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May 1st, 2017, 06:32 PM  #1 
Newbie Joined: Apr 2017 From: PA Posts: 11 Thanks: 0  Finding a cdf
A point P in the unit square has coordinates X and Y chosen at random in the interval [0, 1]. Let D be the distance from P to the nearest edge of the square, and E the distance to the nearest corner. (a) D < 1/4? (b) E < 1/4? find the cumulative distribution F and density f for the random variable D attempt: I thought about using a distance formula to solve this but the forumla had too many coordinates. So then I decided to use P (Y is less than or equal to y) to get P (D < or equal to y), to eventually get P (D < or equal to y  (1/2)), so I eventually came up with the Cdf of F (d) = d  (1/2) on the interval of [1\2,1 and a half] but I got the wrong answer, how should I find the correct cdf? 
May 1st, 2017, 07:34 PM  #2 
Senior Member Joined: Sep 2015 From: CA Posts: 1,199 Thanks: 611 
these will be the regions of $D<c$ and $E<c$ $D$ is pretty clearly $P[D<c] = 1(12c)^2 = 4c(1c),~0\leq c \leq \dfrac 1 2$ $E$ will take a bit more doing since the regions start to overlap. I'll leave that to you. 
May 2nd, 2017, 12:26 AM  #3  
Senior Member Joined: Sep 2015 From: CA Posts: 1,199 Thanks: 611  Quote:
$P[E<1/4]$ then the regions won't overlap and it's really pretty simple.  
May 2nd, 2017, 06:48 PM  #4 
Newbie Joined: Apr 2017 From: PA Posts: 11 Thanks: 0 
Hey Romsek, may I ask how did u get the 12c? Also, thank u for urnhelp!

May 2nd, 2017, 06:48 PM  #5 
Senior Member Joined: Sep 2015 From: CA Posts: 1,199 Thanks: 611  

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cdf, finding 
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