Question

The numbers 1 through 9 are written in separate slips of paper, and the slips are placed into a box. Then, 4 of these slips are drawn at random.
What is the probability that the drawn slips are "1", "2", "3", and "4", in that order?

Answer 1

Let's think of the problem as follows.

Write all the 4-digit numbers that can be formed using the digits from 1 to 9, without repetition, in pieces of paper, and put them in a bag. What is the probability of picking the 4-digit number 1234, among these numbers.

The connection of the 2 problems is as follows:

The 4-digit number, for example 5489, represents drawing first 5, then 4, then 8, then 9 , in the original question.

we did not allow repetition, because for example the number 8918 would represent drawing 8, then 9, then 1 then 8 (again!!), which is not possible, so we lose the connection between the problems.


So there are in total 9*8*7*6= 3024  4-digit numbers, with non-repeating digits.

One of these numbers is 1234 (representing drawing 1, then 2, then 3, then 4)

among these 3024 numbers, the probability of picking 1234 is 
$\frac{1}{3024}= 0.00033$


We could have solved this problem also as :

P(drawing 1, 2, 3, 4 in order)= $\frac{1}{9} * \frac{1}{8} * \frac{1}{7} * \frac{1}{6} = \frac{1}{3024}$

Answer:$\frac{1}{3024}= 0.00033$