Question

1. from a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. what is the probability that at least one of the sides of the triangle is also a side of the octagon?

Explanation:

Step1: Calculate the total number of triangles

The number of ways to choose 3 vertices out of \(n = 8\) vertices of an octagon is given by the combination formula \(C(n,k)=\frac{n!}{k!(n - k)!}\). Here, \(n=8\) and \(k = 3\), so \(C(8,3)=\frac{8!}{3!(8 - 3)!}=\frac{8\times7\times6}{3\times2\times1}=56\).

Step2: Calculate the number of triangles with no - side common with the octagon

We use the formula for the number of triangles formed in a \(n\) - sided polygon with no - side common with the polygon. The number of triangles with no - side common with an \(n\) - sided polygon is \(C(n,k)-n\times(k - 2)\). For \(n = 8\) and \(k=3\), first, we can also calculate it directly.
Let's count the number of triangles with no - side common.
We fix one vertex. Then, we cannot choose its two adjacent vertices. So, for each vertex, the number of non - adjacent vertices to form a triangle with no - side common: If we fix a vertex \(V\), we need to choose 2 vertices from the \(n-4\) non - adjacent vertices. But we will double - count.
The number of triangles with no - side common with the octagon: \(C(8,3)-8\times6=56 - 48 = 8\) (Another way: Let the vertices of the octagon be \(A_1,A_2,\cdots,A_8\). To form a triangle with no - side common, we use the following approach.
We can represent the selection of vertices as \(A_i,A_j,A_k\) such that \(j
eq i + 1,i-1\) and \(k
eq j + 1,j - 1,k
eq i + 1,i-1\) (modulo 8).
The number of triangles with no - side common is \(8\))

Step3: Calculate the probability of the complement event

The probability \(P(\text{no - side common})=\frac{8}{56}=\frac{1}{7}\)

Step4: Calculate the required probability

Using the formula \(P(\text{at least one side common})=1 - P(\text{no - side common})\)
\(P = 1-\frac{1}{7}=\frac{6}{7}\)

Answer:

\(\frac{6}{7}\)