Question

For three events \( A, B, \) and \( C \) of a sample space, if \[ P(\text{exactly one of A or B occurs}) = P(\text{exactly one of B or C occurs}) = P(\text{exactly one of C or A occurs}) = \frac{1}{4} \] and the probability that all three events occur simultaneously is \( \frac{1}{16} \), then the probability that at least one of the events occurs is

• A. \( \frac{3}{16} \)
• B. \( \frac{5}{16} \)
• C. \( \frac{7}{16} \)
• D. \( \frac{7}{32} \)

Hint:

For three events, the probability of at least one occurring is given by: \[ P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(AB) - P(BC) - P(CA) + P(ABC) \] Use this with logical deductions when exact values are not directly given.

Answer 1

The Correct Option is C

Let: - \( P(\text{Exactly one of A or B occurs}) = P(A \oplus B) = \frac{1}{4} \) - Similarly for \( B \oplus C \) and \( C \oplus A \), each is \( \frac{1}{4} \) - Let \( P(A \cap B \cap C) = \frac{1}{16} \) Now, use the identity: \[ P(\text{At least one of } A, B, C) = 1 - P(\text{none of them occur}) \] But directly using the given data and formula: Let the required probability be \( P(A \cup B \cup C) \). From inclusion-exclusion: \[ P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(AB) - P(BC) - P(CA) + P(ABC) \] We won't get exact values for each term, but we can use the idea that exactly one of two occurs means: \[ P(A \oplus B) = P(A \cup B) - 2P(AB) \] So reverse it: \[ P(A \cup B) = \frac{1}{4} + 2P(AB) \] (similar for other pairs) By using symmetry and manipulating these equations and plugging in \( P(ABC) = \frac{1}{16} \), it can be shown (via algebraic simplification) that: \[ P(A \cup B \cup C) = \frac{7}{16} \] Final Answer: \( \boxed{\frac{7}{16}} \)