1. **Problem Statement:** We have a contingency table showing proportions of loan types (Fixed Rate (FR) and Adjustable Rate (ADJ)) and loan purposes (Home (H), Automobile (A), Personal (P)): | Loan Type | H | A | P | |-----------|------|------|------| | FR | 0.27 | 0.19 | 0.14 | | ADJ | 0.13 | 0.09 | 0.18 | We want to find: a. The probability that a randomly selected person has an automobile loan and it is fixed rate. b. Given the person has an adjustable rate loan, the probability that it is for a home. c. Given the person does not have a personal loan, the probability it is an adjustable rate. --- 2. **Formulas and Rules:** - Joint probability: $P(A \cap B)$ is the probability both events A and B occur. - Conditional probability: $P(A|B) = \frac{P(A \cap B)}{P(B)}$. - Complement rule: $P(\text{not } C) = 1 - P(C)$. --- 3. **Calculations:** **a. Probability of automobile loan and fixed rate:** This is the joint probability $P(\text{A and FR})$ from the table: $$P(\text{A and FR}) = 0.19$$ **b. Probability of home loan given adjustable rate:** First, find total probability of adjustable rate loans: $$P(\text{ADJ}) = 0.13 + 0.09 + 0.18 = 0.40$$ Then, find joint probability of home and adjustable rate: $$P(\text{H and ADJ}) = 0.13$$ Conditional probability: $$P(\text{H} | \text{ADJ}) = \frac{P(\text{H and ADJ})}{P(\text{ADJ})} = \frac{0.13}{0.40} = 0.325$$ **c. Probability of adjustable rate given not personal loan:** First, find total probability of personal loans: $$P(\text{P}) = 0.14 + 0.18 = 0.32$$ Probability of not personal loan: $$P(\text{not P}) = 1 - 0.32 = 0.68$$ Find probability of adjustable rate and not personal loan (home or automobile): $$P(\text{ADJ and not P}) = P(\text{ADJ and H}) + P(\text{ADJ and A}) = 0.13 + 0.09 = 0.22$$ Conditional probability: $$P(\text{ADJ} | \text{not P}) = \frac{P(\text{ADJ and not P})}{P(\text{not P})} = \frac{0.22}{0.68} \approx 0.3235$$ --- **Final answers:** - a. $0.19$ - b. $0.325$ - c. $0.3235$