# Distribution of waiting time in bus station

From my window in the office, I can see a bus station, where buses come in average frequency of 15 minutes and the time between arrival of two buses distributed exponentially. One day, I went out my office to the station 10 minutes after a bus arrived. Suppose I walk 6 minutes:

a.What the probability I'll miss a bus?

b. What's the probability I'll miss two buses with independent arrival times?

c.What's the distribution of my waiting time in the station (after I arrived)?

About A+B: since the distribution of time between arrivals distributed exponentially, In A I found $P(X<16)$ Where X is 1st bus arrived at the station. In B I found $P(X\le a,Y\le 16-a)$ and since the arrivals are independent, $=P(X\le a)\cdot P(Y\le16-a)$. We can find all of them but I wonder if I had to express the answer in B without terms which include $a$.

About C:How generally can I find distributions given any details?

I'll be glad for help with C.

EDIT: (A) we had to calculate $P(X<6)$ (from 0 since waiting time is non-negative) equals $=\int_0^6\frac {e^{-\frac x {15}}} {15}=0.33$ . and in B :$P(X+Y<6)=\int_{0}^{6}\frac{e^{\frac{-t}{15}}}{15}\cdot \frac{e^{\frac{-6-t}{15}}}{15}dt=0.018$

• Because I need help deducing C. – user65985 Sep 28 '13 at 14:46
• Question (c) refers to the same quite general property of Poisson processes which was used in questions (a) and (b). Do you see this property? Can you deduce the answer? – Did Sep 28 '13 at 17:27
• Again, I'm too dumb to deudce it. maybe the waiting time in station distributed poisson but I can't see why (time between exponential events?). – user65985 Sep 29 '13 at 1:29

Let $X$ denote the time (measured in minutes) between the last bus which arrived at the bus station before you left your office and the next one. Let $Y$ denote the time (measured in minutes) between the next bus and the next bus after it. By hypothesis, the couple $(X,Y)$ is i.i.d. exponential with known parameter $\lambda=1/15$.

Question (a) asks for $P[X\lt16\mid X\gt10]$. This is not $P[X\lt16]$ but $P[X\lt6]$. Do you see why? Can you compute this value?

Additional hint: If $Z$ is exponentially distributed, then $P[Z\gt x+y\mid Z\gt y]=$ $____$.

Question (b) asks for $P[X+Y\lt16\mid X\gt10]$. The answer is $P[X+Y\lt6]$. Do you see why? Can you compute this value?

Question (c) refers to the same quite general property of Poisson processes which was used in questions (a) and (b). Do you see this property? Can you deduce the answer?

• Im not sure I understood why in a and B we take $X <10$... by the way I think that in b it had to be $P (Y <6)$ no? About C: what do you mea – user65985 Sep 18 '13 at 8:41
• Where do you see [X<10] in the answer? // See Additional hint. – Did Sep 18 '13 at 8:58
• About A: I know this characteristic of the distribution but here we have $[Z<x+y\mid Z>y]$ and same for b. Another issue is I still wonder why the common distribution of $(X,Y)$ is Poissonian and not only X,Y separate.calculating the answer: I think I can manage with a but with b, will it be enough to say it equals $\displaystyle{\sum_{k=1}^6{P(X=k)\cdot P(Y=6-k)}}$? About C: Still I can't see the desired property of Poisson (and in fact I don't know any special property of the distribution). Thanks again. – user65985 Sep 24 '13 at 0:25
• At one point one must stop dancing with words and start computing things, you know. Everything you need is the distribution of (X,Y). I give it in the answer. What is it? No, it is certainly not discrete! Now, to solve (a), compute (for real!) P[10<X<16] and P[X>10] and their ratio... – Did Sep 24 '13 at 5:24
• you are right. about B:I sum continuous probabilities (the $=$ is my mistake of course) that had to be something with double int I have troubles formulating (maybe $\int_{-\infty}^{\infty}\int_{-\infty}^{6-y}$). now am I right. about C: it follows its exponential since loss-of memory or something similar? Does loss-of-memory well defines exponential distribution? about dancing with words: Everyone can do computations but first I want to understand technique (even when it's trivial and tedious). – user65985 Sep 24 '13 at 5:58