M18-10

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


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

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.

can we also solve it this way

let the shorter side be x
then we assume, the longer side will be twice the shorter side i.e. 2x
so the probability of the longer side being twice as long as the shorter side will be 2x/2x+x
which is 2x/3x, which is 2/3 (after cancelling out the x's)

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.