Given $X_1, X_2,X_3$ are independent random variables from $U(0,1)$, find $P(X_1<X_2<X_3)$.
2026-04-04 01:51:59.1775267519
Given independent random variables $X \sim U(0,1)$, find $P(X_1<X_2<X_3)$
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By symmetry, each ordering of $(X_1,X_2,X_3)$ is equally likely, so the probability is $1/3!=1/6$.