Here's the solution to question 1: 1. (i) Probability of Events A and B Given: $P(A' \cap B) = \frac{1}{7}$ and $P(A \cap B) = \frac{2}{9}$. (Note: We assume the given $P(A' \cap D)$ is a typo and should be $P(A' \cap B)$.) a) Calculate $P(B)$. The probability of event B occurring is the sum of the probabilities of (A and B) and (not A and B), since these are disjoint events that make up B. $$P(B) = P(A \cap B) + P(A' \cap B)$$ $$P(B) = \frac{2}{9} + \frac{1}{7}$$ To add these fractions, find a common denominator, which is $9 \times 7 = 63$. $$P(B) = \frac{2 \times 7}{9 \times 7} + \frac{1 \times 9}{7 \times 9}$$ $$P(B) = \frac{14}{63} + \frac{9}{63}$$ $$P(B) = \frac{14 + 9}{63}$$ $$P(B) = \boxed{\frac{23}{63}}$$ b) Calculate $P(A \setminus B)$. The notation $P(A \setminus B)$ typically means $P(A \text{ and not } B)$, which is $P(A \cap B')$. However, with the given information, $P(A \cap B')$ cannot be determined as $P(A)$ is unknown. A common alternative interpretation for $\setminus$ in some contexts, or a possible misprint for $|$, is conditional probability $P(A|B)$. We will proceed with this interpretation as it allows for a solvable problem. $$P(A|B) = \frac{P(A \cap B)}{P(B)}$$ Using the values from the given information and part a): $$P(A|B) = \frac{\frac{2}{9}}{\frac{23}{63}}$$ $$P(A|B) = \frac{2}{9