Question

Let \(A\) and \(B\) be two events such that \(P(A \cup B) = \frac{1}{6}\), \(P(A \cap B) = \frac{1}{4}\) and \(P(\overline{A}) = \frac{1}{4}\), where \(\overline{A}\) stands for the complement of event \(A\). Then, the events \(A\) and \(B\) are:

• A. mutually exclusive and independent
• B. independent but not equally likely
• C. equally likely but not independent
• D. equally likely and mutually exclusive

Hint:

Events are independent if \(P(A \cap B) = P(A)P(B)\). Mutually exclusive if \(P(A \cap B) = 0\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Find \(P(A)\) and \(P(B)\) using given values.

Step 2: Detailed Explanation:
\(P(\overline{A}) = \frac{1}{4} \implies P(A) = 1 - \frac{1}{4} = \frac{3}{4}\).
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
\(\frac{1}{6} = \frac{3}{4} + P(B) - \frac{1}{4}\)
\(\frac{1}{6} = \frac{1}{2} + P(B)\)
\(P(B) = \frac{1}{6} - \frac{1}{2} = \frac{1 - 3}{6} = -\frac{2}{6} = -\frac{1}{3}\) which is impossible.
Given probabilities are inconsistent. There's an error in the data. However, if we assume \(P(A \cup B) = \frac{5}{6}\) instead of \(\frac{1}{6}\), then \(P(B) = \frac{5}{6} - \frac{1}{2} = \frac{5-3}{6} = \frac{1}{3}\). Then \(P(A)P(B) = \frac{3}{4} \times \frac{1}{3} = \frac{1}{4} = P(A \cap B)\). So independent. \(P(A) \neq P(B)\), so not equally likely. Thus independent but not equally likely. Given the options, (B) is the intended answer.

Step 3: Final Answer:
Option (B) independent but not equally likely.