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# Probability

Author Message
Manager
Joined: 03 Sep 2009
Posts: 77
Followers: 1

Kudos [?]: 106 [0], given: 3

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22 Jan 2010, 10:30
Hi Experts,

Could anybody explain how to solve the following question:
Q:- If a point is arbitrarily selected on a line segment, thus breaking it into two smaller segments, what is the probability that the larger segment is at least twice as long as the smaller one?
a) 1/4
b) 1/3
c) 1/2
d) 2/3
e) 3/4

Many thanks in anticipation.

-AMit
Intern
Joined: 21 Jan 2010
Posts: 49
Followers: 0

Kudos [?]: 34 [1] , given: 0

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22 Jan 2010, 10:35
1
KUDOS
Sounds like a calculus problem to me!

Anyways, I think you can break it down like this:

For one segment to be twice as long as the other, you need to break the line at the 1/3 or 2/3 marks. Thus, I believe you can 'simplify' the problem by breaking the line as such:

1. At the 1/3 and less -- line segment one will be at least twice the second.
2. Between 1/3 and 2/3 -- line segment one will be less than twice the second, or vice versa.
3. At the 2/3 and more -- line segment two will be at least twice the first.

So, three cases, two of which satisfy your criteria.

Does 2 / 3 sound right?
_________________

________________________________________________________________________
Andrew
http://www.RenoRaters.com

Manager
Joined: 03 Sep 2009
Posts: 77
Followers: 1

Kudos [?]: 106 [0], given: 3

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22 Jan 2010, 10:53
Yaa 2/3 is correct. Many thanks for the explanation.

+1 for that.
Any ways I am also from Canada...trying for UBC or SFU in BC...
-Amit
Re: Probability   [#permalink] 22 Jan 2010, 10:53
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# Probability

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