Question

A point is chosen at random on a line segment of

length $L$. Interpret this statement, and find the probability

that the ratio of the shorter to the longer segment is

less than $\frac{1}{4}$.

Short answer

The required probability is$\frac{2}{5}$.

Step-by-step solution

Step 1. Given Information.

Here, it is given that the length of a line segment is $L$ on which a point is chosen at random.

Step 2. Find the ratio of shorter to longer segment.

Let the point chosen at random on the given line segment be $X$, which is uniformly distributed over the interval $\left(0,L\right)$.

The probability density function of $X$is given by:

$f\left(x\right)=\left\{\begin{array}{l}\frac{1}{L},0\le x\le L\\ \end{array}\right.$

Ratio of shorter segment to longer segment will be the minimum of the following:

$\frac{X}{L-X},\frac{L-X}{X}$

Step 3. Find the required probability.

The required probability will be:

$$\begin{gathered} P\left\{Min\left(\frac{X}{L-X},\frac{L-X}{X}\right)<\frac{1}{4}\right\} \\ =1-P\left\{Min\left(\frac{X}{L-X},\frac{L-X}{X}\right)>\frac{1}{4}\right\} \end{gathered}$$

Since, minimum of the two is greater than $\frac{1}{4}$.

$$\begin{gathered} 1-P\left(\frac{X}{L-X}>\frac{1}{4},\frac{L-X}{X}>\frac{1}{4}\right) \\ =1-P\left\{4X>L-X,4(L-X)>X\right\} \\ =1-P\left\{5X>L,4L>5X\right\} \\ =1-P\left\{X>\frac{L}{5},X<\frac{4L}{5}\right\} \\ =1-P\left\{\frac{L}{5}<X<\frac{4L}{5}\right\} \\ =1-{\int }_{\frac{L}{5}}^{\frac{4L}{5}}\frac{1}{L}dx \\ =1-\frac{1}{L}\left(\frac{4L}{5}-\frac{L}{5}\right) \\ =1-\frac{3}{5} \\ =\frac{2}{5} \end{gathered}$$

Therefore, the required probability is$\frac{2}{5}.$