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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) $\frac{18}{{}^{20}\mathrm{C}_{3}}$

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Solution

## 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 $\mathrm{Hence},\mathrm{probability}\mathrm{of}\mathrm{consecutive}\mathrm{numbers}=\frac{18}{{}^{20}\mathrm{C}_{3}}\phantom{\rule{0ex}{0ex}}=\frac{18}{\frac{20×19×18×17!}{3!×17!}}\phantom{\rule{0ex}{0ex}}=\frac{18×6}{20×19×18}=\frac{3}{190}$ ∴ 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.

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