Math Core

When a number is chosen "uniformly at random" from an interval, there are infinitely many outcomes, so you can't count them. Instead you measure them: probability becomes a ratio of lengths, areas, or volumes. The trick in almost every AMC problem is to turn the random choices into a point in a square or box, draw the region where the event happens, and compute its area.

From choices to a picture

One number chosen uniformly from $[a, b]$ : a point on a segment. Probability is length over length.
Two numbers chosen independently and uniformly: a point in a rectangle. Probability is area over area.
Three numbers: a point in a box. Probability is volume over volume.

Geometric probability

If a point is chosen uniformly at random in a region $R$ , then the probability it lands in a subregion $S$ is

$P = \frac{\text{area of } S}{\text{area of } R}$

(or length, or volume). For two independent uniform numbers $x$ and $y$ , the point $(x, y)$ is uniform in the rectangle of possible pairs.

A single point, or a line, has zero area. So in geometric probability " $\lt$ " and " $\le$ " give the same answer, and ties have probability $0$ .

Worked example: A corner cut off

Numbers $x$ and $y$ are chosen independently and uniformly from $[0, 1]$ . What is the probability that $x + y \lt \tfrac43$ ?

The point $(x, y)$ is uniform in the unit square. The bad region, $x + y \ge \tfrac43$ , is the triangle in the top-right corner with legs $\tfrac23$ (from $(\tfrac13, 1)$ to $(1, 1)$ to $(1, \tfrac13)$ ). Its area is $\tfrac12 \cdot \tfrac23 \cdot \tfrac23 = \tfrac29$ . So the probability is $1 - \tfrac29 = \tfrac79$ .

Worked example: Meeting at the café

Two friends each arrive at a café at a uniformly random time between noon and 1:00 PM, independently. Whoever arrives first waits $15$ minutes and then leaves. What is the probability that they meet?

Let $x$ and $y$ be the arrival times in minutes after noon. They meet when $|x - y| \le 15$ . This is a band around the diagonal of the $60 \times 60$ square.

Arrival times (x, y). The shaded band |x − y| ≤ 15 is where the friends meet.Open in grapher →

The two unshaded corner triangles each have legs $45$ , so together they have area $45^2 = 2025$ . The square has area $3600$ . The probability they meet is

$1 - \frac{2025}{3600} = 1 - \frac{9}{16} = \frac{7}{16}.$

Worked example: Breaking a stick

A stick of length $1$ is broken at two points chosen independently and uniformly at random. What is the probability that the three pieces can form a triangle?

Three pieces with total length $1$ form a triangle exactly when every piece is shorter than $\tfrac12$ . (If one piece is at least $\tfrac12$ , it's at least as long as the other two combined.)

Let the break points be $x$ and $y$ . By symmetry, consider only $x \lt y$ (half of the square, area $\tfrac12$ ). The pieces are $x$ , $y - x$ , $1 - y$ , and we need $x \lt \tfrac12$ , $y \gt \tfrac12$ , and $y - x \lt \tfrac12$ . That's the triangle with vertices $(0, \tfrac12)$ , $(\tfrac12, \tfrac12)$ , $(\tfrac12, 1)$ , with area $\tfrac18$ . So the probability is $\dfrac{1/8}{1/2} = \dfrac14$ .

Common mistake

"Uniformly at random" must refer to the thing you're measuring. A random point on a segment is not the same as a random angle, and a random chord of a circle depends on how it's chosen. Read the problem carefully and set up coordinates for exactly the quantities that are chosen uniformly.

Worked example: A coin on a grid

A floor is tiled with $5 \times 5$ squares. A coin of radius $1$ is tossed so that its center lands uniformly at random. What is the probability that the coin lies entirely inside one tile?

Look at the tile containing the center. The coin stays inside when its center is at least $1$ unit from every edge, so the center must lie in the central $3 \times 3$ square. The probability is $\dfrac{9}{25}$ .

Tip

For problems with three uniform numbers, the unit cube has volume $1$ , and the region $x + y + z \le 1$ with $x, y, z \ge 0$ is a pyramid with volume $\tfrac16$ . Many 3D questions reduce to cutting off corners like this one.

Practice

Practice 1

A point is chosen uniformly at random inside a unit square. What is the probability that it is less than $\tfrac14$ unit from the nearest side of the square?

Practice 2

A point $(x, y)$ is chosen uniformly at random from the rectangle with $0 \le x \le 4$ and $0 \le y \le 3$ . What is the probability that $x \lt y$ ? Give your answer as a fraction.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

A point is chosen uniformly at random inside a square with side length $2$ . What is the probability that it is within $1$ unit of the center of the square?

Practice 4

A real number $x$ is chosen uniformly at random from $[0, 10]$ . What is the probability that $\lfloor \sqrt{x} \rfloor$ is even?

Practice 5

A point is chosen uniformly at random inside the triangle with vertices $(0, 0)$ , $(6, 0)$ and $(0, 6)$ . What is the probability that its $x$ -coordinate is less than $2$ ? Give your answer as a fraction.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

A stick of length $1$ is broken at two points chosen independently and uniformly at random. What is the probability that all three pieces are longer than $\tfrac15$ ?

Practice 7

Numbers $b$ and $c$ are chosen independently and uniformly from $[0, 1]$ . What is the probability that the equation $x^2 + bx + c = 0$ has real solutions?

Practice 8

Two points are chosen independently and uniformly at random on a circle of radius $1$ . What is the probability that the chord joining them is longer than $1$ ?

Practice 9

Numbers $x$ and $y$ are chosen independently and uniformly from $[0, 2]$ . What is the probability that $xy \le 1$ ?

Real contest practice

2017 AMC 12A, Problem 10: two people choose numbers from different intervals; which one is larger?
2016 AMC 12A, Problem 23: three random numbers as side lengths of a triangle, a volume computation in the unit cube.