+0

0
54
10

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?

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!

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?

Apr 21, 2020
#4
0

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

Apr 22, 2020
#5
0

Melody  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

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)

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
#8
0

LOL

Are you going to give me the address of where I answered it correctly?

Melody  Apr 23, 2020
#9
+1

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