Question:

Given $P(A) = 1/2$ and $P(B) = 2/3$. Events A and B are independent. 1. **Evaluate $P(A \cap B)$:** Since A and B are independent events, the probability of their intersection is the product of their individual probabilities: $$P(A \cap B) = P(A)P(B) = \left(\frac{1}{2}\right)\left(\frac{2}{3}\right) = \frac{2}{6} = \frac{1}{3}$$ Thus, the statement $P(A \cap B) = 1/3$ is TRUE. 2. **Evaluate $P(A \cup B)$:** The probability of the union of two events is given by the formula: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ Using the calculated value for $P(A \cap B)$: $$P(A \cup B) = \frac{1}{2} + \frac{2}{3} - \frac{1}{3} = \frac{3}{6} + \frac{4}{6} - \frac{2}{6} = \frac{3+4-2}{6} = \frac{5}{6}$$ Thus, the statement $P(A \cup B) = 5/6$ is TRUE. 3. **Evaluate $P(\bar{A} \cap \bar{B})$:** Since A and B are independent, their complements $\bar{A}$ and $\bar{B}$ are also independent. First, find the probabilities of the complements: $$P(\bar{A}) = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2}$$ $$P(\bar{B}) = 1 - P(B) = 1 - \frac{2}{3} = \frac{1}{3}$$ Then, for independent events $\bar{A}$ and $\bar{B}$: $$P(\bar{A} \cap \bar{B}) = P(\bar{A})P(\bar{B}) = \left(\frac{1}{2}\right)\left(\frac{1}{3}\right) = \frac{1}{6}$$ Alternatively, using De Morgan's laws and the probability of the union: $$P(\bar{A} \cap \bar{B}) = P(\overline{A \cup B}) = 1 - P(A \cup B) = 1 - \frac{5}{6} = \frac{1}{6}$$ Thus, the statement $P(\bar{A} \cap \bar{B}) = 1/6$ is TRUE. 4. **Evaluate $P(A|B)$:** For independent events A and B, the conditional probability of A given B is simply the probability of A: $$P(A|B) = P(A) = \frac{1}{2}$$ The statement claims $P(A|B) = 1/3$. Since $1/2 \neq 1/3$, this statement is FALSE. The TRUE statements are $P(A \cap B) = 1/3$, $P(A \cup B) = 5/6$, and $P(\bar{A} \cap \bar{B}) = 1/6$.