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 $a \gt 2b.$ )

$\dfrac{1}{12}$

$\dfrac{1}{9}$

$\dfrac{1}{8}$

$\dfrac{1}{6}$

$\dfrac{1}{4}$

Solution:

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 $P = 6\int_0^{1/2} x_2\int_{2x_2}^{1} dx_1\, dx_2 = 6\int_0^{1/2} x_2(1 - 2x_2)\, dx_2.$

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.

2025 AMC 12A Problem 22

Solution:

Problem 22 in Other Years