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

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The probability of randomly selecting a point \((x,y)\) within a unit disc that satisfies the conditions \(|x-y| < 1\) and \(|x+y| < 1\) is calculated. The area defined by these inequalities forms an inscribed square with an area of 2, while the area of the unit circle is \(\pi\). Thus, the probability is the ratio of these areas, resulting in \(2/\pi\). The discussion emphasizes that the region of interest is the interior of the square, as demonstrated by the inclusion of the origin point \((0,0)\). Therefore, the correct answer to the probability question is \(2/\pi\).
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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
 
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So, the answer is indeed \(2/\pi\). I hope this helps!
 
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