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Suppose you choose two numbers $x$ and $y$ independently at random from the interval $[0, 1].$

Given that their sum lies in the interval $[0, 1],$ find the probability that

a) $x^2+y^2<\frac{1}{2},$ and

b) $x>y.$

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    Hint. Draw a picture. Shade the area in the unit square that describes the first condition. Then shade the subset that satisfies(a) and (b). – 2017-02-28
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    I'm voting to close this question as off-topic because user has not shown any evidence of partial solution or any attempt to solve it. – 2017-03-01

1 Answers 1

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The sample space of possible x, y values is the right isosceles triangle with leg 1 on the axes.

The sample space of $x^2+y^2<\frac{1}{2}$ is the origin-centered circle of radius $\frac{\sqrt{2}}{2},$ so the probability here is $\frac{\frac{\pi}{8}}{\frac{1}{2}}=\boxed{\frac{\pi}{4}}.$ Note that all of the circle's area is inside the triangle of area $\frac{1}{2}$ since the circle is tangent to the triangle at $(\frac{1}{2}, \frac{1}{2}).$

For the second one, the sample space is the southeast partition that the line $y=x$ makes. So the intersection with the triangle is exactly $\boxed{\frac{1}{2}}$ of the original triangle sample space (draw these out, it makes it much easier)

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    May I ask where the pi/8 came from? – 2017-02-28
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    The radius of the circle is $\frac{\sqrt{2}}{2},$ and it's a quarter circle so take 1/4 of the area of the whole circle. – 2017-03-01