If \(A\) and \(B\) be two events such that \(P(\bar{A} \cap \bar{B})=\frac{1}{6}, P(A \cap B)=\frac{1}{4}\) and \(P(\bar{A})=\frac{1}{4}\), then (1) \(P(B)=\frac{1}{3}\) (2) \(\mathrm{A}\) and \(\mathrm{B}\) are dependent events (3) \(P(\bar{A} \cup \bar{B})=\frac{1}{4}\) (4) \(\bar{A}\) and \(\bar{B}\) are exhaustive events

The Complement Rule is a useful concept in probability theory. It helps to find the probability of an event not happening. The complement of an event A is denoted as \(\bar{A}\). According to the rule, the probability of the complement of event A is given by: \[ P(\bar{A}) = 1 - P(A) \] In our exercise, we know that \(P(\bar{A}) = \frac{1}{4}\). Therefore, we can find \(P(A)\) as follows: \[ P(A) = 1 - P(\bar{A}) = 1 - \frac{1}{4} = \frac{3}{4} \]

In probability theory, two events A and B are called independent if the occurrence or non-occurrence of one does not affect the occurrence of the other. This is represented by: \[ P(A \cap B) = P(A) \cdot P(B) \] In our example, to determine if events A and B are independent, we need to compare \(P(A \cap B)\) and \(P(A) \cdot P(B)\). We already have the values \(P(A) = \frac{3}{4}\) and \(P(B) = \frac{1}{3}\). Hence, we calculate: \[ P(A) \cdot P(B) = \frac{3}{4} \times \frac{1}{3} = \frac{1}{4} \] Given that \(P(A \cap B) = \frac{1}{4}\), it matches our calculation for independent events. This means A and B are indeed independent.

The Total Probability Theorem is fundamental when dealing with complex probabilities that involve combining multiple events. It states that for any two events A and B: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] In our solution, we use this theorem to find \(P(\bar{A} \cap \bar{B})\) which is 1 minus the probability of \(A \cup B\). Given \(P(\bar{A} \cap \bar{B}) = \frac{1}{6}\), we compute \(P(A \cup B)\) as follows: - \(P(A \cup B) = 1 - P(\bar{A} \cap \bar{B}) = 1 - \frac{1}{6} = \frac{5}{6}\) Using the Total Probability Theorem: - \(\frac{5}{6} = P(A) + P(B) - P(A \cap B)\) - Substituting known probabilities: - \(\frac{5}{6} = \frac{3}{4} + \frac{1}{3} - \frac{1}{4} = \frac{1}{2} + \frac{1}{3}\) Giving the final value for \(P(B)\) as \(\frac{1}{3}\)