Question

Three numbers are chosen at random from $1$ to $20$. The probability that they are consecutive, is

• A. $\frac{1}{190}$
• B. $\frac{1}{120}$
• C. $\frac{3}{190}$
• D. $\frac{5}{190}$

Answer 1

The correct option is C

$\frac{3}{190}$

Find the required probability:

The number of ways of selecting $3$ numbers out of $20$ is given as${}^{20}{C}_3$

Hence the number of total outcomes $={}^{20}{C}_3$

$=\frac{20!}{3!17!}$

$=\frac{20\times 19\times 18}{3\times 2}$

Number of total outcomes $=1140$

Three consecutive numbers from $1$ to $20$ can be selected as $\left\{\left(1,2,3\right),\left(2,3,4\right),\left(3,4,5\right),........,\left(18,19,20\right)\right\}$

Hence, the number of favourable outcomes are $18$.

Hence, required probability $\mathbf{=}\frac{\mathbf{number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{favorable}\mathbf{}\mathbf{outcomes}}{\mathbf{number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{total}\mathbf{}\mathbf{outcomes}}$

$=\frac{18}{1140}$

$=\frac{3}{190}$

Hence, the probability of selecting three consecutive numbers from $1$ to $20$ is $\frac{3}{190}$.

Hence, option (C) is the correct answer.