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M18-10

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M18-10  [#permalink]

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New post 16 Sep 2014, 01:03
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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. \(\frac{1}{4}\)
B. \(\frac{1}{3}\)
C. \(\frac{1}{2}\)
D. \(\frac{2}{3}\)
E. \(\frac{3}{4}\)

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Re M18-10  [#permalink]

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New post 16 Sep 2014, 01:03
Official Solution:

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. \(\frac{1}{4}\)
B. \(\frac{1}{3}\)
C. \(\frac{1}{2}\)
D. \(\frac{2}{3}\)
E. \(\frac{3}{4}\)


The larger segment will be at least twice as long as the smaller one if the breaking point is either in the first or in the last third of the segment. The probability of this is \(\frac{2}{3}\).

The problem can also be solved algebraically. If the length of the original segment is 1 and the length of the first sub-segment is \(x\) then we are looking for \(x\) such that \(\frac{x}{1 - x} \ge 2\) or \(\frac{1 - x}{x} \ge 2\). The first inequality reduces to \(x \ge \frac{2}{3}\), the second to \(x \le \frac{1}{3}\). The probability that either \(x \ge \frac{2}{3}\) or \(x \le \frac{1}{3}\) is \(\frac{2}{3}\).


Answer: D
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Re M18-10  [#permalink]

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New post 03 Aug 2016, 08:50
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Did not understand the solution is explained. Probability is total number of desired outcomes / total outcomes in terms of this usual way of solving can you explain in detail as your explanation at present is not clear to me. Thanks.
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Re: M18-10  [#permalink]

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New post 03 Aug 2016, 11:24
Divide the line into 3 equal segments, lets A, B, C.

After a point is selected, for the longer segment to be 2 times as long as the smaller segment, the point has to be either in segment A or in C.

As length of A = length of B = length of C = 1/3, so, probability that the point is selected in segment A or C is 1/3 + 1/3 = 2/3.

My answer is D.
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New post Updated on: 10 Aug 2016, 03:01
I solved it in the following way:

Say the line is 3x length. What's the probability that the point will break the line into desired segments? It's 2x/3x, since there are two points that satisfy the condition one is at 1x length and the another one at 2x length. Frankly speaking, I didn't notice "at least", but it turns out that I got a correct answer.

In any case, if you divide a line into three segments, there will be two segment on symmetrically opposite ends that satisfy condition.

Originally posted by manlog on 09 Aug 2016, 23:09.
Last edited by manlog on 10 Aug 2016, 03:01, edited 1 time in total.
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Re: M18-10  [#permalink]

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New post 10 Aug 2016, 02:59
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Bunuel wrote:
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. \(\frac{1}{4}\)
B. \(\frac{1}{3}\)
C. \(\frac{1}{2}\)
D. \(\frac{2}{3}\)
E. \(\frac{3}{4}\)


Did it in rather quick and crude method, on a scale of 100 (10 unit intervals):
10:90 - Yes (10X2 = 20)
20:80 - Yes
30:70 - Yes
40:60 - No
50:50 - No
60:40 - No
70:30 - Yes
80:20 - Yes
90:10 - Yes

So probability = 6/9 = 2/3
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New post Updated on: 28 Nov 2016, 23:15
The question has been re-edited stated below

Originally posted by gmatdemolisher1234 on 28 Nov 2016, 22:39.
Last edited by gmatdemolisher1234 on 28 Nov 2016, 23:15, edited 1 time in total.
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New post 28 Nov 2016, 23:13
Bunuel wrote:
Official Solution:

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. \(\frac{1}{4}\)
B. \(\frac{1}{3}\)
C. \(\frac{1}{2}\)
D. \(\frac{2}{3}\)
E. \(\frac{3}{4}\)


The larger segment will be at least twice as long as the smaller one if the breaking point is either in the first or in the last third of the segment. The probability of this is \(\frac{2}{3}\).

The problem can also be solved algebraically. If the length of the original segment is 1 and the length of the first sub-segment is \(x\) then we are looking for \(x\) such that \(\frac{x}{1 - x} \ge 2\) or \(\frac{1 - x}{x} \ge 2\). The first inequality reduces to \(x \ge \frac{2}{3}\), the second to \(x \le \frac{1}{3}\). The probability that either \(x \ge \frac{2}{3}\) or \(x \le \frac{1}{3}\) is \(\frac{2}{3}\).


Answer: D




If the question said - If a point is arbitrarily selected on a line segment, thus breaking it into two smaller segments, what is the probability that the 2 segments are equal-?


Then If the length of the original segment is 1 and the length of the both sub-segment is x/(1-x) = 1, we'll get x= 1/2,

So there is only 1 point 1/2 ..

Now what do we do ?? how should we solve this???
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Re: M18-10  [#permalink]

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New post 28 Mar 2017, 00:40
hi brunel..

after going through the solution i got stuck at the last statement : The probability that either x≥2/3x or x≤1/3x is 2/3

how we can conclude probability is 2/3 and not 1/3.Both the options are given in the answer choices
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Re: M18-10  [#permalink]

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New post 28 Mar 2017, 01:01
sidagar wrote:
hi brunel..

after going through the solution i got stuck at the last statement : The probability that either x≥2/3x or x≤1/3x is 2/3

how we can conclude probability is 2/3 and not 1/3.Both the options are given in the answer choices



Look at the problem in this way.
Consider a line segment AB (any length)

Put two points on the line segment (say C and D), such that 2*AC=CB and 2*DB=AD
(i.e. the two points divide the line segment into 1:2 ratio)


Now, consider the question that you have to arbitrarily select a point on the segment.
Note carefully here that if you chose a point between A and C, or D and B, then it will satisfy the condition, which is that one line segment is greater than the twice of other.

Next, notice that C and D divide the line segment in a ration of 1/3 (i.e., AC=CD=DB=1/3*AB)

We also know that any point between C and D will not satisfy the condition

Therefore Probability of condition not true will be when the point is chosen between C and D, which is 1/3

Hence, probability of condition being true is 2/3 and answer is D

Hope this helps.
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M18-10  [#permalink]

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New post 01 Sep 2017, 10:38
sidagar wrote:
hi brunel..

after going through the solution i got stuck at the last statement : The probability that either x≥2/3x or x≤1/3x is 2/3

how we can conclude probability is 2/3 and not 1/3.Both the options are given in the answer choices



Bunuel

Same question,can u please explain the last step of having probability as 2/3
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New post 01 Sep 2017, 10:44
himanshukamra2711 wrote:
sidagar wrote:
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?


hi brunel..

after going through the solution i got stuck at the last statement : The probability that either x≥2/3x or x≤1/3x is 2/3

how we can conclude probability is 2/3 and not 1/3.Both the options are given in the answer choices



Bunuel

Same question,can u please explain the last step of having probability as 2/3


---

The larger segment will be at least twice as long as the smaller one if we cut anywhere at the two red segments. The probability of that will be 2/3.
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New post 30 Aug 2018, 23:44
The problem can also be solved algebraically. If the length of the original segment is 1 and the length of the first sub-segment is \(x\) then we are looking for \(x\) such that \(\frac{x}{1 - x} \ge 2\) or \(\frac{1 - x}{x} \ge 2\). The first inequality reduces to \(x \ge \frac{2}{3}\), the second to \(x \le \frac{1}{3}\). The probability that either \(x \ge \frac{2}{3}\) or \(x \le \frac{1}{3}\) is \(\frac{2}{3}\).


Bunuel

I do understand the role of this equation
\(\frac{x}{1 - x} \ge 2\)
but I dont understand why we have used the equation given after this where we take up the reciprocal of LHS.
Please explain.
And on what basis do we reject 1/3.
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Re: M18-10   [#permalink] 30 Aug 2018, 23:44
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