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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## A triangle is formed by connecting three randomly chosen ver tagged by: Max@Math Revolution ##### This topic has 3 expert replies and 0 member replies ### GMAT/MBA Expert ## A triangle is formed by connecting three randomly chosen ver ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult [Math Revolution GMAT math practice question] A triangle is formed by connecting three randomly chosen vertices of a hexagon. What is the probability that at least one of the sides of the triangle is also a side of the hexagon? A. 4/5 B. 5/6 C. 7/8 D. 8/9 E. 9/10 _________________ Math Revolution Finish GMAT Quant Section with 10 minutes to spare. The one-and-only Worldâ€™s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. Only$99 for 3 month Online Course
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Max@Math Revolution wrote:
[Math Revolution GMAT math practice question]

A triangle is formed by connecting three randomly chosen vertices of a hexagon. What is the probability that at least one of the sides of the triangle is also a side of the hexagon?

A. 4/5
B. 5/6
C. 7/8
D. 8/9
E. 9/10
Let the 6 vertices = A, B, C, D, E and F.

P(the triangle includes AT LEAST 1 side of the hexagon) = 1 - P(the triangle includes NO sides of the hexagon).

From the 6 vertices, the number of ways to choose 3 to form a triangle = 6C3 = (6*5*4)/(3*2*1) = 20.
Of these 20 triangles, only 2 include no sides of the hexagon:
Triangle ACE
Triangle BDF
Thus:
P(the triangle includes no sides of the hexagon) = 2/20 = 1/10.
P(the triangle includes at least 1 side of the hexagon) = 1 - 1/10 = 9/10.

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Max@Math Revolution wrote:
[Math Revolution GMAT math practice question]

A triangle is formed by connecting three randomly chosen vertices of a regular (or at least convex) hexagon. What is the probability that at least one of the sides of the triangle is also a side of the hexagon?

A. 4/5
B. 5/6
C. 7/8
D. 8/9
E. 9/10

$$? = 1 - {{\,\,\# \,\,\Delta \,\,\,{\rm{no}}\,\,\,{\rm{sides}}\,\,\,{\rm{in}}\,\,\,{\rm{hexag}}\,\,\,} \over {\# \,\,\Delta \,\,{\rm{total}}}} = 1 - {2 \over {C\left( {6,3} \right)}} = 1 - {2 \over {20}} = {9 \over {10}}$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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Since the hexagon has 6 vertices, and we need to choose 3 of them to form the vertices of the triangle, the total number of triangles that can be formed is 6C3 = 20.
Note that only the following 2 triangles do not include an edge of the hexagon:

Therefore, the number of the triangles satisfying the original condition is 18 = 20 - 2.
Thus, the probability that the triangle will include an edge of the hexagon is 18/20 = 9/10.

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