$\begin{array}{1 1}(A)\;\large\frac{186}{190}\\(B)\;\large\frac{187}{190}\\(C)\;\large\frac{188}{190}\\(D)\;\large\frac{18}{20C_3}\end{array} $

Step 1:

Probability that the numbers are not consecutive =1-probability of the numbers being consecutive

Given numbers from 1 to 20

$\therefore$ Sample space =20!

3 numbers consecutively =$18!\times 3!$

Considering 3 numbers as a single digit.

$\therefore$ The numbers will be 18 and those three numbers can be arranged in 3! ways

$\therefore$ Probability =$\large\frac{18!\times 3!}{20!}$

$\Rightarrow \large\frac{3}{190}$

Step 2:

Probability that they are not consecutive

$\Rightarrow$ 1-Probability that they are consecutive

$\Rightarrow 1-\large\frac{3}{190}=\frac{187}{190}$

Hence (B) is the correct answer.

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