# (Simple?) probability question

• January 20th 2010, 12:27 PM
emily922
(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! :)
• January 20th 2010, 02:40 PM
hatsoff
Quote:

Originally Posted by emily922
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 $\frac{\pi}{4}$.
• January 20th 2010, 02:44 PM
emily922
Thank you, that makes lots of sense!