Compute the probability $P(X ≤ Y|\min\{X,Y\}=x)$ Let $X$ and $Y$ be two independent exponential random variables with parameters $\lambda$ and $\mu$, respectively. Compute the probability $P(X ≤ Y|\min\{X,Y\}=x)$.  
Here is what I did:
If $Y=\min\{X,Y\}$, then $Y\lt X$, $P(X ≤ Y|\min\{X,Y\}=x)=0$.
If $X=\min\{X,Y\}$, then $X\lt Y$,
$$
\begin{aligned}
P(X ≤ Y|\min\{X,Y\}=x)&=P(X ≤ Y|X=x)\\
&=\frac{P(Y\gt x)}{f_X(x)}\\
&=\frac{1-P(Y\lt x)}{f_X(x)}\\
&={1-\int_{0}^x\mu e^{-\mu y}dy\over \ \lambda e^{-\lambda x}}
\end{aligned}
$$
Am I right?
 A: $$P(X \le Y \mid \min\{X, Y\} = x) = \frac{P(X \le Y, \min\{X,Y\} \in [x, x+dx])}{P(\min\{X,Y\} \in [x, x+dx])}.$$
The numerator is equal to
$$P(X \le Y, X \in [x, x + dx]) = P(X \in [x, x+dx]) P(Y \ge x) = (\lambda e^{-\lambda x} \, dx)(e^{-\mu x}).$$
The denominator is equal to
$$P(X \le Y, X \in [x, x + dx]) + P(X \ge Y, Y \in [x, x + dx])
= (\lambda e^{-\lambda x} \, dx)(e^{-\mu x}) + (\mu e^{-\mu x} \, dx) (e^{-\lambda x})$$
by similar computations.
Thus the answer is $\frac{\lambda}{\lambda + \mu}$. Note that it is interesting that the answer does not depend on $x$. [You could also quickly answer this question using properties of Poisson processes, but perhaps this is off-topic.]

Response to comment: It actually does work. $$F(x + dx) - F(x) = e^{-\lambda x} - e^{-\lambda (x + dx)} = e^{-\lambda x}(1 - e^{-\lambda \, dx}).$$
Then using $e^z = 1 + z + O(z^2)$, we can plug in $e^{-\lambda dx} = 1 - \lambda \, dx$ to obtain $\lambda e^{-\lambda x} \, dx$.
A: The second part isn't quite right, since $P(X=x)=0$ for any continuous random variable $X$.
You can compute this probability that way,
\begin{align}
P(X<Y|X=x) &= P(x<Y) \\
&= 1 - \int_0^x\mu e^{-\mu y}dy \\
&= 1 - (1-e^{-\mu x}) \\
&= e^{\mu x}
\end{align}
Note that this is not the final answer and you should be careful on how you have split your two cases. 
