# Is this probability method correct?

If x,y,z are selected randomly and independently from the interval [0,1], what is the probability that $x\geq yz$?

The answer is $3/4$, but is there any way to obtain it without using integration?

I thought of using expectation, $E(y)=1/2, E(z)=1/2, E(yz)=1/2*1/2=1/4$, so $P(x\geq yz)=3/4$.

The problem is if I view it the other way round, $E(x)=1/2, P(yz\geq 1/2)$ is not 1/2?

Sincere thanks for any explanation.

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Let $W$ be any $[0,1]$ valued random variable independent of $x$.
Then the probability that $x \geq W$ is the expected value $E[1-W]$ (compare the definitions). For $W = yz$ the rest of the calculation is done in the question.