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Let S be the set of sides and diagonals of a regular pentagon. A pair

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Let S be the set of sides and diagonals of a regular pentagon. A pair  [#permalink]

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New post 20 Mar 2019, 04:45
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A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

20% (03:33) correct 80% (02:38) wrong based on 10 sessions

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Let S be the set of sides and diagonals of a regular pentagon. A pair  [#permalink]

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New post 20 Mar 2019, 05:44
Bunuel wrote:
Let S be the set of sides and diagonals of a regular pentagon. A pair of elements of S are selected at random without replacement. What is the probability that the two chosen segments have the same length?

(A) 2/5
(B) 4/9
(C) 1/2
(D) 5/9
(E) 4/5



A regular pentagon means five sided figure which has all 5 sides equal(RED in color). Now we can draw the diagonals too, and we find there are 5 diagonals ( BLACK in color).
To count total sides is to find ways to select 2 of A, B, C, D, E, that is 5C2 = 10 segments.
So choosing equal sides means choosing two of the five RED sides = 5*4, or choosing two of the five diagonals BLACK lines = 5*4
Total ways to choose 2 sides out of 10 is 10*9

Probability = \(\frac{5*4+5*4}{10*9}=\frac{4}{9}\)

B
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Re: Let S be the set of sides and diagonals of a regular pentagon. A pair  [#permalink]

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New post 20 Mar 2019, 07:23
Bunuel wrote:
Let S be the set of sides and diagonals of a regular pentagon. A pair of elements of S are selected at random without replacement. What is the probability that the two chosen segments have the same length?

(A) 2/5
(B) 4/9
(C) 1/2
(D) 5/9
(E) 4/5


Pentagon has 5 equal sides and 5 diagonals ; total equal sides 10
so case 1: picking 2 sides of equal length; 5/10 * 4/9 = 2/9
case 2 : picking 2 sides of equal diagonal ;5/10 * 4/9 = 2/9
add them = 4/9
IMO B
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Re: Let S be the set of sides and diagonals of a regular pentagon. A pair   [#permalink] 20 Mar 2019, 07:23
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Let S be the set of sides and diagonals of a regular pentagon. A pair

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