Beating a product of uniforms

Let $X, Y, Z \sim \text{Unif}(0,1)$ be independent. Find $\mathbb{P}[X > YZ]$ .

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For $X\sim\text{Unif}(0,1)$ and a fixed $p$ , $\mathbb{P}[X>p]=1-p$ . Condition on $p=YZ$ .
$\mathbb{P}[X>YZ]=1-\mathbb{E}[YZ]$ , and $\mathbb{E}[YZ]=\mathbb{E}[Y]\mathbb{E}[Z]=\tfrac14$ by independence.

Answer

Reveal answer →

0.75 (± 0.001)

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