Results 1 to 3 of 3

Math Help - (Simple?) probability question

  1. #1
    Newbie
    Joined
    Jan 2010
    Posts
    2

    (Simple?) probability question

    I'm just starting a new module of probability, and I get the sense that these are supposed to be revision questions, but I don't seem to be able to work it out. :S

    If X and Y are uniformly distributed continuous random variables on (0,1), then how would I work out P(X^2 + Y^2 <= 1)?

    I've looked through all my past probability work and lecture examples and I can't find anything similar. I know that P(X <= x) = x and P(Y <= y) = y, but how can you combine these, since they depend on each other?

    Thanks for any and all help!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Feb 2008
    Posts
    410
    Quote Originally Posted by emily922 View Post
    I'm just starting a new module of probability, and I get the sense that these are supposed to be revision questions, but I don't seem to be able to work it out. :S

    If X and Y are uniformly distributed continuous random variables on (0,1), then how would I work out P(X^2 + Y^2 <= 1)?

    I've looked through all my past probability work and lecture examples and I can't find anything similar. I know that P(X <= x) = x and P(Y <= y) = y, but how can you combine these, since they depend on each other?

    Thanks for any and all help!
    Well, first you want to identify that x^2+y^2\leq 1 is a disc of radius 1 centered at the origin. Next we observe that the uniform distribution is over the area [0,1]\times[0,1]. So, we are interested in the intersection of these areas, which is the quarter disc in the first quadrant.

    Now, clearly the pdfs of X and Y are both 1. So if R is the region in question, then

    \mathbb{P}(X^2+Y^2\leq 1)=\int_R f_X(x)f_Y(y)dR=\int_RdR,

    which we do not actually need to integrate, since we know that the area of a quarter disc with radius 1 is \frac{\pi}{4}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2010
    Posts
    2
    Thank you, that makes lots of sense!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. simple probability question
    Posted in the Statistics Forum
    Replies: 1
    Last Post: November 2nd 2010, 01:11 AM
  2. Simple probability question
    Posted in the Statistics Forum
    Replies: 2
    Last Post: October 5th 2010, 02:19 AM
  3. Simple Probability Question Help??
    Posted in the Statistics Forum
    Replies: 1
    Last Post: September 20th 2009, 07:16 PM
  4. Help with simple probability question..
    Posted in the Statistics Forum
    Replies: 1
    Last Post: June 8th 2009, 07:52 PM
  5. Simple probability question...
    Posted in the Advanced Statistics Forum
    Replies: 6
    Last Post: April 22nd 2008, 04:04 AM

Search Tags


/mathhelpforum @mathhelpforum