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October 12th, 2018, 06:18 AM  #1 
Senior Member Joined: Dec 2015 From: somewhere Posts: 592 Thanks: 87  Geometric Probability
Probability with two variables x,y If $\displaystyle x\in [0,1] \; $ and $\displaystyle \; y\in [0,1]$ Find probability such that $\displaystyle x+y \leq 1$ Question is why we use division of area created by line $\displaystyle f(t)=t1$ and total area ? Post solution with explanation 
October 12th, 2018, 09:02 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,529 Thanks: 1389 
Draw the line from the point (1,0) to the point (0,1). This is the line described by $y = 1x$ This line divides the positive unit square into 2 triangles. Ordered pairs in the lower triangle (including the line) are such that $x + y \leq 1$ so $P[x+y \leq 1] = \dfrac{\text{area of lower triangle}}{\text{area of entire positive unit square}} = \dfrac{\frac 1 2 }{1} = \dfrac 1 2$ 
October 12th, 2018, 11:54 AM  #3 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 660 Thanks: 87 
Couldn't it be said that x + y is uniformly distributed from 0 to 2 with 0.5 probability of being between 0 and 1 and 0.5 probability of being between 1 and 2?

October 12th, 2018, 01:05 PM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 2,529 Thanks: 1389  
October 13th, 2018, 08:25 AM  #5 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 660 Thanks: 87 
I don't know what 2D and 1D mean, but I figured it out. The numbers closest to 1 will be the most frequent sums. It's like how the sum of two dice has a mean of 7, with 7 as the the mode and the probability decreasing to the extremes of 2 and 12.

October 13th, 2018, 09:52 AM  #6 
Senior Member Joined: Sep 2015 From: USA Posts: 2,529 Thanks: 1389  

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geometric, probability 
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