# Answer to Question #4212 in Abstract Algebra for Bridgette

Question #4212

A piece of string is cut into two pieces at a randomly selected point. What is the probablility that the longer piece is at least x times as long as the shorter piece

Expert's answer

Of course, we should assume that x>=1.Not loosing generality assume that the string is the segment [0,1].

Then the probablility which is asked by the task will be equal to the

length of the set of all

t from [0,1]

that the longer piece is at least x times as long as the shorter piece.

Let t in [0,1] be the point of cutting such that the longer piece is at least x

times as long as the shorter

piece

Consider two cases:1) Suppose 0 <= t <= 1/2.

Then the longer piece has length 1-t, and so we have the inequality:

1-t > tx

whence

t < 1/(x+1)

2) If 1/2 <= t <= 1,

Then the longer piece has length t, and so we have the inequality:

t > (1-t)x

whence

t< x/(1+x)

Thus t belongs to either of the following intervals

[ 0 , 1/(x+1) ] or [ x/(x+1) , 1 ].

The total length of these intervals is

1/(x+1) + 1 - x/(x+1) = 1/(x+1) + 1/(x+1) = 2/(x+1)

Answer. 2/(x+1)

Then the probablility which is asked by the task will be equal to the

length of the set of all

t from [0,1]

that the longer piece is at least x times as long as the shorter piece.

Let t in [0,1] be the point of cutting such that the longer piece is at least x

times as long as the shorter

piece

Consider two cases:1) Suppose 0 <= t <= 1/2.

Then the longer piece has length 1-t, and so we have the inequality:

1-t > tx

whence

t < 1/(x+1)

2) If 1/2 <= t <= 1,

Then the longer piece has length t, and so we have the inequality:

t > (1-t)x

whence

t< x/(1+x)

Thus t belongs to either of the following intervals

[ 0 , 1/(x+1) ] or [ x/(x+1) , 1 ].

The total length of these intervals is

1/(x+1) + 1 - x/(x+1) = 1/(x+1) + 1/(x+1) = 2/(x+1)

Answer. 2/(x+1)

## Comments

## Leave a comment