Answer:
[tex]\dfrac{1}{190}[/tex]
Step-by-step explanation:
This is a problem using combinatorics probability
The total number of combinations for k items out of a set of n items is given by
[tex]C(n, k) = \dfrac{n!}{k! (n-k)!}[/tex]
The left hand side is referred to as n choose k
! represents the factorial function
There are calculators to compute these values
Given 20 numbers the number of ways to choose 3 numbers is given by
[tex]C(10, 3) = \dfrac{20!}{3! (20-3)!}\\= 1140[/tex]
If we are specifically asked for the numbers 1, 2 and 3 in any order the total number of ways of ordering these numbers is 3! == 6 ways
{1, 2, 3}, {1, 3, 2}, 2, 3, 1}, 2, 1, 3}, {3, 1, 2} and {3, 2, 1}
Therefore the desired probability
= Number of ways of arranging 1, 2, 3 ÷ total number of ways in which you can select 3 numbers from 20 numbers
[tex]= \dfrac{6}{1140}\\\\= \dfrac{1}{190}[/tex]
