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#Post#: 222--------------------------------------------------
Will they meet?
By: dinesh Date: November 27, 2012, 8:35 am
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2 people decide to meet between 1 pm and 2pm at a particular
place on a particular day. When one of them arrives there, he
waits for 15 minutes for the other and then leaves. What is the
probability that both will meet?
#Post#: 234--------------------------------------------------
Re: Will they meet?
By: dada786 Date: November 27, 2012, 9:31 am
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If we assume they arrive and leave only at the starting of
minute, I think answer is
They won't meet if,
First person comes at 1pm and second person comes at
1.15/1.16/1.17/...../1.45 (Not after that b'cos both know they
wait for 15 minites and both meet b/w 1 to 2 pm) (30
possibilities) or
First person comes at 1.01pm and second person comes at
1.16/1.17/...../1.45 (29 possibilities)
First person comes at 1.02pm and second person comes at
1.17/...../1.45 (28 possibilities)........................ upto
First person comes at 1.30pm and second person comes at 1.45 (1
possibility)
total 465. Similarly for Second-First pair 465. So total not
possibilities of not meeting are 930.
Total possibilities are 45*45.
The possibility of not meeting is 930/(45*45).
Possibility of meeting is 1-(930/(45*45)).
#Post#: 235--------------------------------------------------
Re: Will they meet?
By: dinesh Date: November 27, 2012, 9:35 am
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Will that be between 0 and 1?
Give a final answer (after calculation)
#Post#: 239--------------------------------------------------
Re: Will they meet?
By: ravusairam Date: November 27, 2012, 9:52 am
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i think it is 1/4 - which is 15/60.
#Post#: 240--------------------------------------------------
Re: Will they meet?
By: srini Date: November 27, 2012, 10:23 am
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is it 1/3.
#Post#: 249--------------------------------------------------
Re: Will they meet?
By: dada786 Date: November 27, 2012, 11:35 pm
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It is 0.54 (approx)
#Post#: 257--------------------------------------------------
Re: Will they meet?
By: dinesh Date: November 28, 2012, 8:34 am
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If they decide to wait only for 10 minutes, the answer is 11/36.
Now can you get the answer for the given problem?
#Post#: 309--------------------------------------------------
Re: Will they meet?
By: dinesh Date: December 3, 2012, 4:03 am
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This question was asked in recent CSIR/ GATE paper...
For the 10 mins problem
Solution method 1: geometric probability
I feel this is the most elegant way to solve the problem.
If we let x denote the time one person arrives at the bar, and y
the other, we can use the notation (x, y) to denote the time in
minutes, after midnight, that each person arrives at the bar.
The trick now is that we can model the situation geometrically
in the Cartesian plane. The x-axis can be labeled from the time
0 minutes until 60 minutes, as can the y-axis.
Now any coordinate in the 60×60 square represents a time that
the pair arrives at the bar. The coordinate (0,9) means one
person shows up at midnight, the second at 12:09, and clearly
they will meet because the times are within 10 minutes of each
other. The coordinate (1, 51), on the other hand, corresponds to
one showing up at 12:01 and the other at 12:51, a time the two
will not meet.
What is the set of coordinates for which the two will meet?
The notation affords a very simple way to describe the event. We
want the two coordinates to be within 10 minutes of each other.
We either want the x coordinate to be 10 units smaller than y,
or 10 units bigger than y. The succinct way of writing that is
we want all coordinates such that | x – y | ≤ 10.
We will draw the lines y = x – 10 and y = x + 10 and shade the
area in between these two lines to denote the event.
The resulting figure is as follows: ( attached )
The probability of meeting is precisely the ratio of the shaded
area to the total square.
This is fairly easy to calculate. The big square is 60×60 =
3,600 in area. Rather than finding the shaded area, let us
calculate the unshaded area and subtract. The unshaded area
consists of two isosceles right triangles with sides of 50. This
means each triangle has an area of (0.5) x (50×50) = 1,250 and
the total unshaded area is double that, 2,500.
The shaded area is found by subtracting the unshaded area from
the total : 1,100 = 3,600 – 2,500.
Thus we conclude the chance the friends will meet is 1,100 /
3,600 = 11/36,
For the 15 mins problem:
Similarly for our problem, the answer = ( 3600 - 45*45 ) / 3600
=
1575/3600
=
7/16
Other methods of proof can be found at
HTML http://mindyourdecisions.com/blog/2011/07/11/math-puzzle-chances-of-meeting-up-with-a-friend-and-game-theory-puzzle-extension/
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