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19 Nov 2022, 03:16
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
35% (02:15) correct
65%
(02:20)
wrong
based on 62
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Three numbers are to be selected at random and without replacement from the five numbers 4, 5, 7, 8 and 11. What is the probability that the three numbers selected could be the lengths of the sides of a triangle?
A. 1/5
B. 2/5
C. 3/5
D. 4/5
E. 1
A. 1/5
B. 2/5
C. 3/5
D. 4/5
E. 1
va95
Joined: 24 Dec 2016
Last visit: 19 Jun 2025
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Location: Armenia
Concentration: Statistics
Schools: Duke Fuqua HBS '26 Wharton '26
GMAT 1: 720 Q49 V40
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19 Nov 2022, 10:25
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I think the answer is D
You can pick any 3 numbers from 5 available in 10 ways (5C3).
The sum of any 2 sides of a triangle has to be greater than the third side. In this case, only 8 combinations follow that rule.
1. 4,5,7
2. 4,5,8
3. 4,7,8
4. 4,8,11
5. 5,7,11
6. 5,8,11
7. 5, 7, 8
8. 7,8,11
So 8 out of 10 (or 4/5) possible combinations can create a triangle.
You can pick any 3 numbers from 5 available in 10 ways (5C3).
The sum of any 2 sides of a triangle has to be greater than the third side. In this case, only 8 combinations follow that rule.
1. 4,5,7
2. 4,5,8
3. 4,7,8
4. 4,8,11
5. 5,7,11
6. 5,8,11
7. 5, 7, 8
8. 7,8,11
So 8 out of 10 (or 4/5) possible combinations can create a triangle.
