Sticky Maths Problem: REP FOR SOLUTION

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elpaw
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#41
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#41
(Original post by meepmeep)
hmmm, you could sketch some kind of graph of it I suppose. The red area is the times when they meet, the blue area the times they don't. I simply worked out the area of the blue triangles then took it from the total area to get 23/144:

PS. The computer is working a little slow tonight - brother decided to put on the virus scan, so sorry about piccie!
:rolleyes:

*slaps forehead*

*promises never to do maths again*
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meepmeep
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#42
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#42
(Original post by elpaw)
:rolleyes:

*slaps forehead*

*promises never to do maths again*
Simple things from a simple mind.
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GH
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#43
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#43
(Original post by meepmeep)
hmmm, you could sketch some kind of graph of it I suppose. The red area is the times when they meet, the blue area the times they don't. I simply worked out the area of the blue triangles then took it from the total area to get 23/144:

PS. The computer is working a little slow tonight - brother decided to put on the virus scan, so sorry about piccie!
I think we are onto a rep worthy winner here. It looks plausible. But can you flesh out the meaning of the graph please. I don't get why, it starts at 5 minutes etc.

Thanks
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meepmeep
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#44
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#44
(Original post by 2776)
I think we are onto a rep worthy winner here. It looks plausible. But can you flesh out the meaning of the graph please. I don't get why, it starts at 5 minutes etc.

Thanks
Sure. Well, the graph represents the hour between 6.00pm and 7.00pm (I think those were the times). The x-axis represents person A and the y-axis person B. Therefore, if person A arrives at 6.23 and person B arrives at 6.43, the point would be (23,43). The black line in the middle represents the set of points which represent the times where the two people arrive at exactly the same time. However, as they stay for 5 minutes, the other person can arrive 5 minutes either way of this time, which is why the red region is 10 minutes wide/high (as if person A arrives at 6.30, person B can arrive between 6.25 and 6.35).

If A arrives at 0, the B can arrive any time between 0 and 5. If B arrives at 0, then A can arrive any time between 0 and 5, which is why (0,5) and (5,0) mark the start of the "red region". The same applies to the other side, giving (60,55) and (55,60). The total area is 60^2 and the two triangles are 55^2/2. Hence:
probability=(60^2-55^2)/60^2=23/144
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GH
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#45
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#45
(Original post by meepmeep)
Sure. Well, the graph represents the hour between 6.00pm and 7.00pm (I think those were the times). The x-axis represents person A and the y-axis person B. Therefore, if person A arrives at 6.23 and person B arrives at 6.43, the point would be (23,43). The black line in the middle represents the set of points which represent the times where the two people arrive at exactly the same time. However, as they stay for 5 minutes, the other person can arrive 5 minutes either way of this time, which is why the red region is 10 minutes wide/high (as if person A arrives at 6.30, person B can arrive between 6.25 and 6.35).

If A arrives at 0, the B can arrive any time between 0 and 5. If B arrives at 0, then A can arrive any time between 0 and 5, which is why (0,5) and (5,0) mark the start of the "red region". The same applies to the other side, giving (60,55) and (55,60). The total area is 60^2 and the two triangles are 55^2/2. Hence:
probability=(60^2-55^2)/60^2=23/144
It all clear now. Trivial really. Well my word is my bond. Hold out for the onslaught of rep. First stop: 55 rep.
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GH
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#46
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#46
And have some more rep for being so helpful. :11 rep.
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meepmeep
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#47
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#47
(Original post by 2776)
It all clear now. Trivial really. Well my word is my bond. Hold out for the onslaught of rep. First stop: 55 rep.
Wow! Thanks a lot. It's appreciated. Don't have to give it if you don't want to next time though.
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theone
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#48
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#48
I think this is a slightly more mathematical explanation of this problem:

Say x arrives before y, we may say this without loss of generality.

The time at which X arrives has a uniform distribution over [0,60] where 0 is 6PM.

Say X arrives at a time C, with C <= 55 we must have
Y arriving in [C, C+5]. The probability of this happening is 5/(60-C).

If X arrives after 55 then Y must arrive before 60 (since Y must come at some point) so our probability is 1 here.

Now integrating 5/60-C between 0 and 55, we get sum of probabilities = integral between 0 and 55 of 5/60-C = (-5ln(60-C)) between 0 and 55. which is -5(ln60-ln5) = 5ln(1/12).

Now our probability in [55,60] = 1. So the overall probability is (55ln(1/12) + 1)/12.

I think this is right, but it's late and I could have made a mistake.
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GH
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#49
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#49
(Original post by theone)
I think this is a slightly more mathematical explanation of this problem:

Say x arrives before y, we may say this without loss of generality.

The time at which X arrives has a uniform distribution over [0,60] where 0 is 6PM.

Say X arrives at a time C, with C <= 55 we must have
Y arriving in [C, C+5]. The probability of this happening is 5/(60-C).

If X arrives after 55 then Y must arrive before 60 (since Y must come at some point) so our probability is 1 here.

Now integrating 5/60-C between 0 and 55, we get sum of probabilities = integral between 0 and 55 of 5/60-C = (-5ln(60-C)) between 0 and 55. which is -5(ln60-ln5) = 5ln(1/12).

Now our probability in [55,60] = 1. So the overall probability is (55ln(1/12) + 1)/12.

I think this is right, but it's late and I could have made a mistake.
I get the answer -11.3058221.
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theone
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#50
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#50
(Original post by 2776)
I get the answer -11.3058221.
Then it's wrong, but i'm certain this is the right approach.
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Juwel
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#51
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#51
(Original post by 2776)
I get the answer -11.3058221.
Either you punched it in the calculator wrong or theone lost it a bit. Let's just applaud meepmeep's quality answer. If only I was like that.
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GH
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#52
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#52
(Original post by ZJuwelH)
Either you punched it in the calculator wrong or theone lost it a bit. Let's just applaud meepmeep's quality answer. If only I was like that.
I used google calculator. Just copy/paste it in.
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me!
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#53
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#53
Hey! because the graph thing has the different regions etc couldn't you do some inequalities or something like that...

x < x +/- 5

that's probably wrong though... it is wrong...

how about x < y +/- 5
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theone
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#54
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#54
I have to go, but i offer rep to anyone who can explain why the basic principle that i've used doesn't work. (and no rep for 'cos you got the wrong answer' )
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meepmeep
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#55
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#55
(Original post by theone)
I think this is a slightly more mathematical explanation of this problem:

Say x arrives before y, we may say this without loss of generality.

The time at which X arrives has a uniform distribution over [0,60] where 0 is 6PM.

Say X arrives at a time C, with C <= 55 we must have
Y arriving in [C, C+5]. The probability of this happening is 5/(60-C).

If X arrives after 55 then Y must arrive before 60 (since Y must come at some point) so our probability is 1 here.

Now integrating 5/60-C between 0 and 55, we get sum of probabilities = integral between 0 and 55 of 5/60-C = (-5ln(60-C)) between 0 and 55. which is -5(ln60-ln5) = 5ln(1/12).

Now our probability in [55,60] = 1. So the overall probability is (55ln(1/12) + 1)/12.

I think this is right, but it's late and I could have made a mistake.
Yeah, something slightly awry there. Comes to -11.3 that. I think you might have to integrate between 0 and 55/60 though. I didn't want to go down this route because it looked a little complex. I think you need to take into account Y arriving before X as well. Anyway, I#m shattered. Goodnight.
PS, no need for rep, especially as this probably isn't the reason why, but if you replace 55 with 55/60, you get a hell of a lot closer to 23/144. I'm not quite sure how you got 5/(60-C)....
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