Length of a line segment

Question

A line segment of length 1 is cut into two pieces at a completely random point. What is the probability that the longer piece is at least three times the length of the shorter piece?

Attempt

Let X be the length of the shorter piece and Y be the length of the longer piece.

X: f(x) = 2, 0 < x < 0.5 or f(x) = 0 otherwise

Y: g(y) = 2, 0.5 < y < 1 or g(y) = 0 otherwise

I think the question is asking for P[Y>= 3 X] but I am getting a negative probability.

Answer 1

If the line was cut at 0.25, then the shorter piece would be 0.25 and the longer piece 0.75, 3 times longer. If the line is cut at a point less than 0.25, the longer piece will be more than 3 times the length of the shorter piece. So I think the answer is $\frac{1}{4}$.

Edit: As rightly pointed out by Badam Baplan, this answer doubles to $\frac{1}{2}$ with symmetry since we can cut the line at a point between 0 and 0.25 or between 0.75 and 1.

Answer 2

$Y=1-X$ so $P(Y \geq 3X)=P(X \leq \frac 1 4)=2\int_0^{1/4}dx=\frac 1 2$.

Answer 3

You just need to have only one probability function .

considering

  y = 3x
  y+x = 1
  3x+x = 1
   x  = 1/4

since the probability of one piece having length less than 1/4 answer is 1/4

Answer 4

The longer piece is at least three times the shorter one when the length of the left piece is in

$$\left[0,\frac14\right]\cup\left[\frac34,1\right].$$

This set has measure $$\frac12.$$

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This is justified by $$\frac l{1-l}\le\frac13\lor3\le\frac l{1-l}$$

or

$$3l\le1-l\lor3-3l\le l$$

or

$$4l\le1\lor3\le 4l.$$

Answer 5

Line segment $AB$ , length $l$, is divided into $4$ equal parts.

From.left to right: $a,b,c,d$, and $a=b=c=d$.

Probability of cutting randomly within $a,b,c$ or $d$ : $(1/4)$.

Out of these equally probable choices only $2$ qualify, $a$, and $d$.

$P(a \lor d)= 1/4+1/4=1/2$