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Question from my book:

Rapid transit trains arrive at a given station every five minutes and depart after stopping at the station for one minute to drop off and pick up passengers. Assuming trains arrive every hour on the hour, what is the probability that a passenger will be able to board a train immediately if he or she arrives at the station at a random instant between 7:54 a.m. and 8:06 a.m.?

I don't even know how to start!

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up vote 1 down vote accepted

Draw a line. Maybe mark $0$ for $8$ o'clock. Then make marks, backward and forward, every minute. Label the forward ones $1$, $2$, $3$, and so on up to $6$. Label the backward ones $59$, $58$, and so on up to $54$.

Now mark the intervals in which the person is able to board immediately. One such interval is the interval between $0$ and $1$. Another is the interval between $5$ and $6$. You can do the one on the other side (there is only one interval there). Now we know fully what's going on.

Divide the total combined length of these intervals by $12$, the total length of time between $7\colon54$ and $8\colon06$.

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Why assume discrete and not continuous? – Inquest Sep 21 '12 at 5:38
I am in fact assuming continuous. The density function of the passenger's arrival time has, we assume, uniform continuous distribution on the interval from $7:54$ to $8:06$. The total combined lengths of the intervals of immediate boarding is $3$. – André Nicolas Sep 21 '12 at 5:42

Trains are in the station for exactly 3 minutes out of the 12 minute time frame: between 7:55 and 7:56, 8:00 and 8:01, and 8:05 and 8:06. Therefore a passanger has a 25% chance $\left(\frac{3}{12}\right)$, of arriving during one of those minutes.

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