1

We need to divide a line segments into two parts by selecting a point at random. Then we have to find the probability that the length of the larger segment is at least 3 times the shorter.

My take to the problem :

Considering a line segment of unit length 1.

Let $X$ denote the length of the segment on the left side. So the pdf can be written as :

$f(x) = \dfrac{1}{2} , 0<x<1$

Now according to the question we need to find $P(X>3(1-X))$ which is $P(X> \dfrac{3}{4})$ , which gives the value $\dfrac{1}{8}$. Is this correct ?

User9523
  • 2,094

1 Answers1

0

Let $Y$ be the number chosen. We will be happy if $Y\le 1/4$ or $Y\ge 3/4$. Computing the probability of this event is easy. It is not $1/8$.

Remark: Your $X$ is my $Y$, and it has pdf $1$ on $(0,1)$. It cannot be $\frac{1}{2}$ on $(0,1)$ and $0$ elsewhere, for that is not a pdf.

With that modification, your calculation can be used to find the probability that the left side is at least $3$ times the right side. It gives answer $\frac{1}{4}$. But our event will also occur if the right side is at least $3$ times the left side. Another $\frac{1}{4}$.

Or else we could let $W$ be the minimum of the two sides. This random variable has density $2$ on $(0,1/2)$ and $0$ elsewhere. Our event occurs if $W\le 1/4$.

André Nicolas
  • 507,029