Question

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. 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?

Answer 1

Answer:

70/5985

Step-by-step explanation:

We know that a quadrilateral needs to have four vertices (or points on the circle). There are always two ways to link the cross — horizontally or vertically. Using my limited knowledge of combinations, we know that choosing four points out of seven equals 35. Multiplying the two ways to connect those lines (again, horizontally and vertically) makes 35*2 = 70 "bow-tie quadrilaterals" that can be formed on the circle using four points. There are 5985 ways four chords can be chosen out of twenty-five chords because C(25,4) equals 5985, so the probability is 70/5985, and then we just need to simplify that fraction.

Answer 2

Answer:

Step-by-step explanation:

Suppose we choose four points on the circle. Then there are exactly two ways that they can form the vertices of a bow-tie.

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Thus, there are exactly 2 times 7c4 = 70 bow-tie quadrilaterals that can be formed among the seven points.

There are 21 chords to choose from, so there are 21c4 = 5985 ways to choose four chords. Therefore, the probability that the four chords form a bow-tie quadrilateral is

70/5985 = 2/171.