The probability is 12\dfrac{1}{2}21​.

Quick Scoop: Core Idea

A three-digit number is formed using the digits 1, 2, 3, and 4 without repetition.
We want:

Probability that this number is divisible by 3.

A number is divisible by 3 if the sum of its digits is divisible by 3.

Step 1: Total possible three-digit numbers

We must form three-digit numbers using 1, 2, 3, 4 without repetition.

  • Number of ways = permutations of 4 digits taken 3 at a time
  • That is 4P3=4!=244P3=4!=244P3=4!=24 total possible three-digit numbers.

Step 2: Check which digit triples give sum divisible by 3

Digits: 1, 2, 3, 4
We choose any 3 of them and look at the sum:

  • 1+2+3=61+2+3=61+2+3=6 β†’ divisible by 3 βœ…
  • 1+2+4=71+2+4=71+2+4=7 β†’ not divisible by 3 ❌
  • 1+3+4=81+3+4=81+3+4=8 β†’ not divisible by 3 ❌
  • 2+3+4=92+3+4=92+3+4=9 β†’ divisible by 3 βœ…

So only these sets work:

  • {1,2,3}\{1,2,3\}{1,2,3}
  • {2,3,4}\{2,3,4\}{2,3,4}

Each such set of 3 distinct digits can be arranged in:

  • 3!=63!=63!=6 different three-digit numbers.

So:

  • From {1,2,3}\{1,2,3\}{1,2,3}: 6 numbers
  • From {2,3,4}\{2,3,4\}{2,3,4}: 6 numbers

Total favourable numbers = 6+6=126+6=126+6=12.

Step 3: Probability

Probability=favourable outcomestotal outcomes=1224=12\text{Probability}=\frac{\text{favourable outcomes}}{\text{total outcomes}} =\frac{12}{24} =\frac{1}{2}Probability=total outcomesfavourable outcomes​=2412​=21​

So, the probability that the three-digit number formed is divisible by 3 is:

12\boxed{\dfrac{1}{2}}21​​

TL;DR: Half of all 24 possible three-digit numbers you can make from 1, 2, 3, 4 without repetition are divisible by 3, so the probability is 1/21/21/2.

Information gathered from public forums or data available on the internet and portrayed here.