So we should show that $P(W^2 > Z > t) = \frac {1}{3} - t + \frac {2}{3} t^{3/2}$ where $t \in (0,1)$ when $Z, W$ are independent $U[0,1]$ random variables.
2026-03-28 08:40:08.1774687208
Get $P(W^2 > Z > t) = \frac {1}{3} - t + \frac {2}{3} t^{3/2}$ where $t \in (0,1)$
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Guide:
Probability is equal to the area when we integrate the pdf, hence the task that I give you is to find the area of the shaded region. Possibly by integration.