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

  1. Jul 24, 2009 #1
    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: [tex]\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}[/tex]

    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. jcsd
  3. Jul 24, 2009 #2


    Staff: Mentor

    Your answer looks fine to me, but the union in this probability
    [tex]\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}[/tex]
    should be an intersection
    [tex]\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}[/tex]
  4. Jul 24, 2009 #3
    Thanks for the correction.
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