Question

If \( A \), \( B \), and \( C \) are any three arbitrary events such that \(P(A) = P(B) = P(C) = \frac{1}{4}\), \( P(A \cap B) = P(B \cap C) = 0 \), and \( P(C \cap A) = \frac{1}{8} \), find the probability that at least one of the events \( A \), \( B \), or \( C \) occur.}

• A. \( \frac{1}{8} \)
• B. \( \frac{3}{8} \)
• C. \( \frac{5}{8} \)
• D. \( \frac{7}{8} \)

Hint:

To find the probability that at least one of several events occurs, use the inclusion-exclusion principle, which accounts for overlaps between events.

Answer 1

The Correct Option is D

We are asked to find the probability that at least one of the events \( A \), \( B \), or \( C \) occur. The probability of at least one of the events occurring is given by the formula: \[ P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C) \] We are given the following information: - \( P(A) = P(B) = P(C) = \frac{1}{4} \) - \( P(A \cap B) = P(B \cap C) = 0 \) - \( P(C \cap A) = \frac{1}{8} \) Now, substitute these values into the formula: \[ P(A \cup B \cup C) = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} - 0 - 0 - \frac{1}{8} + P(A \cap B \cap C) \] We are not given \( P(A \cap B \cap C) \), but since \( P(A \cap B) = 0 \) and \( P(B \cap C) = 0 \), it follows that: \[ P(A \cap B \cap C) = 0 \] Thus, the probability becomes: \[ P(A \cup B \cup C) = \frac{3}{4} - \frac{1}{8} = \frac{6}{8} - \frac{1}{8} = \frac{7}{8} \] Therefore, the probability that at least one of the events \( A \), \( B \), or \( C \) occur is \( \boxed{\frac{7}{8}} \).