Answer:
the probability that an employee takes more than 5 sick days in at most one of the years in a five-year period is approximately 0.946.
Explanation:
Let X be the number of sick days an employee takes in one year. We know that X is uniformly distributed on the values 0, 1, 2, …, 10, which means that the probability of X taking any value in this range is the same.
Let Y be the number of years in the five-year period that the employee takes more than 5 sick days. We want to find the probability that Y is at most 1.
First, we can find the probability that an employee takes more than 5 sick days in any given year. Since X is uniformly distributed on the values 0, 1, 2, …, 10, we have:
P(X > 5) = (10 - 5 + 1) / (10 + 1) = 6 / 11
Next, we can use the binomial distribution to find the probability that an employee takes more than 5 sick days in exactly one year out of five. Let p be the probability that an employee takes more than 5 sick days in any given year (i.e., p = 6/11), and let n = 5 be the number of years. Then:
P(Y = 1) = C(5, 1) * p * (1 - p)^(5-1) = 5 * (6/11) * (5/11)^4
Finally, we can find the probability that an employee takes more than 5 sick days in at most one of the years by adding the probabilities of taking more than 5 sick days in zero years and in one year:
P(Y ≤ 1) = P(Y = 0) + P(Y = 1) = (1 - p)^5 + 5 * (6/11) * (5/11)^4
Therefore, the probability that an employee takes more than 5 sick days in at most one of the years in a five-year period is approximately 0.946.