Question

Problem 5. Buses arrive at 116th and Broadway at the times of a Poisson arrival process with intensity λ arrivals per hour. These may either be M104 buses or M6 buses; the chance that a bus is an M104 is 0.6, while the chance that it is an M6 is 0.4, and the types (M6 or M104) of successive buses are independent. (a) If I wait for an M104 bus, what is the chance that I will wait longer than x hours? (b) What is the probability that two M6 buses and no M104 buses arrive in the first x hours? (c) What is the expected number of hours until the third M6 arrives? (d) What is the variance of the number of hours until the third M6 arrives?

Answer 1

Answer:

Step-by-step explanation:

The problem states that there are only two types of busses - M104 and M6 with probable occurence of 0.6 and 0.4 respectively.

If the average number of busses arriving per hour is λ, the average number of M6 busses per hour is  0.4λ

Now consider a set of 3 M6 busses as an event. The average number of such events per hour will be

μ = 0.4λ / 3

The expected number of hours for the event "THIRD M6 arrives", let's say X is

E[X] = 1 / μ ( exponential distribution) = 3 / 0.4λ

= 7.5 / λ

The variance of event X is =

$Var[x] = \frac{1}{U^2} = \frac{56.25}{\lambda ^2}$