Ideas in Geometry/Instructive examples/Section 3.2 Problem 42 - Lattice Area, Pick's Theorem, and Probability
Terms, Theorems, and Components: 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
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)
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.