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#### CaptainBlack

##### Well-known member

- Jan 26, 2012

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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:

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