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2025 AMC 12A Problem 22

Problem 22 of 25HarderAlgebraCounting & Probability

Three real numbers are chosen independently and uniformly at random between 00 and 1.1. What is the probability that the greatest of these three numbers is greater than 22 times each of the other two numbers? (In other words, if the chosen numbers are abc,a \ge b \ge c, then a>2b.a \gt 2b.)

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Solution

Order the values as x1>x2>x3x_1 \gt x_2 \gt x_3; the joint density of the order statistics is 66 on this region. The event is x1>2x2.x_1 \gt 2x_2. Integrating x3x_3 from 00 to x2x_2 contributes a factor of x2.x_2. Then P=6012x22x21dx1dx2=6012x2(12x2)dx2. \begin{aligned} P &= 6\int_0^{\frac{1}{2}} x_2\int_{2x_2}^{1} dx_1\, dx_2 \\ &= 6\int_0^{\frac{1}{2}} x_2(1 - 2x_2)\, dx_2. \end{aligned} This equals 6(18112)=6124=14.6\left(\dfrac{1}{8} - \dfrac{1}{12}\right) = 6 \cdot \dfrac{1}{24} = \dfrac{1}{4}. Thus, the correct answer is E.

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Concepts: geometric probability · calculus

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.