Probability in forming a triangle

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Amrabdelnaby: Master | Next Rank: 500 Posts; Posts: 137; Joined: Fri Nov 13, 2015 11:01 amFollowed by: 2 members; Thanked: 1 times

by Amrabdelnaby » Tue Nov 17, 2015 4:59 am

Hello guys,

Could you please help me with this?

5. The 4 sticks in a complete bag of Pick-Up Sticks are all straight-line segments of negligible width, but each has a different length: 1 inch, 2 inches, 3 inches, and 4 inches, respectively. If Tommy picks a stick at random from each of 3 different complete bags of Pick-Up Sticks, what is the probability that Tommy CANNOT form a triangle from the 3 sticks?

A. 11/32

B. 13/32

C. 15/32

D. 17/32

E. 19/32

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Source: Beat The GMAT — Problem Solving | Tue Nov 17, 2015 4:59 am

regor60: Master | Next Rank: 500 Posts; Posts: 416; Joined: Thu Oct 15, 2009 11:52 am; Thanked: 27 times

by regor60 » Wed Nov 18, 2015 8:47 am

First, determine total number of permutations = 4^3 or 64. Many of these are the same 3 numbers sequenced differently.

Clearly, there are 4 sequences of 1111-4444 > all of these meet the triangle rule of A+B>C, so 4 CANs

Sequences of 2 digits same, 3rd digit different: 112-114, 221,223-224, 331,332,334, 441-443: 12 sequence that have 3 permutations each equals 36 of 64. 24 of these meet triangle rule, so 24 CANs

Sequences of all 3 digits different: 123, 124, 134,234. 4 sequences with 6 permutations each equals 24 of 64. Only 234 meets triangle rule, with 6 permuations associated with it > 6 CANs

Adding up CANs equals 34, leaving 30/64 or 15/32 probability cannot form triangle

Answer C.

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