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Let $X_1,X_2,X_3$ be independent $Exp(\lambda)$ distributed random variables. I need to find $P(X_1<X_2<X_3)$.

I think we need to use independence, then we can rewrite $P(X_1<X_2<X_3)=P(X_1<X_2)P(X_2<X_3)$, but I don't know how to proceed next. Maybe considering $Y=X_1-X_2$, but then we need to find the distribution of Y, which is hard for me.

Any help would be appreciated, thanks.

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By symmetry,

we have

\begin{align}P(X_1<X_2<X_3)&=P(X_1<X_3<X_2)\\&=P(X_2<X_1<X_3)\\ &=P(X_2<X_3<X_1)\\ &=P(X_3<X_1<X_2)\\ &=P(X_3<X_2<X_1)\\ \end{align}

Since they have to sum up to $1$, $$P(X_1<X_2<X_3)=\frac16$$

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