Geometric Probability

What you need to know

• Probability as a ratio of sizes. When a point is picked at random from a region (every part equally likely), the chance it lands in a part $S$ is $\dfrac{\text{size of } S}{\text{size of the whole region}}$. "Size" means length on a line, area in the plane, or volume in space. Example: a random number between $0$ and $10$ is less than $3$ with probability $\frac{3}{10}$.
• Two random numbers make one random point. Picking $x$ and $y$ at random between $0$ and $L$, independently (neither affects the other), is the same as picking one random point $(x,y)$ in an $L\times L$ square. A condition such as $|x-y|<d$ becomes a region cut out by straight lines.
• Two regions to know. The band $|x-y|\le d$ inside the $L\times L$ square has area $L^2-(L-d)^2$ (for $d\le L$): the square minus two corner triangles, each with legs $L-d$. Example: two friends arrive at random times $x$ and $y$ (in minutes) within one hour, and each waits $15$ minutes. They meet when $|x-y|\le15$, which has probability $\frac{60^2-45^2}{60^2} = \frac{1575}{3600} = \frac{7}{16}$. Also, $x+y\le c$ with $c\le L$ is a right triangle with legs $c$, so its area is $\frac{c^2}{2}$. The worked example uses both regions.
• Dropped coins. A coin of radius $r$ misses every tile edge exactly when its center is at least $r$ from every edge, that is, inside the tile shrunk by $r$ on every side. Example: with $10\times10$ tiles and a coin of radius $1$, the center must land in the middle $8\times8$ square: probability $\frac{64}{100} = \frac{16}{25}$.

How AMC 10 tests it

• Problems 10–18: two people arrive at random times within an hour and each waits $15$ minutes; find the probability they meet (the square with a band along the diagonal).
• Problems 12–20: a random point in a square, circle, or triangle; the probability it is closer to one side, or inside a related shape.
• Problems 15–22: two random points split a segment into three pieces; the probability that the pieces can form a triangle.
• Problems 20–24: a coin dropped on a tiled floor; the probability it does not touch a line.

Standard approaches

1. Name the random numbers and the values each can take. That fixes the whole region.
2. Turn the event into inequalities and sketch the region where it happens.
3. Find that area by cutting it into simple pieces, or by subtracting what you do not want.
4. For dropped objects, shrink the allowed region by the radius and divide the areas.
5. Test extreme cases: pushing the numbers to their limits should push the answer to $1$ or $0$. Example: friends who each wait all $60$ minutes surely meet, and the formula gives $\frac{60^2-0^2}{60^2} = 1$.

Worked example

Real numbers $x$ and $y$ are chosen independently and uniformly at random from $[0, 6]$. What is the probability that $|x - y| \le 2$ and $x + y \ge 4$?

(A) $\frac{1}{3}$ (B) $\frac{7}{18}$ (C) $\frac{4}{9}$ (D) $\frac{1}{2}$ (E) $\frac{5}{9}$

Solution. Here $[0,6]$ means all real numbers from $0$ to $6$, and "uniformly" means every part of it is equally likely.

Step 1: the whole region. The point $(x,y)$ is random in the $6\times6$ square, which has area $36$.

Step 2: the band. $|x-y|\le2$ is the strip between the lines $y = x-2$ and $y = x+2$. Outside it are two corner triangles with legs $6-2 = 4$, each of area $\frac12\cdot4^2 = 8$. So the band has area $36-2\cdot8 = 20$.

Step 3: remove $x+y<4$. The region $x+y<4$ is the triangle with corners $(0,0)$, $(4,0)$, $(0,4)$, of area $\frac12\cdot4\cdot4 = 8$. Its tip near $(4,0)$, below the line $y = x-2$, is outside the band: the triangle with corners $(2,0)$, $(4,0)$, $(3,1)$, with base $2$, height $1$, and area $1$. The tip near $(0,4)$ is its mirror image, also area $1$. So the band loses $8-1-1 = 6$.

Step 4: divide. The region we want has area $20-6 = 14$, so the probability is $\frac{14}{36} = \frac{7}{18}$.

The answer is $\boxed{\textbf{(B)}\ \frac{7}{18}}$.

Pitfalls

• Using lengths when the event depends on two independent random numbers. It must be an area in the square.
• Forgetting to square when scaling. Shrinking every side by a factor $k$ shrinks the area by $k^2$: halving the sides of a square leaves $\frac14$ of the area, not $\frac12$.
• Drawing the band $|x-y|\le d$ with corner triangles whose legs are $d$. The legs are $L-d$ (in the worked example, $4$, not $2$).
• Shrinking the tile by the coin's diameter instead of its radius. Above, the safe square is $8\times8$, not $6\times6$.

Traps that recur

• Handling only the case BP < AP < AB and missing the sliver near D where AB < AP < BP, which leaves pi/6 - sqrt3/8 and cannot even be written in the required form. (2025 AMC 10A #25)
• Treating the three pairwise conditions as independent, or extrapolating the two-variable answer ((n-1)/n)^2, which gives a formula with n-1 instead of n-2. (2012 AMC 10A #25)
• Thinking the third step is skipped or truncated when time runs out (it is completed), or coupling the step lengths to the times. (2022 AMC 10B #23)
• Treating the corner triangles like the diamond, adding quarter-circles or strips along their legs, which lie outside the region the center can occupy. (2021 AMC 10B #23)