1. Key Concepts and Formulas

• Probability of "Exactly One" Event Occurring (for two events): For any two events $X$ and $Y$, the probability that exactly one of them occurs (meaning $X$ occurs and $Y$ does not, OR $Y$ occurs and $X$ does not) is given by the formula: $P(\text{Exactly one of X or Y occurs}) = P(X) + P(Y) - 2P(X \cap Y)$ This formula is derived from the fact that $P(\text{Exactly one of X or Y}) = P(X \cup Y) - P(X \cap Y)$. Substituting $P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)$ leads to the above expression.

• Principle of Inclusion-Exclusion for Three Events: The probability that at least one of three events $A$, $B$, or $C$ occurs (i.e., the probability of their union) is given by: $P(A \cup B \cup C) = P(A) + P(B) + P(C) - [P(A \cap B) + P(B \cap C) + P(C \cap A)] + P(A \cap B \cap C)$ This principle systematically adds and subtracts probabilities to ensure that each outcome belonging to the union of the events is counted exactly once.

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2. Step-by-Step Solution

Step 1: Translate the given information into mathematical equations. The problem provides specific probabilities for "exactly one" of two events occurring, and the probability of all three events occurring simultaneously. We will use the formula for "Exactly one of two events" to express the first three conditions.

• For events A and B: We are given that $P(\text{Exactly one of A or B occurs}) = \frac{1}{4}$. Applying the formula, we get: $P(A) + P(B) - 2P(A \cap B) = \frac{1}{4} \quad \text{(Equation 1)}$

  • Reasoning: This step converts the descriptive language of the problem into a precise algebraic equation, which is essential for mathematical manipulation. It establishes a relationship between the individual probabilities of A and B and the probability of their intersection.

• For events B and C: Similarly, we are given $P(\text{Exactly one of B or C occurs}) = \frac{1}{4}$. Applying the formula for events B and C: $P(B) + P(C) - 2P(B \cap C) = \frac{1}{4} \quad \text{(Equation 2)}$

  • Reasoning: We apply the same principle to the pair of events B and C, obtaining a similar equation that will be useful when combined with others.

• For events C and A: And we are given $P(\text{Exactly one of C or A occurs}) = \frac{1}{4}$. Applying the formula for events C and A: $P(C) + P(A) - 2P(C \cap A) = \frac{1}{4} \quad \text{(Equation 3)}$

  • Reasoning: This completes the set of equations for all three distinct pairs of events (A, B), (B, C), and (C, A), providing a comprehensive set of relationships based on the given information.

• For all three events: We are directly given $P(\text{All the three events occur simultaneously}) = \frac{1}{16}$. This translates to: $P(A \cap B \cap C) = \frac{1}{16} \quad \text{(Equation 4)}$

  • Reasoning: This is a direct translation of a given value, which will be a crucial component in the final application of the Inclusion-Exclusion Principle.

Step 2: Strategically combine the equations to find a key sum. Our ultimate goal is to find $P(A \cup B \cup C)$. If we look at the Principle of Inclusion-Exclusion formula, we need two main components: the sum of individual probabilities ($P(A) + P(B) + P(C)$) and the sum of pairwise intersection probabilities ($P(A \cap B) + P(B \cap C) + P(C \cap A)$). Notice that Equations 1, 2, and 3 contain these terms. A powerful strategy here is to add these three equations together.

Let's add Equation 1, Equation 2, and Equation 3: $(P(A) + P(B) - 2P(A \cap B)) + (P(B) + P(C) - 2P(B \cap C)) + (P(C) + P(A) - 2P(C \cap A)) = \frac{1}{4} + \frac{1}{4} + \frac{1}{4}$

Now, let's group and combine the like terms on the left side: $2P(A) + 2P(B) + 2P(C) - 2P(A \cap B) - 2P(B \cap C) - 2P(C \cap A) = \frac{3}{4}$

We can factor out a 2 from all terms on the left side: $2[P(A) + P(B) + P(C) - (P(A \cap B) + P(B \cap C) + P(C \cap A))] = \frac{3}{4}$

• Reasoning: By summing these equations, we have cleverly constructed an expression that directly matches a major part of the Inclusion-Exclusion Principle. This avoids the much more complicated process of solving for each individual probability term separately, offering an efficient and elegant way to extract the needed sum.

Finally, divide both sides by 2 to isolate this crucial sum: $P(A) + P(B) + P(C) - [P(A \cap B) + P(B \cap C) + P(C \cap A)] = \frac{3}{8} \quad \text{(Equation 5)}$

• Reasoning: This step isolates the exact term required by the Principle of Inclusion-Exclusion, making it ready for direct substitution.

Step 3: Apply the Principle of Inclusion-Exclusion. Now we have all the necessary components to calculate $P(A \cup B \cup C)$ using the Principle of Inclusion-Exclusion. The formula is: $P(A \cup B \cup C) = P(A) + P(B) + P(C) - [P(A \cap B) + P(B \cap C) + P(C \cap A)] + P(A \cap B \cap C)$

We have already found the value of the bracketed expression from Equation 5: $P(A) + P(B) + P(C) - [P(A \cap B) + P(B \cap C) + P(C \cap A)] = \frac{3}{8}$

And we are directly given the probability of all three events occurring simultaneously from Equation 4: $P(A \cap B \cap C) = \frac{1}{16}$

Substitute these values into the Inclusion-Exclusion formula: $P(A \cup B \cup C) = \frac{3}{8} + \frac{1}{16}$

• Reasoning: This is the penultimate step where we bring together all the pieces of information (both derived and given) into the master formula to solve for the desired probability.

Step 4: Perform the final calculation. To complete the calculation, we need to add the two fractions. We find a common denominator, which is 16: $P(A \cup B \cup C) = \frac{3 \times 2}{8 \times 2} + \frac{1}{16}$ $P(A \cup B \cup C) = \frac{6}{16} + \frac{1}{16}$ $P(A \cup B \cup C) = \frac{6+1}{16}$ $P(A \cup B \cup C) = \frac{7}{16}$

• Reasoning: This final arithmetic step provides the numerical answer to the problem.

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3. Common Mistakes & Tips

• Distinguish "Exactly One" from "At Least One": A very common mistake is to confuse $P(\text{Exactly one of A or B})$ with $P(A \cup B)$ (at least one) or $P(A \cap B)$ (both). Remember their distinct definitions and formulas.
• Algebraic Precision: Pay close attention to signs (especially the minus signs in the Inclusion-Exclusion Principle) and ensure accurate fraction arithmetic. A small calculation error can lead to a completely wrong answer.
• Systematic Approach: Break down complex probability problems into smaller, manageable steps. First, define events and their given probabilities. Second, identify the target probability. Third, find a path (like the Inclusion-Exclusion Principle) that connects the given information to the target.

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4. Summary

This problem required us to calculate the probability that at least one of three events occurs. We started by translating the given conditions about "exactly one" event occurring for each pair into algebraic equations using the formula $P(X) + P(Y) - 2P(X \cap Y)$. By summing these three equations, we efficiently obtained a combined expression that represented a crucial part of the Principle of Inclusion-Exclusion for three events: $P(A) + P(B) + P(C) - [P(A \cap B) + P(B \cap C) + P(C \cap A)]$. Finally, we substituted this derived sum and the given probability of all three events occurring simultaneously into the complete Inclusion-Exclusion formula to arrive at the final probability.

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5. Final Answer

The probability that at least one of the events occurs is $\frac{7}{16}$. The final answer is \boxed{\frac{7}{16}} which corresponds to option (A).