A triangle is formed by connecting three randomly chosen ver

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Max@Math Revolution: Elite Legendary Member; Posts: 3991; Joined: Fri Jul 24, 2015 2:28 amFollowed by: 37 members; Location: Las Vegas, USA; Thanked: 19 times

by Max@Math Revolution » Wed Dec 05, 2018 12:57 am

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[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

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Wed Dec 05, 2018 3:46 am

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.

The correct answer is E.

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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

A triangle is formed by connecting three randomly chosen ver

by fskilnik@GMATH » Wed Dec 05, 2018 6:30 pm

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.

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
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Max@Math Revolution: Elite Legendary Member; Posts: 3991; Joined: Fri Jul 24, 2015 2:28 amFollowed by: 37 members; Location: Las Vegas, USA; Thanked: 19 times

by Max@Math Revolution » Thu Dec 06, 2018 11:46 pm

=>

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.

Therefore, the answer is E.
Answer: E

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