I am really stuck on the following question during my exam preparation.

Let $X\_1$, $X\_2$ and $X\_3$ be independent exponential random variables with respective rates $\mu\_1$, $\mu\_2$ and $\mu\_3$. Find:  

a) $P(X\_1 < X\_2 < X\_3)$  

b) $E(X\_2 | X\_1 < X\_2 < X\_3) $

Question a
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For question a) I've tried two approaches but they do not give me the same answer. Why the discrepancy and is one of the two even correct?

### First approach

$P(X\_1 < X\_2 < X\_3) =$

[using the law of total probability]  

$ = P(X\_1 < X\_2 < X\_3 | min X\_i = X\_1) P(min X\_i = X\_1) + P(X\_1 < X\_2 < X\_3 | min X\_i = X\_2) P(min X\_i = X\_2) + P(X\_1 < X\_2 < X\_3 | min X\_i = X\_3) P(min X\_i = X\_3)$

[The second and third term lead to contradictions and therefor are 0]  

$= P(X\_1 < X\_2 < X\_3 | min X\_i = X\_1) P(Xi = min X1)$
$= P(X\_1 < X\_2 < X\_3, min X\_i) = P(X\_1 < X\_2 < X\_3, X\_1 < X\_2, X\_1 < X\_3)$  

$= P(X\_2 < X\_3, X\_1 < X\_2, X\_1 < X\_3)$
[first entry contains redundant information]  

$= P(X\_2 < X\_3)P(X\_1 < X\_2)P(X\_1 < X\_3)$
[They are independent]  

$= u\_2/(u\_2+u\_3) \cdot u\_1/(u\_1+u\_2) \cdot u\_1/(u\_1+u\_3)$

### Second approach

The joint distribution is given by
$f(x\_1,x\_2,x\_3) = f\_1(x\_1)f\_2(x\_2)f\_3(x\_3)$ due to independence

$P(X\_1 < X\_2 < X\_3) = \int\_{x\_2}^{\infty}\int\_0^{\infty}\int\_0^{x\_2}f\_1(x\_1)f\_2(x\_2)f\_3(x\_3)dx\_1 dx\_2 dx\_3 $
$= \int\_0^{\infty}\int\_0^{x\_2}f\_1(x\_1)dx\_1 \int\_{x\_2}^{\infty}f\_3(x\_3)dx\_3 f\_2(x\_2)dx\_2$

$= u\_2/(u\_2+u\_3) - u\_2/(u\_1 + u\_2 + u\_3) $

Question b
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For question b) I want to do the following:

$\int\_{x\_2}^{\infty}\int\_0^{\infty}\int\_0^{x\_2}x\_2\*f\_1(x\_1)f\_2(x\_2)f\_3(x\_3)dx\_1 dx\_2 dx\_3 $
am I on the right track?