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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. \(\frac{3}{4}\)
A. \(\frac{1}{4}\)
B. \(\frac{1}{3}\)
C. \(\frac{1}{2}\)
D. \(\frac{2}{3}\)
E. \(\frac{3}{4}\)
16 Sep 2014, 01:03
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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
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
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03 Aug 2016, 11:24
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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.
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.
