Q.6.44 - A First Course in Probability

Question

If $X_1,X_2,X_3$ are independent random variables that are uniformly distributed over $\left(0,1\right)$, compute the probability that the largest of the three is greater than the sum of the other two. 

Step-by-Step Solution

Answer

There's a good chance that the greatest of the three is bigger than the total of the other two is $\frac{1}{2}$.

Step 1 : Variable :

The numerical value or quantity expressed by an alphabetic letter.

Step 2 : Explanation :

The joint density function is the product of marginal density $\left(0,1\right)$ since $X_1,X_2,X_3$ are independent random variables.

$f_{x_1{x}_2{x}_3}\left(x\right) = \left\{\begin{array}{l}1; 0<x_1,x_2,x_3<1\\ 0; Otherwise \end{array}\right.$

The greatest of the three possibilities is bigger than the total of the two.

$\left(X_1>X_2+X_3\right), (X_2>X_1+X_3)$and $\left(X_3>X_1+X_2\right)$

So because probabilities of the above three terms are all identical due to the symmetric condition, computing the first term is sufficient.

First, find the value of $P(X_1>X_2+X_3)$.

$$\begin{gathered} P(X_1>X_2+X_3)=\underset{\left\{\right(x,y):x_1>x_2+x_3\}}{\int }{f}_{X_1}\left(x_1\right)f_{X_1}\left(x_1\right)f_{X_1}\left(x_1\right)d{x}_{2}d{x}_{3}d{x}_1 \\ ={\int }_0^1{\int }_0^{x_1}{\int }_0^{(x_1-x_3)}1\times 1\times 1 d{x}_{2}d{x}_{3}d{x}_1 \\ ={\int }_0^1{\int }_0^{x_1}{\left(x_2\right)}_0^{(x_1-x_3)}d{x}_{3}d{x}_1 \\ ={\int }_0^1{\int }_0^{x_1}(x_1-x_3)d{x}_{3}d{x}_1 \\ ={\int }_0^1{\left(\frac{x_3^2}{2}\right)}_0^{x_1}d{x}_1 \\ ={\int }_0^1{\left(\frac{x_1^2}{2}\right)}_0^{1}d{x}_1 \\ ={\left(\frac{x_1^3}{6}\right)}_0^1=\frac{1}{6} \end{gathered}$$

As a result, the likelihood of the greatest of the three being more than the total of the other two is : 

$$\begin{gathered} P\left(X_1>X_2+X_3\right)+P (X_2>X_1+X_3)+P(X_3>X_2+X_1) \\ =\frac{1}{6}+\frac{1}{6}+\frac{1}{6} \\ =\frac{3}{6} \\ =\frac{1}{2} \end{gathered}$$

As a result, there's a good chance that the greatest of the three is bigger than the total of the other two is $\frac{1}{2}$.