91Ó°ÊÓ

We are given the probability that a person remains with the company for 10 years or longer, which is \(1/6\). We need to find the probability that the man will work there less than 10 years, which is the complement of the given probability.

To find the probability that the man will work there less than 10 years, we just need to subtract the given probability from \(1\): \(P(\text{man works less than 10 years}) = 1 - P(\text{man works 10 years or more})\) \(P(\text{man works less than 10 years}) = 1 - \frac{1}{6}\) \(P(\text{man works less than 10 years}) = \frac{5}{6}\). b. Probability that both the man and the woman will work there less than 10 years

Since the man's and woman's lengths of service are independent, we can calculate the probability that both work less than 10 years by simply multiplying their individual probabilities: \(P(\text{both work less than 10 years}) = P(\text{man works less than 10 years}) \times P(\text{woman works less than 10 years})\) \(P(\text{both work less than 10 years}) = \frac{5}{6} \times \frac{5}{6}\) \(P(\text{both work less than 10 years}) = \frac{25}{36}\). c. Probability that one or the other or both will work 10 years or longer

We can break this down into three cases: (1) the man works 10 years or longer and the woman works less than 10 years, (2) the woman works 10 years or longer and the man works less than 10 years, and (3) both work 10 years or longer. We then compute the probabilities of these cases separately and sum them up: \(P(\text{one or both work 10 years or more}) = P(\text{man works 10 years or more, woman works less than 10 years})\) \(+ P(\text{woman works 10 years or more, man works less than 10 years})\) \(+ P(\text{both work 10 years or more})\) \(P(\text{one or both work 10 years or more}) = \frac{1}{6} \cdot \frac{5}{6} + \frac{5}{6} \cdot \frac{1}{6} + \frac{1}{6} \cdot \frac{1}{6}\) \(P(\text{one or both work 10 years or more}) = \frac{5}{36} + \frac{5}{36} + \frac{1}{36}\) \(P(\text{one or both work 10 years or more}) = \frac{11}{36}\).