M. Ross problem 12 chapter 5 - Exponential distribution I have a question regarding problem 12(b) and (c) of chapter 5 of M.Ross "Introduction to probability models".
The question is as follows:
If $X_1, X_2, X_3$ are independent exponential random variables with rates $\lambda_i$, $i = 1,2,3$, find 
(b) $P(X_1 < X_2 \mid \max(X_1,X_2,X_3) = X_3)$;
(c) $E[\max X_i \mid X_1 < X_2 < X_3]$.
For (b), my first thought was that $P(X_1 < X_2 \mid \max(X_1,X_2,X_3) = X_3) = P(X_1 < X_2)$ since from my point of few, $\max(X_1,X_2,X_3) = X_3$ does not have to influence $X_1 < X_2$.
According to the answer book of Ross: 
$P(X_1 < X_2 \mid \max(X_1,X_2,X_3) = X_3) = \frac{P(X_1 < X_2 < X_3)}{P(X_1 < X_2 < X_3) + P(X_2 < X_1 < X_3)}$.
Can anyone explain me what mistake I made with my initial reasoning?
For (c) I have no clue how to reason to get the correct answer.

(b) The following might help to understand your mistake.
Let $U_{1},U_{2},U_{3}$ be independent random variables where $P\left(U_{1}=1\right)=P\left(U_{1}=3\right)=\frac{1}{2}$
and $P\left(U_{2}=2\right)=1=P\left(U_{3}=2\right)$.
Then it is evident that: $$P\left(U_{1}<U_{2}\mid\max\left(U_{1},U_{2},U_{3}\right)=U_{3}\right)=1\neq\frac{1}{2}=P\left(U_{1}<U_{2}\right)$$
(Almost) degenerated random variables can be very helpful to examine questions like: "is my intuition correct here?"

(c) 
Under condition $X_{1}<X_{2}<X_{3}$ you are dealing with the original
PDF divided by probability $P\left(X_{1}<X_{2}<X_{3}\right)$. 
To be worked out is the integral:$$\frac{\frac{1}{\lambda_{1}\lambda_{2}\lambda_{3}}\int_{0}^{\infty}\int_{x}^{\infty}\int_{y}^{\infty}ze^{-\lambda_{1}x-\lambda_{2}y-\lambda_{3}z}dzdydx}{\frac{1}{\lambda_{1}\lambda_{2}\lambda_{3}}\int_{0}^{\infty}\int_{x}^{\infty}\int_{y}^{\infty}e^{-\lambda_{1}x-\lambda_{2}y-\lambda_{3}z}dzdydx}$$
Note that the denominator equals $P(X_1<X_2<X_3)$. 
