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Step-by-Step Solution
Step 1: Interpret the Given Conditions
We have three events A, B, and C, along with the following probabilities:
$P(\text{Exactly one of A or B}) = \frac{1}{4}$
$P(\text{Exactly one of B or C}) = \frac{1}{4}$
$P(\text{Exactly one of C or A}) = \frac{1}{4}$
$P(A \cap B \cap C) = \frac{1}{16}$
We need to find $P(A \cup B \cup C)$, which is the probability that at least one of A, B, or C occurs.
Step 2: Express “Exactly One Event” in Terms of Individual and Pairwise Probabilities
For “exactly one of A or B” to occur, the probability can be written as:
$$P(\text{Exactly one of A or B}) = P(A) + P(B) - 2P(A \cap B).$$
We are given that this equals $\frac{1}{4}$:
$$P(A) + P(B) - 2P(A \cap B) = \frac{1}{4} \quad \dots (1)$$
Similarly, we write:
$$P(\text{Exactly one of B or C}) = P(B) + P(C) - 2P(B \cap C) = \frac{1}{4} \quad \dots (2)$$
$$P(\text{Exactly one of C or A}) = P(C) + P(A) - 2P(C \cap A) = \frac{1}{4} \quad \dots (3)$$
Step 3: Add the Three Equations
Adding equations (1), (2), and (3), we get:
$$
[P(A) + P(B) - 2P(A \cap B)]
+ [P(B) + P(C) - 2P(B \cap C)]
+ [P(C) + P(A) - 2P(C \cap A)]
= \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}.
$$
Simplify the left-hand side:
$$
2[P(A) + P(B) + P(C)] - 2[P(A \cap B) + P(B \cap C) + P(C \cap A)]
= \frac{3}{4}.
$$
Divide both sides by 2:
$$
P(A) + P(B) + P(C) - [P(A \cap B) + P(B \cap C) + P(C \cap A)]
= \frac{3}{8}.
$$
Step 4: Use the Inclusion-Exclusion Principle
The probability of at least one of A, B, or C occurring is:
$$
P(A \cup B \cup C)
= P(A) + P(B) + P(C)
- [P(A \cap B) + P(B \cap C) + P(C \cap A)]
+ P(A \cap B \cap C).
$$
From the previous step, we have:
$$
P(A) + P(B) + P(C) - [P(A \cap B) + P(B \cap C) + P(C \cap A)]
= \frac{3}{8}.
$$
And we know $P(A \cap B \cap C) = \frac{1}{16}.$
Step 5: Calculate the Final Probability
Substitute these values into the formula for $P(A \cup B \cup C)$:
$$
P(A \cup B \cup C)
= \frac{3}{8} + \frac{1}{16}
= \frac{6}{16} + \frac{1}{16}
= \frac{7}{16}.
$$
Step 6: State the Result
Therefore, the probability that at least one of the events A, B, or C occurs is
$$
\frac{7}{16}.
$$
