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Bunuel
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RanonBanerjee
The other side must be between 11-7=4, and 11+7=18. That should have 6,9 and 18 from the set.

Therefore probability =3/6=1/2

IMO A

Posted from my mobile device

In my opinion, 18 is invalid, and the answer should be 1/3. Here's why:

The sum of any two sides of the triangle must be greater than the third side and the difference of any two sides of the triangle must be smaller than the third side.

Hence, the third side of the triangle must lie between 11-7=4 and 7+11=18, where 4 and 18 are not included.

So the only possible choices are 6 and 9, which makes the probability 2/6 or 1/3. Hence, option B.
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IMO it is B:

11+7=18, 11-7=4, but these values are excluded. So we are left with: 6,9-> P=2/6 = 1/3
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Bunuel
From the set X = {2, 3, 6, 9, 18, 20}, what is the probability of choosing a number such that it constructs a triangle whose other two sides are 7 and 11?

A. 1/2
B. 1/3
C. 1/4
D. 1/5
E. 1/6


Let, the 3 sides of the triangle are 7, 11, and L.
Property: Sum of two sides of a triangle is greater than the third side.

If L is the longest side, then:
L<7+11
or, L<18

If 11 is the longest side, then:
7+L>11
or, L>4

So, from the set given, only 6 and 9 satisfy both conditions. Hence, probability = 2/6 or 1/3.
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