Question

Three numbers are chosen from 1 to 20. The probability that they are not consecutive is

(a) $\frac{186}{190}$ (b) $\frac{187}{190}$ (c) $\frac{188}{190}$ (d) $\frac{18}{{}^{20}\mathrm{C}_3}$

Answer 1

Since, the set of three consecutive numbers from 1 to 20
are (1, 2, 3) (2, 3, 4) (3, 4, 5) .......... (18, 19, 20)
i.e. 18 is number
$$\begin{gathered} \mathrm{Hence},\mathrm{probability}\mathrm{of}\mathrm{consecutive}\mathrm{numbers}=\frac{18}{{}^{20}\mathrm{C}_3} \\ =\frac{18}{{\displaystyle \frac{20\times 19\times 18\times 17!}{3!\times 17!}}} \\ =\frac{18\times 6}{20\times 19\times 18}=\frac{3}{190} \end{gathered}$$

∴ Probability that three numbers are not consecutive is 1 − probability that three numbers are consecutive
$\mathrm{i}.\mathrm{e}1-\frac{3}{190}=\frac{187}{190}$
Hence, the correct answer is option B.