# Probability that distance from the origin of a uniformly distributed point < x

• slakedlime
In summary, the probability that the point lies within the circle is 1. The area of a circle with a radius of 1 is the same as the unit circle.
slakedlime

## Homework Statement

A point is uniformly distributed within the disk of radius 1.
That is, its density is f(x,y) = C

For 0 ≤ x2 + y2 ≤ 1

Find the probability that its distance from the origin is less than x, 0 ≤ x ≤ 1.

[Note] My book says that the answer is supposed to be x2.

2. The attempt at a solution
Let D be the distance of the point from the origin.
D = x2 + y2, where the x and y values correspond to the point's x and y components respectively.

My interpretation is that the unit circle is centered at origin 0,0.

I can see that D can be GREATER than the point's x component (e.g. as we move away from the x-axis and towards the y-axis). However, how can D be less than x? Isn't this x the same as the circle's x-component? Or is this x independent of the point? For example, we can assume that x is not the same as the circle's x-component (let's denote that as ω). Hence, 0 ≤ x ≤ 1 and 0 ≤ ω ≤ 1, where x may or may not be equal to ω.

If this is the case, do we have to find the probability that ω2 + y2 < x?

Any explanations or guidance on how to proceed with this problem will be immensely appreciated!

Last edited:
What is the area of a circle of radius x?

RGV

$\pi(r^{2})$. I'm not sure how this relates to the question, since I don't understand how D can ever be less than x for a given value of x (unless I'm understanding this incorrectly).

slakedlime said:
$\pi(r^{2})$. I'm not sure how this relates to the question, since I don't understand how D can ever be less than x for a given value of x (unless I'm understanding this incorrectly).

The problem is one of bad notation: the person who wrote the question is using the same symbol (x) to stand for two different things in the same problem. The question should have asked for the probability that the distance is <= d, where 0 < d <= 1.

RGV

As you stated initially, a "uniform" probability density, means that the density function is a constant. Since the integral over the circle is 1, that density function must be one over the area of the disk.

Thank you for your suggestions Ray and HallsofIvy. From what you're saying, I understand the following:

1) The probability that the point lies within the circle is 1.
2) We want to find the probability that the point lies within a distance d of the circle's center, where 0 ≤ d ≤ 1. Hence I divide the area of a circle of radius d by the area of the unit circle.

Can I jump straight to the solution (I get the correct answer) or do I need to back myself up with more logic and precision?

That would be OK with me, but you need to be consistent with the way your instructor/notes/textbook presents the material. Basically, though, a distribution is uniform within a region S if the probability of a subregion of area A is cA, where c is chosen to make P(S) = 1. (The point is that the probability does not vary with the location or shape of the subregion; the only thing that counts is its area.)

RGV

## 1. What is the formula for calculating the probability that the distance from the origin of a uniformly distributed point is less than a given value x?

The formula for calculating this probability is P(D < x) = x^2 / (2a^2), where a is the length of one side of the square in which the points are uniformly distributed.

## 2. How does the probability change as the value of x increases or decreases?

As the value of x increases, the probability decreases. Conversely, as the value of x decreases, the probability increases.

## 3. Can the probability ever be greater than 1?

No, the probability cannot be greater than 1. This is because the total area under the probability distribution curve is always equal to 1.

## 4. How does the shape of the distribution affect the probability?

The shape of the distribution does not affect the probability. As long as the points are uniformly distributed, the probability will remain the same regardless of the shape of the distribution.

## 5. Is the probability affected by the dimensionality of the space in which the points are distributed?

Yes, the probability is affected by the dimensionality of the space. For example, in a two-dimensional space, the probability will be different compared to a three-dimensional space, even if the points are distributed uniformly in both cases.

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