Probability of two buses being in a bus stop at the same time

Question

Question statement:

The bus type of A arrives to a certain bus stop every 30 minutes and waits for 5 minutes. The bus type of B arrives every 60 minutes and waits for 4 minutes. Probabilities of arrival of these two bus types have uniform density, and they are independent of each other.

A man goes to the bus stop at a random time, and starts waiting. What is the probability that two buses be in the bus stops at the same time (i.e.; their waiting time periods intersect with each other)?

This was a question I came across last year when I was studying probability theory. I couldn't solve it then and I am still not able to solve it. I can't even find a starting point. How are the probability questions of this type solved?

Answer 1

Let $t$ be the instant that bus A arrives. Then Bus B must arrive in the interval $t-4, t+5$ if you want both bus at the same time. Bus B arrival is uniformly distributed on on interval of amplitude 64 minutes, hence you have to choose 9 minutes out out 64, which is probability $9/64$.

Answer 2

Within each hourly period, Type A buses arrive twice, 30 minutes apart, and Type B buses once. Consider a period of 1 hour starting from when a Type A bus arrives. For two buses to be at the stop at the same time within that hour, a Type B bus must arrive during one of the following intervals (measured in minutes from the start of the hour): (0, 5), (26, 35), (56, 60). (If it arrives during (56, 60) then the previous Type B bus would have arrived during (-4, 0) so would have overlapped with the first Type A bus at time 0.) That makes a total of 18 minutes within the hour. So (given uniform density and independence) the probability that two buses are (ever) in the stop at the same time is 18/60 or 3/10.

The statement that the man goes to the stop and waits is not relevant to the probability that two buses are (ever) at the stop at the same time. Another question (not asked) would be: once the man starts waiting, what is the probability that a second bus arrives before the first bus to arrive has left? This probability is lower than 3/10 since any overlap of the waiting periods of Type A and Type B buses might relate to the second Type A bus to arrive.

Answer 3

I interpret the problem this way. The probability of the arrival of bus A is $\frac{1}{30}$ while the probability of the arrival of bus B is $\frac{1}{60}$. Since these two events are independent, we get that the joint probability is simply $\frac{1}{1800}$. The extreme cases are when bus B arrives 4 minutes before bus A and when bus B arrives 5 minutes after bus A, so we must sum all the cases in between to get the probability of the event we are interested in. Since the joint is uniform we get that this probability is just $\frac{9}{1800}=\frac{1}{200}$