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Two chinese boys, Aiguo and Chi, arrive at the Dongdan subway station at 8:00 am and wait for the train to Dawanglu, to go to school. Every minute there is a 70% probability for the train to come, but every time the train stops at the station, there is also a 60% probability the two guys miss it, as they are abstracted and playing with their mobile phones.
What is the probability they catch the train by 8:05 am (or, to be more precise, to catch "some" train by 8:05)?
Last edited by salem_ohio (2016-03-13 23:36:15)
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Does the first train come at 8:00 or 8:01?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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We don't know the exact time - it can be any time, with 70% probability.
Does the first train come at 8:00 or 8:01?
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Every minute there is a 70% probability for the train to come
Does this not imply that the trains are expected one a minute?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Not necessarily, there is also a chance it does not come, with a probability of 30%. It may come in the next minute, or even after some minutes.
Every minute there is a 70% probability for the train to come
Does this not imply that the trains are expected one a minute?
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If they miss the first train is another coming?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Yes
If they miss the first train is another coming?
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@bobbym: is it correct to assume that the probability (for the two guys to catch the train) within EVERY minute from 8:00 to 8:05 is independent, since it is uniformly distributed in the course of each minute?
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Hi;
I would think so but the problem is unclear to me.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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So, would it be [(7/10)x(4/10)]^5?
Hi;
I would think so but the problem is unclear to me.
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That sort of assumes that the trains come exactly every minute?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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I don't think it is like this. It says that every minute there is 70% probability that the train comes (within this minute). Right?
That sort of assumes that the trains come exactly every minute?
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That seems to mean there are 5 trains coming, one for each minute.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Yes, but with probability 70% in each minute. I am not sure what exactly this means. If, for example, the first train comes exactly at 8:01, and they miss it, then within the second minute, a second train may come, but with probability 70% (and they may also miss it).
For the first minute, the overall probability is 0.7 x 0.4, but how about in the 5 minutes?
That seems to mean there are 5 trains coming, one for each minute.
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Your interpretation seems to mean lots of trains. But if zillions of trains come... Do we not need to know how many trains are coming?
I am not sure what exactly this means.
I do not understand the question either.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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My understanding is that we have at maximum one train per minute: Since the probability for the train to come is 70% within each minute, this means it may come but there is also a chance it may not come (with 30% probability) within the first minute. Same for the second minute: again the probability for the train to come is 70% and so on.
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I think I understand now.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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but "in mathematics you don't understand things..." - right?
I think I understand now.
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True enough, I can work on it though. I am getting a probability of about .47
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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How??? FYI I don't know the answer
True enough, I can work on it though. I am getting a probability of about .47
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Just did a rough simulation. I think the exact answer is 301 / 625
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Bobby, sorry to sound a bit stupid; what is "simulation"?
Just did a rough simulation. I think the exact answer is 301 / 625
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Hi;
It is often the hardest thing in the world for me to explain my solutions. It is often easier to solve the problem itself! When I was writing articles for a programming journal I always had that difficulty and it still has persisted to this day. Anyway, here is what Wiki says:
A computer simulation is a simulation, run on a single computer, or a network of computers, to reproduce behavior of a system. The simulation uses an abstract model (a computer model, or a computational model) to simulate the system.
In simpler terms I created a small computer program that I believe models the system and ran it a couple of million times to see how the system behaves, what it will do.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi Bobby,
What if we want to solve it using probabilities? (because your solution doesn't help me understand the logic!)
Thank you!
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For the exact answer which I am not certain about I used a Markov chain. That is about as much probability as anybody can get.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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