Question on Conditional Probabilities This one is from Introduction to Probability Models by Sheldon Ross. Not homework, was just trying to solve through some exercises again.
Let $X$ and $Y$ be independent exponential random variables with rates $\lambda$ and $\mu$ respectively. Let $M=\min(X,Y)$.
Find :
\begin{align}
E[MX|M=X]\\E[MX|M=Y]\\Cov(X,M)
\end{align}
What I tried:
\begin{align}
E[MX|M=X]&=E[X^2]\\&=\dfrac{1}{\lambda^2}+\dfrac{1}{\lambda^2}\\&=\dfrac{2}{\lambda^2}
\end{align}
The solutions at the back of the book do:
$$E[MX|M=X]=E[M^2]=\dfrac{2}{(\lambda+\mu)^2}$$
I didn't proceed any further since there was a discrepancy on the first step itself. 
How should I be solving this question?

First of all, $\lambda$ and $\mu$ must be the rates, not the means.
$M$ is an exponential random variable with rate $\lambda + \mu$.  Think of
a Poisson process with rate $\lambda + \mu$, where each occurrence is independently assigned to 
the $X$ process with probability $\lambda/(\lambda+\mu)$, otherwise to the $Y$ process.
$M=X$ means the first occurrence of the process is assigned to the $X$ process.
This event is independent of $M$, but not of $X$.  So $E[MX | M=X] = E[M^2 | M = X] = E[M^2]$.
