Probability/Random variables question

ashah99
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Homework Statement
Finding the probability of randomly choosing a point within the unit square constrained within the quarter circle
Relevant Equations
P(( X, Y ) ∈ A) = ∫∫ fXY ( x, y )dxdy
Hello all, I am wondering if my approach is coreect for the following probability question? I believe the joint PDF would be 1 given that the point is chosen from the unit square. To me, this question can be reduced down to finding the area of 1/4 of a circle with radius 1. Any help is appreciated!
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It's clear from your calculations that you are indeed simply calculating the area. Which is what you would expect for a uniform distribution.
 
PeroK said:
It's clear from your calculations that you are indeed simply calculating the area. Which is what you would expect for a uniform distribution.
Ok, makes sense. Would you agree that my answer is correct? Just want to make sure I understand.
 
ashah99 said:
Ok, makes sense. Would you agree that my answer is correct? Just want to make sure I understand.
Yes, it's just a ratio of areas, as you've calculated.
 
Or use polar coordinates: <br /> \int_0^1 \int_0^{\sqrt{1-x^2}} 1\,dy\,dx = \int_0^{\pi/2}\int_0^1 r\,dr\,d\theta.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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