+22 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?
+22 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!
+118608 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?
+22 No, your solution is not correct, but thank you for trying to help! It's still greatly appreciated!
+118608 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)
you answered this correctly a few months ago
this one is incorrect
+118608 LOL
Are you going to give me the address of where I answered it correctly?
+118608 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
