Sticky Maths Problem: REP FOR SOLUTION

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GH
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#1
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Posted in the Academic subform. Too little traffic.

I have a rather sticky problem which I might tantalize the intelligentia of the forum. Rep for helping.

There are 2 people, who want to meet up. Between the hours of 6.oopm to 7.oo pm (ie. 1 hour). Each will stay for 5 minutes before going away. So, they must meet each other within the 5 minutes.

They will both arrive at random times between the hours of 6-7pm.

So, what is the probability of them meeting together?
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Muse
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(Original post by 2776)
Posted in the Academic subform. Too little traffic.

I have a rather sticky problem which I might tantalize the intelligentia of the forum. Rep for helping.

There are 2 people, who want to meet up. Between the hours of 6.oopm to 7.oo pm (ie. 1 hour). Each will stay for 5 minutes before going away. So, they must meet each other within the 5 minutes.

They will both arrive at random times between the hours of 6-7pm.

So, what is the probability of them meeting together?
I checked with someone else and they said it was 1/144 as well.
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curryADD
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(Original post by 2776)
Posted in the Academic subform. Too little traffic.

I have a rather sticky problem which I might tantalize the intelligentia of the forum. Rep for helping.

There are 2 people, who want to meet up. Between the hours of 6.oopm to 7.oo pm (ie. 1 hour). Each will stay for 5 minutes before going away. So, they must meet each other within the 5 minutes.

They will both arrive at random times between the hours of 6-7pm.

So, what is the probability of them meeting together?
1/24th because 5/60 is for both people. since you want both people to be there for the same time and they are arriving randomly its 1/24th!
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GH
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(Original post by timeofyourlife)
I checked with someone else and they said it was 1/144 as well.
Oh, tell me who. I know for a fact that ?/144 is teh correct answer. But I was more think about (5/12)^2. Do you have any Maths peeps in your uni accomadtion area?
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Muse
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(Original post by 2776)
Oh, tell me who. I know for a fact that ?/144 is teh correct answer. But I was more think about (5/12)^2. Do you have any Maths peeps in your uni accomadtion area?
Just asked someone else online so not particularly accurate! If it was 5/12 x 5/12 then it would be 25/144 which surely is too high a probability?
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me!
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5/60 X 5/60

= 1/12 X 1/12

= 1/144
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GH
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(Original post by timeofyourlife)
Just asked someone else online so not particularly accurate! If it was 5/12 x 5/12 then it would be 25/144 which surely is too high a probability?
Hmm, but the longest time period possible for a meeting is 9.9999 minutes, as one person just catches the waiting one at the end of his waiting period.

And that is 1/6 of the time from 6-7pm.
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Sire
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I think you'll find that it is 1/210. I could be wrong, but I just did it long hand match up and it seems to work.
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me!
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(Original post by 2776)
Hmm, but the longest time period possible for a meeting is 9.9999 minutes, as one person just catches the waiting one at the end of his waiting period.

And that is 1/6 of the time from 6-7pm.
wouldn't it be 4.999?
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Sire
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(Original post by Sire)
I think you'll find that it is 1/210. I could be wrong, but I just did it long hand match up and it seems to work.
That is to say. One chance in two hundred and ten.
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GH
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(Original post by Sire)
I think you'll find that it is 1/210. I could be wrong, but I just did it long hand match up and it seems to work.
Can you show me the working out and the train of thought please, cheers.
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Muse
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(Original post by 2776)
Hmm, but the longest time period possible for a meeting is 9.9999 minutes, as one person just catches the waiting one at the end of his waiting period.

And that is 1/6 of the time from 6-7pm.
I know what you mean, but isn't that an obscure way of looking at it. the simple method just looks at the chance of two people meeting in a particular time period. unless it's some calculation for a degree or something, all other ways of working it out seem over-complicated.
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GH
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(Original post by me!)
wouldn't it be 4.999?
Ahh, yes. I think that is correct. Mental block. Sorry.
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TK
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this thread discrimintaes against people who are crap at maths (like me)
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GH
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(Original post by timeofyourlife)
I know what you mean, but isn't that an obscure way of looking at it. the simple method just looks at the chance of two people meeting in a particular time period. unless it's some calculation for a degree or something, all other ways of working it out seem over-complicated.
Well, apparently it is very complicated, but the teacher did give me a clue. He said to try and draw an array type. ie time on x axis. I first thought of like an area type graph, and then intergrating it. But it is supposedly much more simple than that.
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Sire
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(Original post by 2776)
Can you show me the working out and the train of thought please, cheers.
Right. Draw two lines of three dots running parallel to each other on a piece of paper. Then using the right hand side. Match all three on the right hand side to the bottom left one. That gives you three. Now do the same to the middle left one, and the upper left one. You should have nine matches. Now reverse it, but take into account any match ups that have occurred and discount this. In all you should have 12 possible match ups. I just did this with 12 match ups as there are 12 lots of 5 in 60minutes. I'm pretty damn sure I got it right.
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(Original post by 2776)
Posted in the Academic subform. Too little traffic.

I have a rather sticky problem which I might tantalize the intelligentia of the forum. Rep for helping.

There are 2 people, who want to meet up. Between the hours of 6.oopm to 7.oo pm (ie. 1 hour). Each will stay for 5 minutes before going away. So, they must meet each other within the 5 minutes.

They will both arrive at random times between the hours of 6-7pm.

So, what is the probability of them meeting together?
if they are clever enough, the probability will be 1. person A will arrive between 6.05 and 6.55, knowing that person B will arrive between 6 and 7 (they have that 5 minute overlap). but this is all relative (ie A and B can be interchanged without making a difference) so person B will likewise arrive between 6.05 and 6.55. A thinks ahead and arrives (by the same logic) between 6.10 and 6.50, and because of the interchangable logic, so does B. A then again thinks ahead dot dot dot arriving:
6.15 - 6.45
6.20 - 6.40
6.25 - 6.35

so they will meet at 6.30
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me!
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#18
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(Original post by 2776)
Well, apparently it is very complicated, but the teacher did give me a clue. He said to try and draw an array type. ie time on x axis. I first thought of like an area type graph, and then intergrating it. But it is supposedly much more simple than that.
What about factor trees? might take sometime though...
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(Original post by Tinykates)
this thread discrimintaes against people who are crap at maths (like me)
"iz it coz i iz crap?"
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GH
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#20
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(Original post by Sire)
Right. Draw two lines of three dots running parallel to each other on a piece of paper. Then using the right hand side. Match all three on the right hand side to the bottom left one. That gives you three. Now do the same to the middle left one, and the upper left one. You should have nine matches. Now reverse it, but take into account any match ups that have occurred and discount this. In all you should have 12 possible match ups. I just did this with 12 match ups as there are 12 lots of 5 in 60minutes. I'm pretty damn sure I got it right.
I tried this, but only get 9 matches, as if you reverse it, then there are already matches. Unless you mean something different with the "reverse it"

Can you tell me how you got the parallel dots thing. I don't really see teh mathematical concept behind it.
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