14. Luggage at the airport. 200 suitcases were loaded onto the plane. After the flight, passengers meet their suitcases on the conveyor belt. As soon as the conveyor belt started, the loader began placing suitcases on it one by one every two seconds in a random order (as they were brought). Among the passengers is a group of businessmen who are waiting for their 10 suitcases.

a) (from 8th grade, 3 points). Find the probability that the businessmen will have to wait exactly two minutes for their last suitcase after the conveyor belt starts.

b) (from 9th grade, 3 points). Find the expected value of the random variable "the time the businessmen will have to wait for their last suitcase."
Solution. a) To have exactly two minutes elapse, the last suitcase must be the 60th in the count. Among the first 59 suitcases, we select 9 spots for the 9 suitcases from the group we need, and place the 10th suitcase from this group in the 60th position. This results in exactly $C_{59}^{9}$ ways to choose 10 spots for 10 suitcases such that the last one is the 60th in the count. In total, there are $C_{200}^{10}$ equally possible ways to choose spots on the conveyor belt for 10 suitcases out of 200.

b) Let's imagine adding one more suitcase (the 11th businessman's suitcase, who did not fly) to the sequence of suitcases on the conveyor belt and consider this suitcase to be the last one on the belt. Then the entire sequence of 201 suitcases is divided into 11 groups, with the end of each group being a suitcase of one of the businessmen. By symmetry, the expected length of each group is $\frac{201}{11}$. Therefore, the expected length of 10 groups is $\frac{2010}{11}$, and the last suitcase will be expected to arrive after $\frac{4020}{11} \approx 365.45$ seconds, which is slightly more than 6 minutes.

Answer: a) $\frac{C_{59}^{9}}{C_{200}^{10}} ;$ b) $\frac{4020}{11}$ seconds.