# I Random variable and probability

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1. Mar 2, 2016

### Imagin_e

Hi!
I'm searching for guidance and help since I don't know how to solve this problem. Here it is:

a) The two-dimensional random variable (ξ,η) is uniformly distributed over the square
K={(x,y): 0≤x≤1 , 0≤y≤1} . Let ζ=√ξ22 me the distance between the origo and the point (ξ,η) . Calculate the probability P(ζ<1)

b) The three-dimensional random variable ξ=(ξ1,ξ2,ξ3) is uniformly distributed on a sphere, which has origo as it's center and a radius of 1. Calculate the probability that ξ ends in the area A which is given by:
A={(x1,x2,x3): 1/3 <√x12+x22+x32≤2/3} . You should, in other words, calculate :
P(1/3<ζ≤2/3) , where ζ=√ξ12+ξ22+ξ32

Note: √ means square root of.. I literarily have no idea where to start.

2. Mar 2, 2016

### Buzz Bloom

Hi Imagin_e:

A key point about these problems is "uniformly distributed".

For (a) this means the probability P is the area of a part of the unit square ζ<1. This square has total area = 1.

For (b) this means P will be a volume of a part of a sphere (1/3<ζ≤2/3) divided by the volume of the whole sphere.

Hope this helps.

Regards,
Buzz

3. Mar 2, 2016

### Aakash Gupta

(a) Since the probability distribution is uniform in the square K(take it as ABCD here), there P(ξ,η) = 1 for all points inside the square(why? : because ∫∫P(ξ,η)dξdη with 0<ξ<1 & 0<η<1 should be 1 and marginal distribution for ξ and η are uniform)
Now we need to find probability of all those points lying below the arc ADB(to the opposite of e, poly1 side), since the distribution is uniform,
therefore, we will have P(ζ<1) = Area below Arc / Area of square = π/4 (area of square is 1)

(b)Now I hope that you can do this as its similar to (a), but if you face problem, feel free to reply

Last edited: Mar 2, 2016
4. Mar 2, 2016

### Buzz Bloom

Hi Aakash:

Your answer is flawed. What is the total area of the circle of which a part is inside the square? What is the fraction of the whole circle that is inside the square?

Regards,
Buzz

5. Mar 2, 2016

### Aakash Gupta

Oh sorry that must be pi/4, is there anything wrong apart from that?

6. Mar 2, 2016

### Buzz Bloom

Hi Aakash:

That's OK now.

Regards,
Buzz