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A point is arbitrarily selected on a line segment thus

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Manager
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A point is arbitrarily selected on a line segment thus [#permalink] New post 27 Aug 2007, 20:27
A point is arbitrarily selected on a line segment thus breaking it into two smaller segments.What is the probability that the bigger segment is at least twice as long as the smaller.

1/4

1/3

1/2

2/3

3/4
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Re: Segment Probab [#permalink] New post 27 Aug 2007, 20:33
dreamgmat1 wrote:
A point is arbitrarily selected on a line segment thus breaking it into two smaller segments.What is the probability that the bigger segment is at least twice as long as the smaller.

1/4

1/3

1/2

2/3

3/4


2/3

i would say this is one of the best questions.
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 [#permalink] New post 27 Aug 2007, 23:38
Fistail, could you please share with us your way of doing this Q?
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Re: Segment Probab [#permalink] New post 28 Aug 2007, 03:18
dreamgmat1 wrote:
A point is arbitrarily selected on a line segment thus breaking it into two smaller segments.What is the probability that the bigger segment is at least twice as long as the smaller.

1/4

1/3

1/2

2/3

3/4


Think of the number line with the segment between 0 and 1, which point x chosen as cutting point.
The left segment is more than twice as long as the right if x > 2(1-x) i.e. x > 2/3
Thus the probability that the left segment is more than twice as long as the right is 1/3, as is the probability that the right is more than twice as logn as the left. So the required probability is 1/3 + 1/3 =2/3
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Re: Segment Probab [#permalink] New post 28 Aug 2007, 06:41
Whatever wrote:
Fistail, could you please share with us your way of doing this Q?


nothing to add on what kevin said.

but still my point is: when the point is placed at or above 2/3 and at or below 1/3, then the resulting parts have this relationship.
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 [#permalink] New post 28 Aug 2007, 17:27
Thanks a lot, but could you please explain the step when we have x > 2/3 and from this we are get probability of 1/3 that the left segment is twice as log a the right? 1/3 means that we have 3 segments in a line then?..

Edited: ok, I got it, thanks.

Last edited by Whatever on 28 Aug 2007, 21:42, edited 1 time in total.
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 [#permalink] New post 28 Aug 2007, 19:21
Answer is 2/3 but could someone elaborate i have the same doubts as the last post
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 [#permalink] New post 28 Aug 2007, 21:17
let the line segments be l1 (shorter end) and l2 (longer end).

Then l2 >= 2l1, So l2/l1 >= 2/1

So the ratio of l2:l1 must be at least 2:1. if the length of the segment is l, then l2 = 2l/3 and l1 = l/3 (at the very least). The probability is therefore 2l/3 * 1/l = 2/3.
  [#permalink] 28 Aug 2007, 21:17
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