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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 ?

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    $\begingroup$ Here are some questions to nudge you along - how many lines are there in $S$? How many different lengths of line are there in $S$? How can I use the symmetry of the situation to find all the lines in $S$ parallel to a given line of a particular length? Mostly questions like this are about organising your thinking. $\endgroup$ – Mark Bennet Aug 31 '13 at 13:23
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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.

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  • $\begingroup$ This seems to be the right answer, thank you. But how can we be sure that a certain line, for example, 1, which is "one side away" from an edge is parallel? I think I get the picture you are saying, and the answer would be correct. $\endgroup$ – user87611 Aug 31 '13 at 20:44
  • $\begingroup$ To clarify my follow-up question: there are eleven sides. Let's number them from 1 to 11. The line given by 1-2 would be parallel to 11-3 according to what you said. This seems right, but how can we be sure? Thank you, $\endgroup$ – user87611 Aug 31 '13 at 20:45
  • $\begingroup$ @user87611 That is because (and only because) it is a regular $11$-gon. The line from the middle of side $1$-$2$ to vertex $7$ is a symmetry axis, whose reflection interchanges $1$ and $2$, and also $3$ and $11$, so those lines are perpendicular to this symmetry axis. $\endgroup$ – Marc van Leeuwen Aug 31 '13 at 20:58
  • $\begingroup$ I see now, thanks $\endgroup$ – user87611 Aug 31 '13 at 21:06
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$S$ 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.

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  • $\begingroup$ Actually I don't think inclusion/exclusion is very useful here. To compute the number of triples with at least one parallel pair one might either compute the number of triples with exactly one parallel pair and add the number of entirely parallel triples, or one could take the complement of the number of triples with three distinct directions. Neither way one uses inclusion/exclusion. $\endgroup$ – Marc van Leeuwen Aug 31 '13 at 20:04
  • $\begingroup$ @MarcvanLeeuwen I consider the "triples with exactly one parallel pair and add entirely parallel triples" as PIE. The reason for that is when you count the exactly one parallel pair, you have to exclude all the remaining parallel lines, and then add that back in. $\endgroup$ – Calvin Lin Aug 31 '13 at 23:19
  • $\begingroup$ What you mention is just the "principle of disjoint union", which is a far more basic principle than inclusion-exclusion. The latter involves counting (the complement of) a non-disjoint union by correcting the sum of the sizes of the parts by terms for mutual intersections. $\endgroup$ – Marc van Leeuwen Sep 1 '13 at 5:48
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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 = $0.04251$.

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  • $\begingroup$ "no edge has more than 1 parallel line associated with it" just isn't so. Consider the edge between verts 5 and 6; then lines 4-7, 3-8, etc. are all parallel to that edge. $\endgroup$ – Steven Stadnicki Aug 31 '13 at 17:11
  • $\begingroup$ @steven, thanks for that... $\endgroup$ – Sumedh Aug 31 '13 at 19:05
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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

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