At Least Two Points in the Circle

Suppose points are randomly generated so that they can land anywhere within the square with equal probability.

If three such points are generated, what is the probability that at least two of them will lie inside a circle inscribed within the square?

Note: Give your answer as a number between 0 0 and 1 1


The answer is 0.8816.

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1 solution

The probability that any one of the three points lands inside the circle is C = area of the circle area of the square = π r 2 ( 2 r ) 2 = π 4 C=\frac{\mbox{area of the circle}}{\mbox{area of the square}}=\frac{\pi r^2}{(2r)^2}=\frac{\pi}{4} Since each point position is independent of the next one, we have: P ( at least 2 lie inside the circle ) = P ( 3 lie inside the circle ) + P ( 2 lie inside the circle and 1 lie outside the circle ) = C 3 + 3 C 2 ( 1 C ) 0.8816 P(\mbox{at least 2 lie inside the circle})=P(\mbox{3 lie inside the circle})+P(\mbox{2 lie inside the circle and 1 lie outside the circle})=C^3+3C^2(1-C)\approx\boxed{0.8816}

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