Three numbers chosen from 1 to 20. Probability that they are not consecutive is:

Correct option is B

Given:
Total numbers = 20
Number of numbers chosen = 3
Formula used:
Total ways = $$\binom{20}{3}$$​​
Number of ways to choose 3 consecutive numbers from 1 to 20
= 20 − 3 + 1 = 18
Probability = $$\frac{\text{favourable cases}}{\text{total cases}}$$​​
Solution:
Total possible selections:
​$$\binom{20}{3} = \frac{20 \cdot 19 \cdot 18}{6} = 1140$$​​
Number of selections where numbers are consecutive:
18
Number of selections where numbers are not consecutive:
​$$1140 − 18 = 1122$$​​
Required probability:
​$$= \frac{1122}{1140}= \frac{187}{190}$$​​
The correct answer is (b) $$\frac{187}{190}.$$​​