Question

Three points X₁, X₂, X₃ are selected at random on a line L. What is the probability that X₂ lies between X₁ and X₃?

Short answer

The probability that X₂ lies between X₁ and X₃ is$\frac{1}{3}$

Step-by-step solution

Content Introduction 

Three points X₁ , X₂ , X₃ are selected at a random on a line L.

Number of ways of arranging and things taking all at a time is n!. Total number of points is 3.

So, the number of ways of arranging 3 points on a line is 3! which is equal to 6.

Content Explanation

Possible arrangements are:

$$\begin{gathered} X_1{X}_2{X}_3 \\ {X}_1{X}_3{X}_2 \\ {X}_2{X}_1{X}_3 \\ {X}_2{X}_3{X}_1 \\ {X}_3{X}_1{X}_2 \\ {X}_3{X}_1{X}_2 \end{gathered}$$

Out of these X₂ lies between X₁ and X₃in two arrangements.

Therefore, Probability that X₂lies between X₁and X₃is$$\begin{gathered} =\frac{2}{6} \\ =\frac{1}{3} \end{gathered}$$