91Ó°ÊÓ

To determine the distribution of the time it takes for each rider to get off the cable car, we note that the time between stops follows an exponential distribution with rate λ, and there are n stops before the last rider gets off. The sum of n independent exponential random variables with the same rate has a Gamma distribution. Therefore, the distribution of the time at which the ith rider gets off the car is a Gamma distribution with parameters i (shape) and λ (rate).

As we found in step 1, the distribution for the time at which the last (nth) rider gets off the car is a Gamma distribution with parameters n and λ. So, the distribution of the time at which the last rider departs the car is: \(T_n \sim Gamma(n, \lambda)\)

Now let's find the probability that all the other riders are home by the time the last rider departs, i.e., given that the last rider departs at time t. To do this, we look at the time taken by all the other riders to walk home, and find the probability that their walking time is less than or equal to the time t at which the last rider departs. The walking time of each rider has an exponential distribution with rate μ, which are independent of each other and the departure process. If we condition on the last rider departing at time t, and given that the i-th rider departs at time \(T_i\), then the probability of the i-th rider reaching home before the last rider departs can be expressed as: \(P(W_i \leq t - T_i)\) Here, \(W_i\) is the time taken by i-th rider to walk home. To find the probability that all other riders are home by the time the last rider departs, we need to find the joint probability of all riders reaching home before time t. That is the product of the probabilities for each rider given that the departure process is independent of the walking time process: \(P(W_1 \leq t - T_1, W_2 \leq t - T_2, ..., W_{n-1} \leq t - T_{n-1} | T_n = t)\) Since the riders’ walking times are independent, this probability can be written as the product of individual probabilities: \(P(W_1 \leq t - T_1 | T_n = t)P(W_2 \leq t - T_2 | T_n = t) ... P(W_{n-1} \leq t - T_{n-1} | T_n = t)\) Now using the property of the exponential distribution, we have: \(P(W_i \leq t - T_i | T_n = t) = 1-e^{-\mu(t - T_i)}\) Thus, the probability that all other riders are home by the time last rider departs is given by: \(\prod_{i=1}^{n-1} (1-e^{-\mu(t - T_i)})\) There is one final step -- we need to average this probability over the last rider's departure time distribution (Gamma distribution).