(B)}, \quad \text{provided } P(B) > 0$ Explanation: We are interested in the proportion of times $A$ occurs within the instances where $B$ occurs. The numerator is the probability of both occurring, and the denominator is the probability of the condition occurring.

Step-by-Step Solution

Our goal is to calculate $P\left(A \mid B^{\prime}\right)+P\left(B \mid A^{\prime}\right)$ using the given probabilities: $P(A)=\frac{1}{3}$, $P(B)=\frac{1}{5}$, and $P(A \cup B)=\frac{1}{2}$.

Step 1: Calculate the Probability of the Intersection, $P(A \cap B)$

• What we are doing: We need $P(A \cap B)$ because it's a component for calculating difference events and is derived directly from the given union probability.
• Why: The difference events $P(A \cap B')$ and $P(B \cap A')$ (which are the numerators for our conditional probabilities) depend on $P(A \cap B)$.
• Using the Addition Rule for Probabilities: $P(A \cap B) = P(A) + P(B) - P(A \cup B)$
• Substitute the given values: $P(A \cap B) = \frac{1}{3} + \frac{1}{5} - \frac{1}{2}$
• Find a common denominator (30) to perform the arithmetic: $P(A \cap B) = \frac{10}{30} + \frac{6}{30} - \frac{15}{30}$ $P(A \cap B) = \frac{10 + 6 - 15}{30} = \frac{16 - 15}{30} = \frac{1}{30}$ So, $P(A \cap B) = \frac{1}{30}$.

Step 2: Calculate Probabilities of Complementary Events, $P(A')$ and $P(B')$

• What we are doing: We need the probabilities of the complements of A and B, which will serve as the denominators for the conditional probabilities $P(B \mid A')$ and $P(A \mid B')$, respectively.
• Why: The definition of conditional probability $P(X \mid Y) = \frac{P(X \cap Y)}{P(Y)}$ requires $P(Y)$, which in our case are $P(A')$ and $P(B')$.
• Using the Probability of a Complementary Event formula, $P(E') = 1 - P(E)$:
  • For $P(A')$: $P(A') = 1 - P(A) = 1 - \frac{1}{3} = \frac{2}{3}$
  • For $P(B')$: $P(B') = 1 - P(B) = 1 - \frac{1}{5} = \frac{4}{5}$

Step 3: Calculate Probabilities of Difference Events, $P(A \cap B')$ and $P(B \cap A')$

• What we are doing: We calculate the probabilities of $A$ occurring without $B$, and $B$ occurring without $A$. These will be the numerators for our conditional probabilities.
• Why: The definition of conditional probability $P(X \mid Y) = \frac{P(X \cap Y)}{P(Y)}$ requires $P(X \cap Y)$, which in our case are $P(A \cap B')$ and $P(B \cap A')$.
• Using the Probability of a Difference Event formula, $P(X \cap Y') = P(X) - P(X \cap Y)$:
  • For $P(A \cap B')$: $P(A \cap B') = P(A) - P(A \cap B) = \frac{1}{3} - \frac{1}{30}$ $P(A \cap B') = \frac{10}{30} - \frac{1}{30} = \frac{9}{30} = \frac{3}{10}$
  • For $P(B \cap A')$: $P(B \cap A') = P(B) - P(A \cap B) = \frac{1}{5} - \frac{1}{30}$ $P(B \cap A') = \frac{6}{30} - \frac{1}{30} = \frac{5}{30} = \frac{1}{6}$

Step 4: Calculate the Conditional Probabilities, $P(A \mid B')$ and $P(B \mid A')$

• What we are doing: We now have all the components to calculate the two conditional probabilities requested in the problem.
• Why: These are the terms we need to sum to get the final answer.
• Using the Conditional Probability formula, $P(X \mid Y) = \frac{P(X \cap Y)}{P(Y)}$:
  • For $P(A \mid B')$: $P\left(A \mid B^{\prime}\right) = \frac{P(A \cap B')}{P(B')} = \frac{\frac{3}{10}}{\frac{4}{5}}$ $P\left(A \mid B^{\prime}\right) = \frac{3}{10} \times \frac{5}{4} = \frac{15}{40} = \frac{3}{8}$
  • For $P(B \mid A')$: $P\left(B \mid A^{\prime}\right) = \frac{P(B \cap A')}{P(A')} = \frac{\frac{1}{6}}{\frac{2}{3}}$ $P\left(B \mid A^{\prime}\right) = \frac{1}{6} \times \frac{3}{2} = \frac{3}{12} = \frac{1}{4}$

Step 5: Sum the Conditional Probabilities

• What we are doing: This is the final step as required by the question.
• Why: We are asked to find the sum of these two probabilities.
• Add the two calculated conditional probabilities: $P\left(A \mid B^{\prime}\right) + P\left(B \mid A^{\prime}\right) = \frac{3}{8} + \frac{1}{4}$
• Find a common denominator (8) to add the fractions: $= \frac{3}{8} + \frac{1 \times 2}{4 \times 2} = \frac{3}{8} + \frac{2}{8} = \frac{3+2}{8} = \frac{5}{8}$

Common Mistakes & Tips

• Fraction Arithmetic: A common source of errors in probability problems is mistakes in adding, subtracting, multiplying, or dividing fractions. Always ensure your fraction calculations are accurate and simplify fractions whenever possible.
• Misinterpreting Notation: Understand the precise meaning of $P(A \cup B)$, $P(A \cap B)$, $P(A')$, and especially $P(A \mid B)$. A conditional probability statement like $P(A \mid B')$ is not the same as $P(A \cap B')$ or $P(A \cup B')$.
• Venn Diagrams: For expressions involving unions, intersections, and complements (like $A \cap B'$), visualizing with a Venn diagram can help confirm the correct formula. For instance, $P(A \cap B')$ represents the area of circle A that is outside circle B.

Summary

We systematically solved the problem by first calculating the probability of the intersection of events $A$ and $B$. Then, we found the probabilities of the complementary events $A'$ and $B'$. These values allowed us to determine the probabilities of the difference events ($A$ only, $B$ only) which form the numerators for our conditional probabilities. Finally, we calculated the two required conditional probabilities and summed them to arrive at the final answer.

The final answer is $\boxed{\frac{5}{8}}$, which corresponds to option (B).