Question

Consider a 9 sided polygon, then, the probability that if three vertices are selected at random, the triangle so formed will contain the centre of the polygon is \[\dfrac{k}{14}\] the value of \[k\] is

Answer 1

Hint: We solve this problem by considering the possibilities for a fixed vertex.That is we assume any one vertex of the polygon as the vertex of the required triangle.Here, have the condition that the gap between the other two vertices must be at most 4 that is the gap should be 4 or less than 4.
This is because in order to get the triangle containing the centre of polygon the triangle should be acute angled triangle which gives that the gap should be at most 4
We have the formula of probability as
\[P=\dfrac{\text{number of possible outcomes}}{\text{total number of outcomes}}\]
We are given with a 9 sided polygon.

Complete answer:
Let us assume that the given polygon as follows

Now, let us assume that one vertex of the required triangle as 0
Now, let us find the number of triangles by considering the number of pairs of remaining sides of the triangle.
Let us assume that the remaining vertices of the triangle as \[a,b\] where, \[a,b\] represents the vertex number
We are asked to find the number of triangles that contain the centre of the polygon.
We know that for a 9 sided polygon the gap between the other two vertices must be at most 4, that is the gap should be 4 or less than 4 in order to get the centre of the polygon in the triangle.
Here, we considered the first vertex as 0
By using the above condition the possible vertices of \[a\] are 1, 2, 3 and 4
Now, let us take the possibilities of \[b\] for each vertex of \[a\]
(1) For \[a=1\] the possible vertex of \[b\] is 5
(2) For \[a=2\] the possible vertex of \[b\] is 5 or 6
(3) For \[a=3\] the possible vertex of \[b\] is 5 or 6 or 7
(4) For \[a=4\] the possible vertex of \[b\] is 5 or 6 or 7 or 8
Here, we can see that there are total of 10 triangles that contain the centre of polygon in the triangle
Let us assume that the number of total number of outcomes as \[n\] then we get
\[\Rightarrow n=10\]
Let us assume that the total number of outcomes as \[N\]
Here, we can see that we have already selected one vertex as 0 then the number of ways of selecting the remaining 2 vertices from 8 vertices is given as
\[\begin{align}
  & \Rightarrow N={}^{8}{{C}_{2}} \\
 & \Rightarrow N=\dfrac{8!}{2!\left( 8-2 \right)!} \\
 & \Rightarrow N=\dfrac{8\times 7\times 6!}{2\times 6!}=28 \\
\end{align}\]
We know that the formula of probability as
\[P=\dfrac{\text{number of possible outcomes}}{\text{total number of outcomes}}\]
By using the above formula we get the required probability as
\[\begin{align}
  & \Rightarrow P=\dfrac{n}{N} \\
 & \Rightarrow P=\dfrac{10}{28} \\
 & \Rightarrow P=\dfrac{5}{14} \\
\end{align}\]
We are given that the required probability as \[\dfrac{k}{14}\]
By comparing the two probabilities we get
\[\Rightarrow k=5\]
Therefore, we can conclude that the value of \[k\] is 5

Note:
We have a direct formula of this problem.
We have the formula that the probability of the triangle formed from the vertices of \[2n+1\] sided polygon such that the triangle contains the centre of polygon is given as
\[P=\dfrac{n+1}{4n-2}\]
We are given that the polygon has 9 sides
Here, we can see that the number 9 can be represented in form \[2n+1\] as
\[\Rightarrow 9=2\left( 4 \right)+1\]
Therefore we can conclude that \[n=4\]
By using the above formula we get the required probability as
\[\begin{align}
  & \Rightarrow P=\dfrac{4+1}{4\left( 4 \right)-2} \\
 & \Rightarrow P=\dfrac{5}{14} \\
\end{align}\]
We are given that the required probability as \[\dfrac{k}{14}\]
By comparing the two probabilities we get the value of \[k\] is 5