Ken's Question from Yahoo Answers: Probability Question For Verification?

  • Context: MHB 
  • Thread starter Thread starter CaptainBlack
  • Start date Start date
  • Tags Tags
    Probability
Click For Summary
SUMMARY

The probability that a randomly chosen point \({\rm{P}}(x,y)\) within a unit disc satisfies the conditions \(|x-y| < 1\) and \(|x+y| < 1\) is definitively \(2/\pi\). This conclusion is derived from the area of the inscribed square, which has an area of \(2\), compared to the area of the unit circle, which is \(\pi\). The analysis confirms that the required region is the interior of the square, as demonstrated by evaluating the point \((0,0)\) against the inequalities.

PREREQUISITES
  • Understanding of basic probability concepts
  • Familiarity with geometric shapes, specifically circles and squares
  • Knowledge of area calculations for geometric figures
  • Ability to interpret inequalities in a two-dimensional context
NEXT STEPS
  • Study the properties of inscribed shapes within circles
  • Learn about uniform distribution in probability theory
  • Explore advanced probability topics, such as conditional probability
  • Investigate visualizations of probability distributions using scatter plots
USEFUL FOR

Mathematicians, statisticians, educators, and students interested in probability theory and geometric probability applications.

CaptainBlack
Messages
801
Reaction score
0
Question:
"A point \({\rm{P}}(x,y)\) is chosen at random in a unit disc, centred at \((0,0)\).

The probability required is that the point chosen is such that both \(| x -y| \lt 1\) and \(|x+y| \lt 1\) .

Is the answer \(2/\pi\) or \(1-2/\pi\)?

Thank you."Answer:
I take disc to be a disc of radius 1 centred at the origin.

The region defined by the inequalities \(|x-y| \lt 1\) and \(|x+y| \lt 1\) is an inscribed square to the circle, which has side \(\sqrt{2}\) and hence area \(2\). The area of the circle is \(\pi\), so the probability that a point sampled uniformly on the unit disc satisfies the inequalities is the ratio of these two area: \(2/\pi\).

To convince yourself that the required region is the interior of the square rather than the exterior consider the point \((0,0)\), does it satisfy the inequalities. It it does then you want the interior of the square rather than the exterior.

Below is a scatter plot showing random points uniformly sampled on the unit disc and in black those satisfying the inequalities:

https://lh3.googleusercontent.com/AsrqIRhjcPwGKPW6RSzDwZRoH0ryjndkugx09Ohv2VkvdbS60GwQ4Gtv2A4qZZSiWoBqxPZVPw
 
Last edited:
Physics news on Phys.org
4LhKv1R4M3w1jH8l0t3A6LJW1z2kQlW4oLJfKQe2-3vW4uw3RrZ-DeQ2k6n1U

So, the answer is indeed \(2/\pi\). I hope this helps!
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
11K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
6K
Replies
2
Views
2K