+0  
 
0
109
2
avatar

How this work

 

We define a bow-tie quadrilateral as a quadrilateral where two sides cross each other.  An example of a bow-tie quadrilatera is shown below.  distinct points are chosen on a circle. We draw all  chords that connect two of these points. Four of these chords are selected at random. What is the probability that these four chosen chords form a bow-tie quadrilateral?

 Feb 21, 2023
 #1
avatar-9 
+1

Consider the set of all possible pairs of points that can be chosen from the given set of distinct points on the circle. There are $\binom{n}{2}$ such pairs, where $n$ is the total number of distinct points on the circle. Each pair of points corresponds to a chord that can be drawn between them.

Now, we want to find the probability that when we choose four chords at random, they form a bow-tie quadrilateral. To do this, we can count the total number of ways to choose four chords and the number of ways to choose four chords that form a bow-tie quadrilateral.

The total number of ways to choose four chords is $\binom{\binom{n}{2}}{4}$, which is the number of ways to choose 4 pairs of points from the set of $\binom{n}{2}$ pairs.

To count the number of ways to choose four chords that form a bow-tie quadrilateral, we first choose two chords that intersect. There are $\binom{4}{2} = 6$ ways to do this. Once we have chosen the intersecting chords, we need to choose two more chords that do not intersect either of the first two chords. There are $\binom{\binom{n}{2}-4}{2}$ ways to do this.

Therefore, the number of ways to choose four chords that form a bow-tie quadrilateral is $6 \times \binom{\binom{n}{2}-4}{2}$.

Hence, the probability that four randomly chosen chords form a bow-tie quadrilateral is:

$$ \frac{6 \times \binom{\binom{n}{2}-4}{2}}{\binom{\binom{n}{2}}{4}} $$

Simplifying the expression, we get:

$$ \frac{3(n-4)}{(n-1)(n-2)(n-3)} $$

where $n$ is the total number of distinct points on the circle.

 Feb 21, 2023

3 Online Users