Calculate $P(X>Y^2)$

Question

Calculate $P(X>Y^2)$ given that $$f(x,y)=6(x-y)$$ for all $(x,y)$ such as $$0\le y\le x\le1$$

My solution: $$6\cdot\int_{0}^{y^2}\int_{0}^{x}(x-y)dydx$$ which in turn results in : $y^6$. However I have no way of knowing whether or not I have actually solved this correctly and hence arrived at the right answer. Is my solution correct? and how do I verify my answer?

Answer 1

Refer to the graph:

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The total probability is: $$P(\underbrace{0\le Y\le X\le 1}_{\text{the gray region}})=P(0\le Y\le X \ \cap \ 0\le X\le 1)=\int_0^1\int_0^x6(x-y)dydx=1.$$ The required probability is: $$P(X>Y^2)=P(Y^2<X)=P(0\le Y^2<X\le 1)=P(\underbrace{0\le Y<\sqrt{X}\le 1}_{\text{the gray and orange regions altogether}})=\\ P(\underbrace{0\le Y\le X\le 1}_{\text{the gray region}})+P(\underbrace{0\le \color{red}{X< Y}\le \sqrt{X}\le 1}_{\text{the orange region}})=1+0=1,$$ because: $$f(x,y)=0, \color{red}{X<Y} \Rightarrow P(X<Y)=0.$$

Answer 2

This probability is equal to $1$. This is because $$\Pr\{X>Y\}=1\overbrace{\implies}^{0\le Y^2\le Y\le 1} \Pr\{X>Y^2\}=1 $$