Lesson 120: Using Geometric Formulas to Find the Probability of an Event

New Concept

The same definition applies to geometric probability when working with the area of geometric shapes.

What’s next

Next, you’ll apply area formulas for rectangles and circles to calculate the probability of events in various geometric scenarios.

Property

To find the probability of an event using geometry, use the ratio of the favorable area to the total possible area:

Explanation

Imagine a bird randomly landing in a big garden. Its chance of hitting the tomato patch is the patch's area divided by the whole garden's area. Geometric probability simply compares the 'favorable' space to the 'total' space to find the odds. It’s a space race, but for points, not planets!

Examples

• A 20 ft by 10 ft pool has a 5 ft by 4 ft raft. The probability a ball hits the raft is
  $$P(\text{hit}) = \frac{5 \cdot 4}{20 \cdot 10} = \frac{20}{200} = \frac{1}{10}$$
  .
• A target has a 10-inch radius outer circle and a 2-inch radius inner circle. The probability of hitting the inner circle is
  $$P(\text{inner}) = \frac{\pi(2)^2}{\pi(10)^2} = \frac{4\pi}{100\pi} = \frac{1}{25}$$
  .

A rectangular field is 20 feet by 30 feet, with a small sandbox inside that is 4 feet by 5 feet. Let's find the chance a randomly thrown toy lands in the sandbox.

Example Problem

What is the probability a toy lands in the sandbox area of the rectangular field?

Step-by-Step

1. The probability is the ratio of the favorable area (the sandbox) to the total area (the entire field).

2. Use the formula $A = l \cdot w$ to find the area of each rectangle.

3. Multiply to find the areas.

4. Simplify the fraction to find the final probability.

Property

The complement of an event is all the outcomes in the sample space that are not included in the event.

Explanation

Why calculate the tricky part when you can calculate the easy part and just subtract? To find the odds of not hitting a target, find the probability of hitting it and subtract that from 1. It’s a clever shortcut that saves you from messy calculations and makes you look like a probability wizard!

Examples

• A circular town (radius 20 miles) has a square park (side 4 miles). The probability a raindrop misses the park is
  $$\operatorname{P}(\text{miss}) = 1 - \frac{4^2}{\pi(20)^2} = 1 - \frac{16}{400\pi} \approx 0.987$$
  .
• A square wall (side 8 ft) has a triangular target (base 3 ft, height 2 ft). The probability of missing it is
  $$\operatorname{P}(\text{miss}) = 1 - \frac{\frac{1}{2} \cdot 3 \cdot 2}{8^2} = 1 - \frac{3}{64} = \frac{61}{64}$$
  .

Sometimes it's easier to calculate the probability of an event not happening. Let's find the chance a dart misses the bullseye on a target.

Example Problem

A target consists of two concentric circles. The outer circle has a radius of 10 inches, and the inner bullseye has a radius of 2 inches. What is the probability a dart that hits the target does not land in the bullseye?

Step-by-Step

1. We will find the probability of the complement event. The key idea here is that the probability of not hitting the bullseye is $1$ minus the probability of hitting it.

2. First, find the probability of hitting the bullseye, which is the ratio of the bullseye's area to the total target area.

3. Use the area formula for a circle, $A = \pi r^2$.

4. Simplify the powers and the fraction. The $\pi$ terms cancel out.

5. Now, subtract this probability from $1$ to find the probability of not hitting the bullseye.

Property

When finding the ratio of the areas of two circles, leave the areas in terms of $\pi$ to make simplification easier.

Explanation

Think of $\pi$ as a tagalong in your calculations. By keeping it in your numerator and denominator, you can cancel it out instantly, just like a common factor. This trick avoids messy decimals and simplifies your fraction-crunching life. Work smarter, not harder, and let pi cancel itself out of the equation for a clean finish!

Examples

• A target has a 15-inch radius outer circle and a 5-inch radius inner circle. The probability of hitting the inner circle is
  $$\operatorname{P}(\text{inner}) = \frac{\pi(5)^2}{\pi(15)^2} = \frac{25\pi}{225\pi} = \frac{1}{9}$$
  .
• A school zone has a 3-mile radius. Students within a 1-mile radius can walk. The probability a student is a walker is
  $$\operatorname{P}(\text{walk}) = \frac{\pi(1)^2}{\pi(3)^2} = \frac{1\pi}{9\pi} = \frac{1}{9}$$
  .