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Three numbers are chosen at random from 1 to 20 . The probability that they are consecutive is
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$\frac{3}{190}$
Hints: Total number of cases ; ${ }^{20} \mathrm{C}_3=\frac{20 \times 19 \times 18}{2 \times 3}=20 \times 19 \times 3=1140$
Total number of favourable cases $=18$
$\therefore$ Required probability $=\frac{18}{1140}=\frac{3}{190}$
Total number of favourable cases $=18$
$\therefore$ Required probability $=\frac{18}{1140}=\frac{3}{190}$
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