Geometric Probability

In classical probability, we often count distinct, separate outcomes (like rolling a die or flipping a coin). However, when outcomes are continuous—such as choosing a random point on a line or throwing a dart at a board—we use geometric probability.

Geometric probability uses measurements like length, area, or volume to find the likelihood of an event. The fundamental formula is:

$P(E) = \frac{\text{Measure of the Favorable Region}}{\text{Measure of the Total Region}}$

Depending on the problem, the "measure" could be a 1D length, a 2D area, or a 3D volume.

1D Geometric Probability (Lengths)

If a point is chosen at random on a line segment, the probability that it falls within a specific smaller segment is the ratio of their lengths.

Example: A piece of string is 10 cm long. If you cut it at a completely random point, what is the probability that the cut is made within 2 cm of either end?

• Total length: $10$ cm.
• Favorable length: The first 2 cm and the last 2 cm. Total favorable length $= 2 + 2 = 4$ cm.
• Probability: $P = \frac{4}{10} = \frac{2}{5}$.

2D Geometric Probability (Areas)

When dealing with flat surfaces, probability is the ratio of the favorable area to the total area.

Example 1: The Dartboard

A dart is thrown randomly at a circular board of radius 10. Assuming the dart always hits the board, find the probability it lands within 3 units of the center.

1. Total Region: The entire dartboard is a circle with radius $R = 10$. $\text{Total Area} = \pi R^2 = \pi (10)^2 = 100\pi$
2. Favorable Region: A smaller circle in the center with radius $r = 3$. $\text{Favorable Area} = \pi r^2 = \pi (3)^2 = 9\pi$
3. Probability: $P = \frac{9\pi}{100\pi} = \frac{9}{100} = 0.09$

Example 2: The Meeting Problem

Two people agree to meet at a cafe between 12:00 and 1:00. Each person will arrive at a random time within that hour and wait exactly 15 minutes for the other person. If the other person does not arrive, they leave. What is the probability that they meet?

This classic problem is best solved by graphing the times as coordinates $(x, y)$ on a 2D plane.

1. Let $x$ be the arrival time of Person A (in minutes past 12:00), and $y$ be the arrival time of Person B.
2. Total Region: Both $x$ and $y$ are between $0$ and $60$. This forms a square on a graph. $\text{Total Area} = 60 \times 60 = 3600$
3. Favorable Region: They meet if the difference between their arrival times is 15 minutes or less. Mathematically, this is $|x - y| \le 15$, which gives two boundary lines: $y \le x + 15$ and $y \ge x - 15$.
4. Calculating the Area: Instead of finding the complex middle shape directly, find the area of the two "missed connection" triangles in the top-left and bottom-right corners of the square.
   • Each triangle has a base and height of $60 - 15 = 45$.
   • Area of one triangle $= \frac{1}{2} \times 45 \times 45 = 1012.5$.
   • Total area of both triangles $= 2025$.
5. Favorable Area: $\text{Favorable Area} = 3600 - 2025 = 1575$
6. Probability: $P = \frac{1575}{3600} = \frac{7}{16}$

There is a $\frac{7}{16}$ (or $43.75\%$) chance that the two people will successfully meet.