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

1. Jul 24, 2009

### cse63146

1. The problem statement, all variables and given/known data

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

Determine P(Y>0.5|Y>X^2)
2. Relevant equations

3. The attempt at a solution

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

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)

Last edited: Jul 24, 2009
2. Jul 24, 2009

### Staff: Mentor

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

3. Jul 24, 2009

### cse63146

Thanks for the correction.