Based on the information provided, here are the step-by-step calculations:Let the events be defined as:$A_1$: Person has good health$A_2$: Person has average health$A_3$: Person has poor health$E$: Person contracts the diseaseFrom the data given:$P(A_1) = \frac{700}{1000} = 0.7$$P(A_2) = \frac{200}{1000} = 0.2$$P(A_3) = \frac{100}{1000} = 0.1$The conditional probabilities of contracting the disease are:$P(E|A_1) = 25\% = 0.25$$P(E|A_2) = 35\% = 0.35$$P(E|A_3) = 50\% = 0.50$(i) Probability that a randomly selected person has contracted the diseaseWe use the Law of Total Probability:$$P(E) = P(E|A_1)P(A_1) + P(E|A_2)P(A_2) + P(E|A_3)P(A_3)$$Substitute the values:$$P(E) = (0.25 \times 0.7) + (0.35 \times 0.2) + (0.50 \times 0.1)$$$$P(E) = 0.175 + 0.07 + 0.05$$$$P(E) = 0.295$$Answer: The probability that the person has contracted the disease is 0.295 (or 29.5%).(ii) Probability that the person is from category $A_2$, given that they have not contracted the diseaseLet $E'$ be the event that the person has not contracted the disease.We need to find $P(A_2|E')$.First, calculate the probability of not contracting the disease, $P(E')$:$$P(E') = 1 - P(E) = 1 - 0.295 = 0.705$$Next, calculate the probability of not contracting the disease given the person is in category $A_2$:$$P(E'|A_2) = 1 - P(E|A_2) = 1 - 0.35 = 0.65$$Now, apply Bayes' Theorem:$$P(A_2|E') = \frac{P(E'|A_2) \cdot P(A_2)}{P(E')}$$Substitute the values:$$P(A_2|E') = \frac{0.65 \times 0.2}{0.705}$$$$P(A_2|E') = \frac{0.13}{0.705}$$$$P(A_2|E') = \frac{130}{705} = \frac{26}{141}$$$$P(A_2|E') \approx 0.1844$$Answer: The probability that the person is from category $A_2$ given they have not contracted the disease is $\frac{26}{141}$ or approximately 0.184.