# Ideas in Geometry/Instructive examples/Section 3.2 Problem 42 - Lattice Area, Pick's Theorem, and Probability

## Terms, Theorems, and Components: edit

### Lattice edit

The plane of black dots on which the objects are drawn is referred to as a lattice or Lattice (group). A lattice consists of a set of dots which lie in perfect rows and have a distance of 1 to any dot directly above, below, left, or right of itself. The lattice acts as a guide for drawing objects.

### Pick's Theorem edit

Pick’s theorem is a tool used to determine the area of any object constructed on the lattice.

The formula looks like this: A=b/2+n-1

Where:

A = Area (in units^2)

b = The number of lattice points on the border of the object.

n = The number of lattice points inside the object (do not include border points)

### Regions edit

Small region: The red-ish area inside the larger region

Large region: The pink/purple-ish area inside formed on the lattice.

## Finding the probability of selecting a random point within the large region which is also contained in the small region: edit

To do this, I have selected the following method:

1. Find the area of the larger region using Pick’s Theorem.

2. Find the area of the smaller region using Pick’s Theorem.

3. Create a ratio of the Area of the smaller region and the larger region.

4. Determine the probability of the likelihood a random point will fall in the smaller region and the larger region.

### Step 1: Find the Area of the Larger Region edit

By simple counting we can determine that there are 9 border points and 9 internal points.

Now we plug them into Pick’s Theorem:

A = 9/2+9-1

A = 4.5 + 9 – 1

A = 12.5 units^2

### Step 2: Find the Area of the Smaller Region edit

By counting we can determine that there are 5 border points and 0 internal points.

Now for Pick’s Theorem:

A = 5/2 + 0 – 1

A = 2.5 – 1

A = 1.5 units^2

### Step 3: Find the ratio of the two regions edit

From Pick’s Theorem, we have determined that the Area of the larger region is 12.5 units^2 and that the Area of the smaller region is 1.5 units^2. This gives us a ratio of 1.5/12.5. We are comparing the amount of space in the smaller portion to the amount of area in the larger region. This helps us compare the area in a simple way.

Step 4: Find the probability of selecting a random point that is in the large region and the small region===

From our ratio, we found that of 12.5 units^2 in the larger region, 1.5 units^2 belong to both the larger region and the smaller region. To determine the probability, we need to compare the chances of picking a point in the smaller region with the availability of the entire region.

Because 1.5 units^2 are contained in the 12.5 units^2, we can say that when selecting a point in the larger area of 12.5, there is a chance we will pick a point in the smaller region of 1.5. So there is a 1.5 chance in 12.5 of selecting a point in the smaller region. Our initial probability is 1.5/12.5. However, mathematicians rarely like decimals contained in fractions for a final answer, so we can increase the numbers by dividing by a common factor of .5. After doing this, we arrive at our final probability of 3/25.

## Final Answer: edit

In sum, the chance of selecting a point in the larger region that is also in the smaller region is 3/25.