I stared at this for a bit because writing it in terms of max and min felt clunky.
Recognize that L and S are the order statistics of two independent uniform variables, so their joint density is constant (2) over the triangle 0 ≤ S ≤ L ≤ 1. Then compute the probability by integrating this joint density over the region where L² > S, i.e., S < L².
Pro tip: After solving, sanity-check your answer by considering extreme cases: if the condition were L > S (always true), probability should be 1; if L² > 1 (never true), probability should be 0. Also, note that the answer is a simple fraction, which is typical for uniform order statistic problems.
Since X and Y are independent uniform on [0,1], the joint density of (L, S) is f(l, s) = 2 for 0 ≤ s ≤ l ≤ 1, and 0 otherwise.
The event L² > S corresponds to the region in the (l, s) plane where s < l², within the triangle 0 ≤ s ≤ l ≤ 1.
Compute P(L² > S) = ∫∫_{s < l²} 2 ds dl over the region. Determine the limits: for a given l, s ranges from 0 to min(l, l²). Since l² ≤ l for 0 ≤ l ≤ 1, the upper limit is l².
Integrate with respect to s first: ∫_{0}^{1} ∫_{0}^{l²} 2 ds dl = ∫_{0}^{1} 2l² dl = 2/3.
Check that the result is between 0 and 1 and makes intuitive sense. Optionally, derive using order statistics or geometric probability for confirmation.
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