Question

3. You are to wait for a bus to arrive. The bus arrives every 30 minutes, but you dont know the exact time it will arrive. Thus, you can wait any time between 0 and 30 minutes, and you believe all wait times are equally likely. Let X be the random variable of the amount of time in minutes vou will wait for the bus.

a) Say you wait for the bus on two independent days. What is the probability that you wait more than 20 minutes on both days? What about the probability of waiting more than 20 minutes on just one of the days?

Answer 1

Let X denote the waiting time for the bus Also, the waiting time is between 0 and 30 minutes, and all wait times are equally likely. Therefore, X has continuous uniform distribution over (0, 30) i.e. X ~ U(0, 30) Then probability density function of X is f_x(x) = { 1/30: 0 lessthanorequalto x lessthanorequalto 30 0: otherwise. Now, Let X: wait time for the bus on one day and Y: wait time for the bus on second day (other day) (another day) Here, X ~ U(0, 30) and Y ~ U(0, 30) Also, X and Y are independent. Now, f_x(x) = 1/30: 0 lessthanorequalto x lessthanorequalto 30 f_y (y) = 1/30: 0 lessthanorequalto y lessthanorequalto 30 prob. (you wait more than 20 minutes on both days) = P(X > 20, Y > 20) = P(X > 20) * P(Y > 20) -(since, X and Y are independent) = (Integral_20^30 1/30 dx) (Integral_20^30 1/30 dy) = (1/30 [x]_20^30) (1/30 [y]_20^30)