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. $\frac{18}{{}^{20}{C}_3}$

Answer 1

The correct option is B

$\frac{187}{190}$

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)

Number of ways to choose three

$=\frac{\text{consecutive number from 1 to 20}}{\text{Number of ways to choose three}}$

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)

$=1-\frac{3}{190}=\frac{187}{190}$

Hence, the correct answer is option (b).