QUESTION: A person and a bus will arrive at a bus stop at any random time between $12$ pm to $1$ pm. The person will wait at the bus stop for 20 minutes, and the bus will wait for 5 minutes. Find the probability that the person catches the bus.
MY ANSWER:. I know that this can be easily solved by geometry, but I want to know where hides the mistake in my approach. Here is how I have done it$→$
Starting from 12:00 to 01:00, we have got exactly 61 ways for each to arrive $\{0,1,2,…..,60\}$, (where each number represents the minute at which one arrives. $0$ represents $12:00$ and therefore $60$ represents $01:00$) .
Now observe, that if the person arrives at $0$ then the bus must arrive between $0$ to $20$ and therefore has exactly $21$ ways to do so.
If he arrives at $1$ then bus may arrive between $0$ to $21$ and therefore has exactly $22$ ways to do so (since bus will wait for $5$ minutes).
This continues till he arrives at $6$ and therefore the bus may arrive between $1$ and $26$ and therefore has exactly $26$ ways to do so.
And the number of ways for the bus to arrive remains to be $26$ until he arrives at $40$.
When he arrives at $41$, the bus may arrive between $36$ and $60$ and therefore has exactly $25$ ways to do so (including the way $36$).
Now the number of ways for the bus to arrive goes on decreasing (since neither the bus nor the person can arrive after $01:00$) until it reaches the value $6$ when the person arrives at $60$.
Now compiling this result we obtain a simple expression that looks like $→$
$\frac{21}{61} + \frac{22}{61} + \frac{23}{61} + \frac{24}{61} +\frac{25}{61} + (\frac{26}{61}*36) + \frac{25}{61} + \frac{24}{61} + \frac{23}{61}+…..+ \frac{6}{61}$
(Note: we have multiplied $36$ since from 5 to 40 we have 36 ways, including both)
This expression can be easily solved ( by using, $\frac{n(n+1)}{2}$ or some other way) and the answer comes out to be $\frac{1361}{61}$.
Since with every term of the expression, we have considered that the person arrives at some exact minute, each term should have been multiplied by the way in which the person can arrive at that minute, which is nothing but $\frac{1}{61}$ .
Therefore we arrive at the answer, $\frac{1361}{61}. \frac{1}{61}$ which is $0.365761892$.
But the answer given is $0.3576388889$ or $\frac{103}{288}$, which clearly is missed by a narrow margin..
Where is the mistake in my answer? I mean, where is the flaw in the logic we have implemented… Can anyone please help me out?
Thank you so much.
Best Answer
The problem is continuous and must be treated as continuous. For example, if the person arrives at the earliest time 12, he can wait 20 minutes, which is $\frac{1}{3}$ of the bus's duration, but in your solution, he can wait for $\frac{21}{61}\neq \frac{1}{3}$, so there will be a small discrepancy.