How Do You Calculate P(Y>0.5|Y>X^2) for Uniformly Distributed Variables?

[cse63146]

Homework Statement

X and Y are independant and uniformly distributed on the unit interval (0, 1)

Determine P(Y>0.5|Y>X^2)

Homework Equations

The Attempt at a Solution

I set it up as the following: \frac{P(Y &gt; 0.5 \cup Y &gt; X^2)}{P(Y &gt; X^2)}<br /> = \frac{\int^1_{1/2} \int^{\sqrt{y}}_0 dx dy}{\int^1_{0} \int^{\sqrt{y}}_0 dx dy}

for the final answer I got ~0.6465. Did I make any mistakes?

The only thing that's bugging me is whether or not x is properly bounded in the integral. In the question y > x^2, so I assumed the square root of y is also greater than x. And since x is between 0 and 1, I believe that 0<x<sqrt(y)

 

[Mark44]

Your answer looks fine to me, but the union in this probability
\frac{P(Y &gt; 0.5 \cup Y &gt; X^2)}{P(Y &gt; X^2)} = \frac{\int^1_{1/2} \int^{\sqrt{y}}_0 dx dy}{\int^1_{0} \int^{\sqrt{y}}_0 dx dy}
should be an intersection
\frac{P(Y &gt; 0.5 \cap Y &gt; X^2)}{P(Y &gt; X^2)} = \frac{\int^1_{1/2} \int^{\sqrt{y}}_0 dx dy}{\int^1_{0} \int^{\sqrt{y}}_0 dx dy}

 

[cse63146]

Thanks for the correction.