We define a bow-tie quadrilateral as a quadrilateral where two sides cross each other. An example of a bow-tie quadrilateral is shown below.

The Image I cannot upload, but it is of a circle with 7 points on it, and two of each of the points on 2 sides of the triangle are connected, making a bow-tie shape.

Seven distinct points are chosen on a circle. We draw all \(\binom{7}{2} = 21\) chords that connect two of these points. Four of these \(21\) chords are selected at random. What is the probability that these four chosen chords form a bow-tie quadrilateral?

awoafdwawda Apr 19, 2020

#2**+1 **

I appreciate the answer, but I don't think it is correct. I want an explanation along with the answer, because I really want to learn how to solve these types of probability problems. Thanks!

awoafdwawda Apr 20, 2020

#3**+1 **

Here is the question as I understand it.

there are 7 distinct points on a circle. ANY 4 lines are drawn. each one connecting one point to a different point.

What is the probability that these 4 lines form a bowtie?

I think there are 7C4 ways to choose 4 points. = 35 ways

For each of these there are 2 ways they can be joined to make a bow tie. So that is 70 ways.

There are 21 chords so I think there must be 21*20*19*18 ways to choose 4 chords. 6840ways

So maybe if my logic is right the prob of forming a bow tie is

70 / 6840 = **7/684**

**Can you check if this is correct or not?**

Melody Apr 21, 2020

#4**0 **

No, your solution is not correct, but thank you for trying to help! It's still greatly appreciated!

awoafdwawda Apr 22, 2020

#6**+1 **

ok I have a different idea.

There are 7C2 = 21 possible cords in total

I want 4 of them so there are 15C4 = 1365 ways to select 4 chords

To make a bow tie I need to chose just 4 points from the 7 so that is 7C4=35 ways

With each of these 35 sets of 4 points there is 2 ways to join them to make a bow tie, so that is 70 bow ties.

So maybe the answer is 70/1365 = 2/39

**Can you please check that answer.**

I am assuming that you have the ability to plug numbers into a box and be told right or wrong. Can you do that?

(rotations are __not__ counted as the same)

Melody Apr 23, 2020

#7**+1 **

you answered this correctly a few months ago

this one is incorrect

Guest Apr 23, 2020

edited by
Guest
Apr 23, 2020

#10**0 **

**Thanks very much guest for finding that.**

My logic is actually the same both times. It is nice to know that I did learn something.

It was just a careless error this time.

I actually do not know where i got the 15 from, it obviously should have been 21.

So the denominator should be 21C4 = 5985

Answer 70/5985

Melody
Apr 23, 2020