1. **State the problem:** We have a total of 10,000 boxes of bananas from Ecuador and Honduras with the following counts: - Ecuador: 6,000 boxes (200 damaged, 840 overripe) - Honduras: 4,000 boxes (365 damaged, 295 overripe) We need to find probabilities related to damaged and overripe fruit. 2. **Calculate total damaged and overripe boxes:** $$\text{Total damaged} = 200 + 365 = 565$$ $$\text{Total overripe} = 840 + 295 = 1135$$ 3. **(a) Probability a box contains damaged fruit:** $$P(\text{damaged}) = \frac{\text{Total damaged}}{\text{Total boxes}} = \frac{565}{10000} = 0.0565$$ Probability a box contains overripe fruit: $$P(\text{overripe}) = \frac{1135}{10000} = 0.1135$$ 4. **(b) Probability the box is from Ecuador or Honduras:** There are only two origins so: $$P(\text{Ecuador or Honduras}) = P(\text{Ecuador}) + P(\text{Honduras}) = \frac{6000}{10000} + \frac{4000}{10000} = 1$$ 5. **(c) Probability that given overripe fruit, box is from Honduras:** Use conditional probability: $$P(\text{Honduras} | \text{overripe}) = \frac{P(\text{Honduras} \cap \text{overripe})}{P(\text{overripe})} = \frac{295/10000}{1135/10000} = \frac{295}{1135} \approx 0.2597$$ 6. **(d) Probability box contains damaged or overripe fruit:** - If mutually exclusive (no overlap): $$P(\text{damaged or overripe}) = P(\text{damaged}) + P(\text{overripe}) = 0.0565 + 0.1135 = 0.17$$ - If not mutually exclusive, and assuming damaged and overripe counts could overlap, total damaged or overripe is: $$P(\text{damaged or overripe}) = P(\text{damaged}) + P(\text{overripe}) - P(\text{damaged} \cap \text{overripe})$$ Since no overlap data provided, overlap is unknown. If overlap is zero, same as above. Otherwise, subtract overlap probability.