Three numbers are chosen at random from numbers 1 to 20 . The probability that they are consecutive is

The sample space consists of all 3-element subsets of $S = \{1, 2, \dots, 20\}$, yielding $\binom{20}{3} = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140$ equally likely outcomes.

Consecutive triplets correspond to sets of the form $\{x, x+1, x+2\}$ where $1 \leq x \leq 18$. There are 18 such sets, as the smallest starting value is 1 and the largest is 18.

The probability is thus $\frac{18}{1140} = \frac{3}{190}$, matching option C.