$\Pr\{\max(X,Y,Z)=X\}$, where $X,Y,Z$ are independently distributed exponential random variables Let $X,Y,Z$ be independent but not identically distributed *exponential random variables.
What is the probability that $X$ is the maximum of $X,Y,Z$, i.e. $\Pr\{\max(X,Y,Z)=X\}$?
My approach was that:
\begin{align*}
\Pr\{\max(X,Y,Z)=X\}
&= \Pr(\{X>Y\}\cap\{X>Z\})\\ 
&= \Pr\{X>Y\}\Pr\{X>Z\}\\
&= \Pr\{\min(X,Y)=Y\}\Pr\{\min(X,Z)=Z\}\\
&=\frac {x}{x+y} \cdot \frac{z}{x+z},
\end{align*}
where $x$,$y$,$z$ are the rates of $X$,$Y$,$Z$ respectively.
Where is the error in this approach? 
Thanks.
 A: Let $X,Y,Z$ be independent exponentials with rates $\lambda, \mu, \nu$.
First, 
recall (prove for yourself) that if $X_1, X_2$ are independent exponentials with rates $\lambda_1, \lambda_2$, then
$$\min\{X_1,X_2\}\sim\text{Exp}(\lambda_1+\lambda_2).\tag{$\ast$}$$
Notice that
$$\{X = \max(X,Y,Z)\}\iff\{\max(Y,Z)<X\}.$$
Then this can only happen in two ways: either $\{Y<Z<X\}$ or $\{Z<Y<X\}$.
Thus, 
\begin{align*}
P(X = \max(X,Y,Z))&= P(\max(Y,Z)<X)\\
&=P(Y<Z<X)+P(Z<Y<X)\\
&=P(Y<\min\{Z,X\})P(Z<X)+P(Z<\min\{Y,X\})P(Y<X)\tag{1}\\
&=\frac{\mu}{\lambda+\mu+\nu}\frac{\nu}{\lambda+\nu}+\frac{\nu}{\lambda+\mu+\nu}\frac{\mu}{\mu+\lambda}\tag{2}\\
&=\frac{\mu\nu}{\lambda+\mu+\nu}\left[\frac{1}{\lambda+\nu}+\frac{1}{\lambda+\mu}\right]
\end{align*}
where (1) is true by the memoryless property, and (2) is true by competing exponentials and the use of $(\ast)$. 
A: The events $X>Z$ and $X>Z$ are not independent, so the probability of their simultaneous occurrence is not the product of the individual probabilities.
You need to integrate the joint density function $\rho(x,y,z)$ over an appropriate volume bounded by $x,y,z$ being positive and $x$ being greater than $y$ and $z$.
I got confused because the title says exponential but the text no longer mentions this. In the case of exponential variables the integral works out nicely.
