+0

# Six points on a circle are given. Four different chords joining pairs of the six points are selected at random. What is the probability that

0
3398
6

Six points on a circle are given. Four different chords joining pairs of the six points are selected at random. What is the probability that the four chords form the sides of a convex quadrilateral?

May 15, 2015

#2
+15

Six points on a circle are given. Four different chords joining pairs of the six points are selected at random. What is the probability that the four chords form the sides of a convex quadrilateral ?

Let the six points 1, 2, 3, 4, 5, 6

All convex quadrilateral  are all selection(4 Numbers)  in ascending order from the set 1, 2, 3, 4, 5, 6

$$\rm{convex~quadrilateral } = \bordermatrix{Points& 1 & 2 & 3 & 4 & 5 & 6 \cr 1. &1 &2 &3 &4 & & \cr 2. &1 &2 &3 & &5 & \cr 3. &1 &2 &3 & & &6 \cr 4. &1 &2 & &4 &5 & \cr 5. &1 &2 & &4 & &6 \cr 6. &1 &2 & & &5 &6 \cr 7. &1 & &3 &4 &5 & \cr 8. &1 & &3 &4 & &6 \cr 9. &1 & &3 & &5 &6 \cr 10.&1 & & &4 &5 &6 \cr 11.& &2 &3 &4 &5 & \cr 12.& &2 &3 &4 & &6 \cr 13.& &2 &3 & &5 &6 \cr 14.& &2 & &4 &5 &6 \cr 15.& & &3 &4 &5 &6 \cr} = \binom64$$

Point-connections:

$$\small{\text{ \begin{array}{|c|c|c|c|c|c|c|} \hline &1&2&3&4&5&6 \\ \hline 1&x&1\Rightarrow2&1\Rightarrow3&1\Rightarrow4&1\Rightarrow5&1\Rightarrow6 \\ 2& &x&2\Rightarrow3&2\Rightarrow4&2\Rightarrow5&2\Rightarrow6 \\ 3& & &x&3\Rightarrow4&3\Rightarrow5&3\Rightarrow6 \\ 4& & & &x&4\Rightarrow5&4\Rightarrow6 \\ 5& & & & &x&5\Rightarrow6 \\ 6& & & & & &x \\ \hline \end{array} }} = \binom62=15$$

Four  chords from 15 pont-connections

$$=\binom{\binom62}{4}=\binom{15}{4}$$

the probability that the four chords form the sides of a convex quadrilateral

$$\small{\text{ \begin{array}{rcl} &=&\dfrac{\binom64}{\binom{15}{4}} =\dfrac{ \dfrac{6!}{4!\cdot(6-4)!} }{ \dfrac{15!}{4!\cdot(15-4)!} } =\dfrac{6!}{2!} \cdot \dfrac{11!}{15!} =\dfrac{3\cdot 4\cdot 5\cdot 6}{ 12\cdot 13 \cdot 14 \cdot 15} =\dfrac{36}{3276} =0.01098901099 \end{array} }}$$ May 17, 2015

#1
+13

I don't know Mellie, your questions are hard.  :))

Fristly any quadrilateral that is formed will be a convec quadrilateral because opposite angels in a cyclic quads are supplementary - so none can be greater than 180 degrees. First lets see how many diagonals that there are

From S there are 5, then from A there are 4, then 3 then 2 then 1 = 5+4+3+2+1=15 diagonal

How many ways can we choose 4 of these 15 diagonals.    15C4

$${\left({\frac{{\mathtt{15}}{!}}{{\mathtt{4}}{!}{\mathtt{\,\times\,}}({\mathtt{15}}{\mathtt{\,-\,}}{\mathtt{4}}){!}}}\right)} = {\mathtt{1\,365}}$$         ways

I'm going to attempt to list all the cyclic quad possibilities

SABCS,SABDS,SABES, SACDS, SACES, SADES, SBCDS, SBCES, SBDE, SCDES      that is 10

Now  I think I have exhasted all the ones with the point S in them

ABCDA, ABCEA, ABDEA, ACDEA     that is 4

Now  I think I have exhasted all the ones with the point S or A in them.

BCDEB  that is the only one left

Total = 10+4+1 = 15

So I get     $${\frac{{\mathtt{15}}}{{\mathtt{1\,365}}}} = {\frac{{\mathtt{1}}}{{\mathtt{91}}}} = {\mathtt{0.010\: \!989\: \!010\: \!989\: \!011}}$$

That is my best guess at the moment.   It might even be correct May 17, 2015
#6
+1

Nice Work! This is correct. Although your wording was a little bit confusing but still, great!!!!!!!! Robin+Jonathan  May 30, 2016
#2
+15

Six points on a circle are given. Four different chords joining pairs of the six points are selected at random. What is the probability that the four chords form the sides of a convex quadrilateral ?

Let the six points 1, 2, 3, 4, 5, 6

All convex quadrilateral  are all selection(4 Numbers)  in ascending order from the set 1, 2, 3, 4, 5, 6

$$\rm{convex~quadrilateral } = \bordermatrix{Points& 1 & 2 & 3 & 4 & 5 & 6 \cr 1. &1 &2 &3 &4 & & \cr 2. &1 &2 &3 & &5 & \cr 3. &1 &2 &3 & & &6 \cr 4. &1 &2 & &4 &5 & \cr 5. &1 &2 & &4 & &6 \cr 6. &1 &2 & & &5 &6 \cr 7. &1 & &3 &4 &5 & \cr 8. &1 & &3 &4 & &6 \cr 9. &1 & &3 & &5 &6 \cr 10.&1 & & &4 &5 &6 \cr 11.& &2 &3 &4 &5 & \cr 12.& &2 &3 &4 & &6 \cr 13.& &2 &3 & &5 &6 \cr 14.& &2 & &4 &5 &6 \cr 15.& & &3 &4 &5 &6 \cr} = \binom64$$

Point-connections:

$$\small{\text{ \begin{array}{|c|c|c|c|c|c|c|} \hline &1&2&3&4&5&6 \\ \hline 1&x&1\Rightarrow2&1\Rightarrow3&1\Rightarrow4&1\Rightarrow5&1\Rightarrow6 \\ 2& &x&2\Rightarrow3&2\Rightarrow4&2\Rightarrow5&2\Rightarrow6 \\ 3& & &x&3\Rightarrow4&3\Rightarrow5&3\Rightarrow6 \\ 4& & & &x&4\Rightarrow5&4\Rightarrow6 \\ 5& & & & &x&5\Rightarrow6 \\ 6& & & & & &x \\ \hline \end{array} }} = \binom62=15$$

Four  chords from 15 pont-connections

$$=\binom{\binom62}{4}=\binom{15}{4}$$

the probability that the four chords form the sides of a convex quadrilateral

$$\small{\text{ \begin{array}{rcl} &=&\dfrac{\binom64}{\binom{15}{4}} =\dfrac{ \dfrac{6!}{4!\cdot(6-4)!} }{ \dfrac{15!}{4!\cdot(15-4)!} } =\dfrac{6!}{2!} \cdot \dfrac{11!}{15!} =\dfrac{3\cdot 4\cdot 5\cdot 6}{ 12\cdot 13 \cdot 14 \cdot 15} =\dfrac{36}{3276} =0.01098901099 \end{array} }}$$ heureka May 17, 2015
#3
+5

Very nice, Melody and heureka......!!!

This WAS a tough one   !!!!

One observation......  if we laid out ABCDES in a "line"......we only need to choose 4 of any of the 6 letters. Then, putting each resulting set in rotational order will  give us all the possible cyclic quads.   May 17, 2015
#4
0

That is how I apprached it and Heureka too I think.

I tried to use the table fuction on the forum for layout.  But it does not work anywhere good enough and it looks like a mess.

I expect Heureka's and mine are very similar but mine is probably riddled with errors. :)

May 18, 2015
#5
-1

Not to be rude or anything, but this problem is hard because it is from week 12 of introduction to counting and probability online class. It is an aops class so it is supposed to be pretty hard. I think the whole purpose of it is to learn, not get answers off the web. I think mellie has asked the answers to all the problems. Why are you here then, you ask? Well I wanted to see if you could actually get the answer from simply googling the exact problem. And I guess you can. Anyways I guess you are just wasting you/your parent's money if you paid for a class and just asked the answers in a forum.

May 28, 2016