Counting & ProbabilityDifficulty 30-85123 tagged problems

Geometric Probability

Geometric Probability is the habit of computing a probability as a ratio of lengths, areas, or volumes. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.

Practice this technique Match my level

The Key Result

$$P(\text{event}) = \frac{\text{measure of favorable region}}{\text{measure of whole sample space}}$$

When outcomes are spread uniformly over a length, area, or volume, the probability of an event is the size of its favorable region divided by the size of the whole region.

Why It Works

1
Model the random outcome as a point chosen uniformly from a region $\Omega$ (an interval, a square, a solid), so every sub-region is equally likely per unit of measure.
2
Uniformity means probability is proportional to measure (length, area, or volume), with the constant fixed by $P(\Omega) = 1$.
3
Hence for a favorable sub-region $F \subseteq \Omega$, $P(F) = \dfrac{\operatorname{measure}(F)}{\operatorname{measure}(\Omega)}$.
4
The whole task reduces to drawing the region, shading the favorable part, and computing the two measures.

Worked Examples

Example 1

Two friends each arrive at a cafe at a uniformly random time between 3:00 and 4:00, independently. Each waits $15$ minutes for the other. What is the probability they meet?

1
Let $x, y \in [0, 60]$ be their arrival minutes; the sample space is the $60 \times 60$ square of area $60^2 = 3600$.
2
They meet exactly when $|x - y| \le 15$, the band between the lines $y = x + 15$ and $y = x - 15$.
3
The two corner triangles where they miss ($|x-y| > 15$) are right triangles each with legs $60 - 15 = 45$, so each has area $\tfrac{1}{2}\cdot 45^2$, and the two together total $45^2 = 2025$.
4
Favorable area $= 3600 - 2025 = 1575$, so $P(\text{meet}) = \dfrac{1575}{3600} = \dfrac{7}{16}$.

Answer:

$\dfrac{7}{16}$

Warm-up

A point is chosen uniformly at random on the segment $[0, 10]$. What is the probability it lands within distance $2$ of the midpoint $5$?

1
The sample space is the interval $[0, 10]$ with total length $10$.
2
'Within distance $2$ of $5$' is the sub-interval $[3, 7]$, of length $4$.
3
Probability $= \dfrac{\text{favorable length}}{\text{total length}} = \dfrac{4}{10} = \dfrac{2}{5}$.

Answer:

$\dfrac{2}{5}$

Contest level

Two numbers $x, y$ are chosen independently and uniformly at random from $[0, 1]$. What is the probability that $x + y \le 1$ AND $x \le 2y$?

1
The sample space is the unit square $[0,1]^2$ of area $1$. Shade the region satisfying both inequalities.
2
$x + y \le 1$ is the lower-left triangle below the line $y = 1 - x$. Within it, impose $x \le 2y$, i.e. $y \ge x/2$ (above the line $y = x/2$).
3
The two lines $y = 1 - x$ and $y = x/2$ meet where $1 - x = x/2 \Rightarrow x = \tfrac{2}{3}, y = \tfrac{1}{3}$. The favorable region is the triangle with vertices $(0,0)$, $(0,1)$, and $(\tfrac{2}{3}, \tfrac{1}{3})$.
4
Its area (base along the $y$-axis of length $1$, horizontal 'height' $\tfrac{2}{3}$) is $\tfrac{1}{2} \cdot 1 \cdot \tfrac{2}{3} = \tfrac{1}{3}$, so the probability is $\tfrac{1}{3}$.

Answer:

$\dfrac{1}{3}$

Olympiad / Challenge

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

1
Let the two break points be $x, y$, uniform on $[0,1]^2$ (area $1$). The three piece lengths must each be less than $\tfrac{1}{2}$ for the triangle inequality to hold (the longest piece must be under half the total).
2
Consider the case $x < y$ (area $\tfrac{1}{2}$); the pieces are $x$, $y - x$, $1 - y$. The triangle conditions become $x < \tfrac{1}{2}$, $y - x < \tfrac{1}{2}$, and $1 - y < \tfrac{1}{2}$ (i.e. $y > \tfrac{1}{2}$).
3
In the triangle $\{0 < x < y < 1\}$, the favorable sub-region is $x < \tfrac12,\ y > \tfrac12,\ y - x < \tfrac12$, a smaller triangle of area $\tfrac{1}{8}$ (it has vertices $(0,\tfrac12), (\tfrac12,\tfrac12), (\tfrac12,1)$).
4
By symmetry the case $x > y$ contributes equally, so the total favorable area is $2 \cdot \tfrac{1}{8} = \tfrac{1}{4}$, and the probability is $\tfrac{1}{4}$.

Answer:

$\dfrac{1}{4}$

Going Deeper

The framework $P = \frac{\text{measure of favorable}}{\text{measure of whole}}$ works in any dimension and with any uniform measure (length, area, volume); the hard part is correctly DRAWING the favorable region from the constraints and computing its measure with geometry or integration. Conditions like $|x - y| \le d$, $x + y \le c$, or $\max < \tfrac{1}{2}$ all translate into lines/planes cutting the sample space.
Geometric probability appears on AMC/AIME as 'two people arrive at random times,' 'break a stick,' 'random points in a square/triangle,' and 'random chord' problems. It is also the continuous limit of discrete uniform counting — the same favorable-over-total ratio, with area replacing a count.
Pitfall: the answer depends entirely on WHICH quantity is chosen uniformly — Bertrand's paradox shows a 'random chord' has different probabilities under different (all reasonable) uniform models. Pin down the sample space and its measure precisely before computing, and watch boundaries (a measure-zero edge does not change a continuous probability, but a wrong region does).

Spot the Signal

Use it when outcomes are continuous and the favorable set is a region.
You can describe the hard part as computing a probability as a ratio of lengths, areas, or volumes, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by modeling the sample space as a region and measuring the favorable part of it.
Name the relation that makes Geometric Probability legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Choosing the wrong dimension for the sample space, or mismeasuring the favorable region.
Applying Geometric Probability because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Draw the sample space as a region; the probability is just the favorable fraction of it.

Try it on:

Find a probability that reduces to a ratio of geometric measures.

Back to techniques

Counting & ProbabilityDifficulty 30-85123 tagged problems

Geometric Probability

Practice this technique Match my level

The Key Result

$$P(\text{event}) = \frac{\text{measure of favorable region}}{\text{measure of whole sample space}}$$

When outcomes are spread uniformly over a length, area, or volume, the probability of an event is the size of its favorable region divided by the size of the whole region.

Why It Works

1
Model the random outcome as a point chosen uniformly from a region $\Omega$ (an interval, a square, a solid), so every sub-region is equally likely per unit of measure.
2
Uniformity means probability is proportional to measure (length, area, or volume), with the constant fixed by $P(\Omega) = 1$.
3
Hence for a favorable sub-region $F \subseteq \Omega$, $P(F) = \dfrac{\operatorname{measure}(F)}{\operatorname{measure}(\Omega)}$.
4
The whole task reduces to drawing the region, shading the favorable part, and computing the two measures.

Worked Examples

Example 1

Two friends each arrive at a cafe at a uniformly random time between 3:00 and 4:00, independently. Each waits $15$ minutes for the other. What is the probability they meet?

1
Let $x, y \in [0, 60]$ be their arrival minutes; the sample space is the $60 \times 60$ square of area $60^2 = 3600$.
2
They meet exactly when $|x - y| \le 15$, the band between the lines $y = x + 15$ and $y = x - 15$.
3
The two corner triangles where they miss ($|x-y| > 15$) are right triangles each with legs $60 - 15 = 45$, so each has area $\tfrac{1}{2}\cdot 45^2$, and the two together total $45^2 = 2025$.
4
Favorable area $= 3600 - 2025 = 1575$, so $P(\text{meet}) = \dfrac{1575}{3600} = \dfrac{7}{16}$.

Answer:

$\dfrac{7}{16}$

Warm-up

A point is chosen uniformly at random on the segment $[0, 10]$. What is the probability it lands within distance $2$ of the midpoint $5$?

1
The sample space is the interval $[0, 10]$ with total length $10$.
2
'Within distance $2$ of $5$' is the sub-interval $[3, 7]$, of length $4$.
3
Probability $= \dfrac{\text{favorable length}}{\text{total length}} = \dfrac{4}{10} = \dfrac{2}{5}$.

Answer:

$\dfrac{2}{5}$

Contest level

Two numbers $x, y$ are chosen independently and uniformly at random from $[0, 1]$. What is the probability that $x + y \le 1$ AND $x \le 2y$?

1
The sample space is the unit square $[0,1]^2$ of area $1$. Shade the region satisfying both inequalities.
2
$x + y \le 1$ is the lower-left triangle below the line $y = 1 - x$. Within it, impose $x \le 2y$, i.e. $y \ge x/2$ (above the line $y = x/2$).
3
The two lines $y = 1 - x$ and $y = x/2$ meet where $1 - x = x/2 \Rightarrow x = \tfrac{2}{3}, y = \tfrac{1}{3}$. The favorable region is the triangle with vertices $(0,0)$, $(0,1)$, and $(\tfrac{2}{3}, \tfrac{1}{3})$.
4
Its area (base along the $y$-axis of length $1$, horizontal 'height' $\tfrac{2}{3}$) is $\tfrac{1}{2} \cdot 1 \cdot \tfrac{2}{3} = \tfrac{1}{3}$, so the probability is $\tfrac{1}{3}$.

Answer:

$\dfrac{1}{3}$

Olympiad / Challenge

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

1
Let the two break points be $x, y$, uniform on $[0,1]^2$ (area $1$). The three piece lengths must each be less than $\tfrac{1}{2}$ for the triangle inequality to hold (the longest piece must be under half the total).
2
Consider the case $x < y$ (area $\tfrac{1}{2}$); the pieces are $x$, $y - x$, $1 - y$. The triangle conditions become $x < \tfrac{1}{2}$, $y - x < \tfrac{1}{2}$, and $1 - y < \tfrac{1}{2}$ (i.e. $y > \tfrac{1}{2}$).
3
In the triangle $\{0 < x < y < 1\}$, the favorable sub-region is $x < \tfrac12,\ y > \tfrac12,\ y - x < \tfrac12$, a smaller triangle of area $\tfrac{1}{8}$ (it has vertices $(0,\tfrac12), (\tfrac12,\tfrac12), (\tfrac12,1)$).
4
By symmetry the case $x > y$ contributes equally, so the total favorable area is $2 \cdot \tfrac{1}{8} = \tfrac{1}{4}$, and the probability is $\tfrac{1}{4}$.

Answer:

$\dfrac{1}{4}$

Going Deeper

The framework $P = \frac{\text{measure of favorable}}{\text{measure of whole}}$ works in any dimension and with any uniform measure (length, area, volume); the hard part is correctly DRAWING the favorable region from the constraints and computing its measure with geometry or integration. Conditions like $|x - y| \le d$, $x + y \le c$, or $\max < \tfrac{1}{2}$ all translate into lines/planes cutting the sample space.
Geometric probability appears on AMC/AIME as 'two people arrive at random times,' 'break a stick,' 'random points in a square/triangle,' and 'random chord' problems. It is also the continuous limit of discrete uniform counting — the same favorable-over-total ratio, with area replacing a count.
Pitfall: the answer depends entirely on WHICH quantity is chosen uniformly — Bertrand's paradox shows a 'random chord' has different probabilities under different (all reasonable) uniform models. Pin down the sample space and its measure precisely before computing, and watch boundaries (a measure-zero edge does not change a continuous probability, but a wrong region does).

Spot the Signal

Use it when outcomes are continuous and the favorable set is a region.
You can describe the hard part as computing a probability as a ratio of lengths, areas, or volumes, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by modeling the sample space as a region and measuring the favorable part of it.
Name the relation that makes Geometric Probability legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Choosing the wrong dimension for the sample space, or mismeasuring the favorable region.
Applying Geometric Probability because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Draw the sample space as a region; the probability is just the favorable fraction of it.

Try it on:

Find a probability that reduces to a ratio of geometric measures.