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ABCDEFGH is an octagon. Two vertices of the octagon are selected

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ABCDEFGH is an octagon. Two vertices of the octagon are selected  [#permalink]

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New post 19 Jun 2019, 07:29
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Question Stats:

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ABCDEFGH is an octagon. Two vertices of the octagon are selected at random. Find the probability that these vertices form a diagonal of the octagon.

A. \(\frac{1}{7}\)

B. \(\frac{2}{7}\)

C. \(\frac{3}{7}\)

D. \(\frac{4}{7}\)

E. \(\frac{5}{7}\)
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ABCDEFGH is an octagon. Two vertices of the octagon are selected  [#permalink]

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New post Updated on: 19 Jun 2019, 09:17
Number of ways you can select 2 verices= 8C2=28
Number of ways these vertices form a diagonal= nC2-n=8C2-8=20

the probability that these vertices form a diagonal of the octagon. = 20/28=5/7

firas92 wrote:
ABCDEFGH is an octagon. Two vertices of the octagon are selected at random. Find the probability that these vertices form a diagonal of the octagon.

A. \(\frac{1}{7}\)

B. \(\frac{2}{7}\)

C. \(\frac{3}{7}\)

D. \(\frac{4}{7}\)

E. \(\frac{5}{7}\)

Originally posted by nick1816 on 19 Jun 2019, 08:54.
Last edited by nick1816 on 19 Jun 2019, 09:17, edited 1 time in total.
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Re: ABCDEFGH is an octagon. Two vertices of the octagon are selected  [#permalink]

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New post 19 Jun 2019, 09:11
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nick1816 wrote:
Number of ways you can select 2 verices= 8C2=28
Number of ways these vertices form a diagonal= (1/2)*(8C1 * 1)=4

the probability that these vertices form a diagonal of the octagon. = 4/28=1/7


A diagonal of a polygon is any line connecting two vertices that is not an edge, so an octagon has quite a few diagonals -- I think you're only counting lines connecting vertices that are strictly opposite each other.

Here we can select a pair of vertices in 8*7/2! = 28 ways. If we draw a line between our pair of vertices, it will be a diagonal unless it's one of the 8 edges of the octagon. So we can draw 28 - 8 = 20 diagonals, and the answer is 20/28 = 5/7.
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Re: ABCDEFGH is an octagon. Two vertices of the octagon are selected  [#permalink]

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New post 19 Jun 2019, 09:15
My bad!! Gotta edit my solution. ;)
IanStewart wrote:
nick1816 wrote:
Number of ways you can select 2 verices= 8C2=28
Number of ways these vertices form a diagonal= (1/2)*(8C1 * 1)=4

the probability that these vertices form a diagonal of the octagon. = 4/28=1/7


A diagonal of a polygon is any line connecting two vertices that is not an edge, so an octagon has quite a few diagonals -- I think you're only counting lines connecting vertices that are strictly opposite each other.

Here we can select a pair of vertices in 8*7/2! = 28 ways. If we draw a line between our pair of vertices, it will be a diagonal unless it's one of the 8 edges of the octagon. So we can draw 28 - 8 = 20 diagonals, and the answer is 20/28 = 5/7.
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Re: ABCDEFGH is an octagon. Two vertices of the octagon are selected  [#permalink]

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New post 21 Jun 2019, 04:23
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The total no of diagonals for a n-sided polygon can be given by the formula- (n(n-3))/2.
so in this case total no of diagonals= 8(8-3)/2=20 diagonals

Total no. of ways of selecting 2 points out of 8= 8C2=28 ways

Thus required probability=20/28=5/7

Thus E is the right answer
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Re: ABCDEFGH is an octagon. Two vertices of the octagon are selected  [#permalink]

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New post 09 Aug 2019, 12:39
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Octagon has 8 vertices. Probability of picking the first vertex is 8/8 (doesn't matter which one).
Then there are 7 other vertices (x/7). Can't pick the one adjacent left or adjacent right as those are not diagonal.
(7-2)/7=5/7
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Re: ABCDEFGH is an octagon. Two vertices of the octagon are selected   [#permalink] 09 Aug 2019, 12:39
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