Probability that $\max(X,Y)> a \min(X,Y)$ where $X$ and $Y$ are independent and uniformly distributed on $[0,1]$ Two independent random variable $X$ and $Y$ having probability density functions uniform in the interval [0,1]. When $a \geqslant 1$, what is the probability that $\max(X,Y)> a \min(X,Y)$,  in terms of $a$?
I am not able to get the concept and principle behind this question.
 A: Since events $\{X \geqslant Y\}$ and $\{X < Y\}$ are disjoint:
$$ \begin{eqnarray}
  \Pr(\max(X,Y) > a \min(X,Y)) &=& \Pr(\max(X,Y) > a \min(X,Y), X \geqslant Y) \\ && + \Pr(\max(X,Y) > a \min(X,Y), X < Y) \\ &=& \Pr(X > a Y, X \geqslant Y) + \Pr(Y > a X, X < Y)
\end{eqnarray}
$$
Now proceed graphically. Since $a>1$, the event $\{X > a Y, X \geqslant Y\} = \{ X > a Y \}$ is the portion of unit square $[0,1]^2$ that has a shape of a triangle, and similarly for the other event $$\{Y>a X, X<Y\} = \{Y> a X, Y>X\} = \{Y > a X\}$$ The areas are these regions (which correspond to the probabilities) are the same. 
Figure out their areas and add them up:
$$
   \Pr(\max(X,Y) > a \min(X,Y))  = 2 \Pr\left(X > a Y\right) = \frac{1}{a}
$$
The following picture might be helpful:

A: Conditioning on whether $X$ or $Y$ is the maximum of the two random variables you can write the required probability as: $$\begin{align*}P(\max\{X,Y\}>a\min\{X,Y\})&=P(X>aY\mid X\ge Y)P(X\ge Y)\\[0.2cm]&+P(Y>aX\mid Y\ge X)P(Y\ge X)\end{align*}$$ which due to symmetry of $X,Y$ can be simplified to: $$P(\max\{X,Y\}>a\min\{X,Y\})=2\cdot P(X>aY\mid X\ge Y)P(X\ge Y)$$ Again due to symmetry $P(X\ge Y)=\frac12$ and therefore it remains to calculate the probability $$P(X>aY\mid X\ge Y)=\frac{P(X>aY, X\ge Y)}{P(X\ge Y)}=\frac{P(X>aY)}{\frac12}=2P(X>aY)$$ Conditioning on $Y$, the probability on the RHS becomes $$P(X>aY)=\int_{Y}P(X>ay)f_Y(y)dy=\int_{0}^{\frac1a}\frac{1-ay}{1-0}\cdot1\,dy=\left.\frac{2y-ay^2}{2}\right|_0^{\frac1a}=\frac{1}{2a}$$ were the integration limits were selected so that $0<ay<1$. Thus $$P(X>aY \mid X\ge Y)=2\cdot \frac{1}{2a}=\frac{1}{a}$$ and therefore $$P(\max\{X,Y\}>a\min\{X,Y\})=\not 2\cdot\frac{1}{a}\cdot \frac{1}{\not 2}=\frac{1}{a}$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{a \geq 1}$.

There is a ${\sf\underline{straightforward}}$ answer:

\begin{align}&\color{#66f}{\large\left.\int_{0}^{1}\int_{0}^{1}\dd x\,\dd y\,
\right\vert_{\,\max\pars{x,y}\ >\ a\min\pars{x,y}}}
=\left.\int_{0}^{1}\int_{0}^{1}\dd x\,\dd y\,
\,\right\vert_{x\ <\ y \atop {\vphantom{\Large A}y\ >\ ax}} +
\left.\int_{0}^{1}\int_{0}^{1}\dd x\,\dd y
\,\right\vert_{x\ >\ y \atop {\vphantom{\Large A}x\ >\ ay}}
\\[5mm]&=\left.\int_{0}^{1}\int_{0}^{1}\dd x\,\dd y\,\right\vert_{\,x\ <\ y/a}
+\left.\int_{0}^{1}\int_{0}^{1}\dd y\,\dd x\,\right\vert_{\,y\ <\ a/x}
=2\int_{0}^{1}\int_{0}^{y/a}\,\dd x\,\dd y
=2\int_{0}^{1}{y \over a}\,\dd y=\color{#66f}{\Large{1 \over a}}
\end{align}
