Question

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

Options

• \[\frac{186}{190}\]

   

• \[\frac{187}{190}\]

   

• \[\frac{188}{190}\]

   

• \[\frac{18}{^{20}{}{C}_3}\]

   

Answer 1

Number of ways to choose three numbers from 1 to 20 = \[^{20}{}{C}_3\] = 1140

Now, the set of three consecutive numbers from 1 to 20 are (1, 2, 3), (2, 3, 4), (3, 4, 5), ...., (18, 19, 20).
So, the number of ways to choose three numbers from 1 to 20 such that they are consecutive is 18.
P(three numbers choosen are consecutive) =\[\frac{\text{ Number of ways to choose three consecutive numbers from 1 to 20} }{\text{ Number of ways to choose three numbers from 1 to 20} } = \frac{18}{^{20}{}{C}_3} = \frac{18}{1140} = \frac{3}{190}\]

∴ P(three numbers choosen are not consecutive) = 1 − P(three numbers choosen are consecutive) = \[1 - \frac{3}{190} = \frac{187}{190}\]