Three diagonals of a regular 11-sided regular polygon are chosen ; probability of parallelism Could someone help me with this?
Suppose P is an 11-sided regular polygon and S is the set of all lines that contain two distinct vertices of P. If three lines are randomly chosen from S, what is the probability that the chosen lines contain a pair of parallel lines ?
 A: Hint: Let $L$ be the number of lines (which you can compute easily). As each line is parallel to a unique side of the $11$-gon, they are partitioned into$~11$ classes of size $L/11$ according to their direction. To count how many among the $\binom L3$ triples contain at least one parallel pair, you can subtract from $\binom L3$ the number of choices that avoid any parallel pairs. The number is obtained by multiplying the number $\binom{11}3$ of choices of $3$ distinct directions by the number $(L/11)^3$ of choices of one line from each direction class.
A: Ok, so its a 11 sided polygon. 
so there are 11 vertices. 
 S is the set of all lines that contain two distinct vertices of P.
so, total lines in S, or you can say total elements in S are :
$$
    \binom{11}2 \qquad\text{no of ways of choosing $2$ vertices out of $11$} = 55
$$
total pairs of lines = $\binom{55}2 = 1485$
Now, number of pairs of parallel lines for every edge possible = 
$$
    \binom{\lfloor n/2\rfloor - 1}2 = \binom42= 6.
$$ 
so for $11$ edges, = $66$
note that for even number of sides in polygon, the number of pairs of parallel lines will be half of those calculated by above formula, because every pair will be repeated for parallel edges.  
probability that the chosen pair will be parallel, is $66/1485 = 0.0444444$
Use Poisson's distribution with $\lambda$ as $.044444$, probability = $0.04251$. 
A: $S$ is a set of ${11 \choose 2}$ lines.
The probability space that you are working in is ${ {11 \choose 2} \choose 3} $ set of triples of lines (order not relevant).
Hint: Use the Principle of Inclusion and Exclusion to count the number of triples which have at least 1 pair of parallel lines.
Hint: Calculate the probability as success / outcomes.
A: But you are choosing three lines.  The probability that you have calculated is for choosing a pair of parallel lines.  The question asked is you randomly choose 3 lines and that it contain exactly one parallel pair.  I would agree with Steven that the probability space is 55C3.  Each edge will have 4 parallel lines and only that.  The lines made up of other vertices are not parallel to any other.  
Thus, the answer that I think is
11C1*5C2*50C1/55C3 = 100/477.  The rational is 11C1  is the choice that you make of 11 sets of 5 parallel lines (including the edge).  You choose 2 of these 5 and hence 5C2 and the other line you choose among 50 lines that are not parallel to the chosen pair.
Let me know if my reasoning is correct, Mark.
Thanks
Satish
