a) State in words the relationship between two events R and S when P(R \cap S) = 0 The events A and B are independent with P(A) = \frac{1}{4} and P(A \cup B) = \frac{2}{3} Find b) P(B) c) P(A \cap B) d) P(B|A) - Edexcel - A-Level Maths Statistics - Question 2 - 2012 - Paper 1

To find P(B), we can use the formula for the probability of the union of independent events: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ Since A and B are independent, we have: $P(A \cap B) = P(A) \cdot P(B)$ Substituting the given values: $\frac{2}{3} = \frac{1}{4} + P(B) - \left( \frac{1}{4} \cdot P(B) \right)$ This simplifies to: $\frac{2}{3} = \frac{1}{4} + P(B)(1 - \frac{1}{4})$ Thus, $\frac{2}{3} = \frac{1}{4} + \frac{3}{4}P(B)$ Now, isolating P(B): $P(B) = \frac{2/3 - 1/4}{3/4} = \frac{8/12 - 3/12}{3/4} = \frac{5/12}{3/4} = \frac{5}{9}$ Therefore, P(B) = \frac{5}{9}.

To find P(A ∩ B), we can use the independence property of the events: $P(A \cap B) = P(A) \cdot P(B)$ We already calculated P(B) as \frac{5}{9}, so: $P(A \cap B) = \frac{1}{4} \cdot \frac{5}{9} = \frac{5}{36}.$

To find the conditional probability P(B|A), we use the formula: $P(B|A) = \frac{P(A \cap B)}{P(A)}$ From part (c), we have: $P(A \cap B) = \frac{5}{36}$ Substituting the value of P(A): $P(B|A) = \frac{\frac{5}{36}}{\frac{1}{4}} = \frac{5}{36} \cdot \frac{4}{1} = \frac{20}{36} = \frac{5}{9}.$