In my exam preparation I'm currently stuck at this task:
Let $X_1,..,X_5$ be i.i.d random variables each having uniform distributions in the interval (0, 1)
Find the probability that $X_1$ is the minimum and $X_5$ is the maximum among these random variables.
$P(X_1 \leq$ min{$X_2$,…,$X_5$)
=$P(X_1 \leq X_2, X_1 \leq X_3, …, X_1 \leq X_5)$
=$P(X_1 \leq X_2)\cdot … \cdot P(X_1 \leq X_5)$
=$P(X_1 – X_2 \geq 0)\cdot … \cdot P(X_1 – X_5 \geq 0)$
=$\frac{1}{2}^4$
This is my (wrong) calculation for now. I made the maximum with the same flawed idea in mind and I just cannot think of the correct answer*. Thanks for any hint!
*Per answersheet it is $\frac{1}{20}$. But I need to understand how to get there.
Best Answer
Your method fails because the events $X_1<X_2$ and $X_1<X_3$ are not independent.
Per the given distribution and independence, the probability that any two of the random vars are equal is $0$. Hence one of them is the minimal and another is the maximal among them. By mere symmetry, the probability that $\min =X_1$ and $\max = X_5$ is the same as the probability of $\min=X_i$, $\max=X_j$ for any choice of $i\ne j$. As there are $20$ such choices, we arrive at $\frac1{20}$.