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Let $X$ and $Y$ be independent random variables, each of which is uniformly distributed between $0$ and $1$.

Find the probability that $(X−\frac 1 2)^2+(Y−\frac 1 2)^2\leq \frac 1 9$. Give at least 8 correct digits after the decimal point. Hint: don't compute any integrals; think about this problem geometrically.

I understand that the uniform distribution means each value has an equal chance of occurring but it threw me off when saying don't compute any integrals.

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Make a picture. A square and within it a disk. –  drhab Apr 30 at 11:32
    
Ahhh - the good old homework factory back at work. Same question from same class posted here yesterday: stats.stackexchange.com/questions/95688/… ... There really should be a time delay on answering these things. –  wolfies Apr 30 at 12:59

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up vote 2 down vote accepted

The vector $(X,Y)$ is uniformly distributed on $[0,1]^2$, i.e. $$ P((X,Y)\in A)=\frac{|A|}{|[0,1]^2|}=|A|,\quad A\subseteq [0,1]^2. $$ Now, express the probability that $(X-\tfrac12)^2+(Y-\tfrac12)^2\leq \tfrac19$ as $P((X,Y)\in A)$ for a suitable $A\subseteq [0,1]^2$ and compute.

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i'm sorry I'm a little confused still –  Jilly Apr 30 at 11:28
    
Confused by what? –  Stefan Hansen Apr 30 at 11:29
    
How to express the expression as P(X,y) for a suitable A –  Jilly Apr 30 at 11:31
    
Stefan means, draw a graph of $(x-\frac 12)^2 + (y-\frac 12)^2 = \frac 19$ and find out how it cuts the square $[0..1]^2$. |A| is an expression of area. –  Graham Kemp Apr 30 at 11:32
    
@GrahamKemp $x-\frac{1}{2}$ must replace $x+\frac{1}{2}$ here. –  drhab Apr 30 at 11:34

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