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655-705 Level|   Probability|      
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Bunuel
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If a point is arbitrarily selected on a line segment, thus breaking it 26 Sep 2020, 10:15
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55% (01:14) correct 45% (01:30) wrong based on 559 sessions

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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. \(\frac{3}{4}\)


M18-10

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Bunuel
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If a point is arbitrarily selected on a line segment, thus breaking it 27 Sep 2020, 10:15
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Bunuel
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. \(\frac{3}{4}\)


M18-10
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Check the diagram below:



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

This problem can also be approached algebraically. If we consider the length of the original segment as 1 and the length of the first sub-segment as \(x\), then we are looking for values of \(x\) that satisfy \(\frac{x}{1 - x} \ge 2\) or \(\frac{1 - x}{x} \ge 2\). Solving the first inequality, we get \(x \ge \frac{2}{3}\), and for the second, we find \(x \le \frac{1}{3}\). Therefore, the probability that either \(x \ge \frac{2}{3}\) or \(x \le \frac{1}{3}\) holds is \(\frac{2}{3}\).


Answer: D
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nick1816
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If a point is arbitrarily selected on a line segment, thus breaking it 26 Sep 2020, 10:42
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Suppose if the point E lies in region AB, then 2*AE<=ED and if point E lies in region CD, then AE>=2*ED.

Hence, the length of favorable region=x+x=2x

Total length of line segment= 3x

Probability = 2x/3x = 2/3


Bunuel
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. \(\frac{3}{4}\)


M18-10
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Sonia2023
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If a point is arbitrarily selected on a line segment, thus breaking it 28 Nov 2023, 10:07
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Bunuel
Bunuel
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. \(\frac{3}{4}\)


M18-10

Check the diagram below:



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

This problem can also be approached algebraically. If we consider the length of the original segment as 1 and the length of the first sub-segment as \(x\), then we are looking for values of \(x\) that satisfy \(\frac{x}{1 - x} \ge 2\) or \(\frac{1 - x}{x} \ge 2\). Solving the first inequality, we get \(x \ge \frac{2}{3}\), and for the second, we find \(x \le \frac{1}{3}\). Therefore, the probability that either \(x \ge \frac{2}{3}\) or \(x \le \frac{1}{3}\) holds is \(\frac{2}{3}\).



Answer: D
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Can you please confirm whether my understanding is correct?
When you say that x<=1/3, we cover 1/3rd area of the number line and then when x>=2/3, we cover another 1/3rd area so overall 2/3 probability?
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If a point is arbitrarily selected on a line segment, thus breaking it 28 Nov 2023, 10:31
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Sonia2023
Bunuel
Bunuel
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. \(\frac{3}{4}\)


M18-10

Check the diagram below:



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

This problem can also be approached algebraically. If we consider the length of the original segment as 1 and the length of the first sub-segment as \(x\), then we are looking for values of \(x\) that satisfy \(\frac{x}{1 - x} \ge 2\) or \(\frac{1 - x}{x} \ge 2\). Solving the first inequality, we get \(x \ge \frac{2}{3}\), and for the second, we find \(x \le \frac{1}{3}\). Therefore, the probability that either \(x \ge \frac{2}{3}\) or \(x \le \frac{1}{3}\) holds is \(\frac{2}{3}\).



Answer: D

Can you please confirm whether my understanding is correct?
When you say that x<=1/3, we cover 1/3rd area of the number line and then when x>=2/3, we cover another 1/3rd area so overall 2/3 probability?
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Yes, that's correct.
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If a point is arbitrarily selected on a line segment, thus breaking it 02 Jul 2025, 07:22
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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?

solution:-

consider this a line segment:- ___________________

Now break this in 2 parts:- ____(a)____|_____(b)______

a = smaller segment
b = larger segment

given b = 2a
asked:- what fraction of the larger segment is at least twice as long as the smaller one ?

b/b+a = 2a/2a+a
= 2a/3a
= 2/3

Option D
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