# Three numbers are chosen from 1 to 20.

Question:

Three numbers are chosen from 1 to 20. Find the probability that they are not consecutive

A. $\frac{186}{190}$

B. $\frac{187}{190}$

C. $\frac{188}{190}$

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

Solution:

B. 187/190

Explanation:

Since, the set of three consecutive numbers from 1 to 20 are (1, 2, 3), (2, 3, 4), (3, 4, 5), …, (18,19,20)

Considering 3 numbers as a single digit

∴ the numbers will be 18

Now, we have to choose 3 numbers out of 20. This can be done in 20C3 ways

∴ n(S) = 20C3

The desired event is that the 3 numbers are choose must consecutive. So,

$\mathrm{P}$ (numbers are consecutive) $=\frac{18}{{ }^{20} \mathrm{C}_{3}}$

$=\frac{18}{\frac{20 !}{3 !(20-3) !}}\left[\because{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=\frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}) ! \mathrm{r} !}\right]$

The above equation can be written as

$=\frac{18}{\frac{20 \times 19 \times 18 \times 17 !}{3 \times 2 \times 1 \times 17 !}}$

$=\frac{\frac{18}{20 \times 19 \times 18}}{6}$

$=\frac{6}{20 \times 19}$

On simplifying we get

$=\frac{3}{190}$

$P($ three number are not consecutive $)=1-\frac{3}{190}$

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

$P($ three number are not consecutive $)=1-\frac{3}{190}$

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

$=\frac{187}{190}$

Hence, the correct option is (B).

$\mathrm{P}$ (three number are not consecutive) $=1-\frac{3}{190}$

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

$=\frac{187}{190}$

Hence, the correct option is (B).