2025 AMC 12A Problem 22

Question

Three real numbers are chosen independently and uniformly at random between

0

and

1.

What is the probability that the greatest of these three numbers is greater than

2

times each of the other two numbers? (In other words, if the chosen numbers are

a \ge b \ge c,

then

a \gt 2b.

)

\dfrac{1}{12}

\dfrac{1}{9}

\dfrac{1}{8}

\dfrac{1}{6}

\dfrac{1}{4}

With order statistics

x_1 \ge x_2 \ge x_3,

the joint density is

6

on the region

x_1 \gt x_2 \gt x_3

Integrate over

x_3 \in (0, x_2),

x_1 \in (2x_2, 1),

and

x_2 \in (0, \tfrac{1}{2})

Order the values as

x_1 \gt x_2 \gt x_3

; the joint density of the order statistics is

6

on this region. The event is

x_1 \gt 2x_2.

Integrating

x_3

from

0

to

x_2

contributes a factor of

x_2.

Then

\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\left(\dfrac{1}{8} - \dfrac{1}{12}\right) = 6 \cdot \dfrac{1}{24} = \dfrac{1}{4}.

Thus, the correct answer is E.

Problem 22 in other years