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October 12th, 2018, 07:18 AM   #1
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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 $\displaystyle f(t)=t-1$ and total area ?
Post solution with explanation
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October 12th, 2018, 10:02 AM   #2
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Draw the line from the point (1,0) to the point (0,1).

This is the line described by $y = 1-x$

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 $P[x+y \leq 1] = \dfrac{\text{area of lower triangle}}{\text{area of entire positive unit square}} = \dfrac{\frac 1 2 }{1} = \dfrac 1 2$
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October 12th, 2018, 12:54 PM   #3
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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?
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October 12th, 2018, 02:05 PM   #4
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Quote:
Originally Posted by EvanJ View Post
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?
The joint probability density of (X,Y) is 2D uniform.

Their sum won't be 1D uniform. Work it out and see.
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October 13th, 2018, 09:25 AM   #5
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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.
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October 13th, 2018, 10:52 AM   #6
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Quote:
Originally Posted by EvanJ View Post
I don't know what 2D and 1D mean
2 dimensional and 1 dimensional.

(X,Y) has a 2 dimensional joint distribution

X+Y has a 1 dimensional distribution.
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