Find $P(Y> \max\{X,-X\})$ Given:
$$f(x,y)=\frac{1}{π}; x^2+y^2<1$$
Find:$P(Y> \max\{X,-X\})$
I understood the the range is that of a circle but I have no clue where to begin. Kindly help.
 A: Most simply put, $\max\{X, -X\} = |X|$ and the angle at the cusp of the absolute value function is $\pi/2$, or $\frac 14$ of the disk. Thus the area above $|X|$ is $\frac 14$ of the area of the disk, so
$$
P(Y > \max\{X, -X\}) = \frac 14.
$$
A: I think the best way to approach this problem is to think about it geometrically. One way to think of this is "what fraction of the area enclosed by the unit circle satisfies $Y < \max(-X,X)$ (or equivalently $Y < |X|)?"
First, if $Y < 0$, then we see that $Y < -\max(-X,X)$ because we know that $\max(-X,X)$ must be non-negative. We know that $Y < 0$ occurs for half of the points of the unit circle, so now we know the conditional probability $P(Y < \max(-X,X) | Y < 0) = 1/2$.
If $Y > 0$, then there are two cases to consider. If $X > 0$, then having $Y < \max(-X,X) = X$ requires that the point $(X,Y)$ is located below the line $Y = X$ and above the $X$-axis, and, geometrically, this line emanates from the center of the circle at a 45 degree angle with the $X$-axis. Therefore the sector of points satisfying the condition $Y < \max(-X,X)$ constitute a fraction of $\frac{45}{360} = \frac{1}{8}$ of the circle.  
If $Y > 0$ and $X < 0$, then the condition $Y < \max(-X,X)$ reduces to $Y < |X|$ which (for negative $X$) means that points $(X,Y)$ must lie between the line $Y = -X$ and the $X$-axis. These satisfactory points also occupy a fraction of $\frac{45}{360} = \frac{1}{8}$.
A point $(X,Y)$ satisfies $Y < \max(-X,X)$ if and only if $(X,Y)$ lies in one of the regions describe above, which collective account for a fraction of $\frac{1}{2} + \frac{1}{8} + \frac{1}{8} = \frac{3}{4}$ of the total area of the circle, and we see $$ P(Y < \max(-X,X) = \frac{3}{4}$$
If you want the probability of the complementary event, you could then conclude that $$P(Y > \max(-X,X)) = 1 - P(Y > \max(-X,X)) = 1 - \frac{3}{4} \Longrightarrow P(Y > \max(-X,X)) = \frac{1}{4}$$
