By combinatory:

To pick three different numbers between 10 and 20 inclusive: \(C^3_{11}=\frac{11*10*9}{3*2*1}=165\)

To pick 12 (1 possibility), 15 (1 possibility) and one another: 11-2=9 possibilities

So the probability is \(\frac{9}{165}=\frac{3}{55}\)

The correct answer is B.

By probability:

The probability to pick first number 12: 1/11

The probability to pick second number 15: 1/10

The probability to pick third number different from 12 and 15: 9/9

Since, we can permutate numbers 12,15, and third in 3! possibilities the final probability is: \(6*\frac{1}{11}*\frac{1}{10}*\frac{9}{9}=\frac{3}{55}.\)

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