To solve the problem in the image:

Problem Analysis

We need to determine the probability that a randomly chosen number from a set $M$ is divisible by 3. The set $M$ consists of all single-digit, two-digit, and three-digit numbers that can be formed using the digits $1, 2, 3$, without repetition in any number.

Step 1: Total Count of Numbers

1. Single-digit numbers: There are 3 possibilities ($1, 2, 3$).
2. Two-digit numbers: Using the digits $1, 2, 3$ without repetition, the count is: $P(3, 2) = \frac{3!}{(3-2)!} = 6$
3. Three-digit numbers: All permutations of $1, 2, 3$, the count is: $P(3, 3) = 3! = 6$

Thus, the total number of elements in $M$ is: $3 + 6 + 6 = 15$

Step 2: Numbers Divisible by 3

A number is divisible by 3 if the sum of its digits is divisible by 3. We calculate this for each subset:

1. Single-digit numbers:

   • Only $3$ is divisible by 3.

2. Two-digit numbers:

   • Possible numbers: $12, 13, 21, 23, 31, 32$
   • Check the sums:
     • $12: 1+2=3$ (divisible by 3)
     • $21: 2+1=3$ (divisible by 3)
     • Other sums ($4, 5$) are not divisible by 3.
   • Divisible numbers: $12, 21$ (2 numbers).

3. Three-digit numbers:

   • Possible numbers: $123, 132, 213, 231, 312, 321$
   • Check the sums:
     • $123: 1+2+3=6$ (divisible by 3)
     • $132: 1+3+2=6$ (divisible by 3)
     • $213: 2+1+3=6$ (divisible by 3)
     • $231: 2+3+1=6$ (divisible by 3)
     • $312: 3+1+2=6$ (divisible by 3)
     • $321: 3+2+1=6$ (divisible by 3)
   • Divisible numbers: All 6 numbers.

Thus, the total count of numbers divisible by 3 is: $1 (\text{single-digit}) + 2 (\text{two-digit}) + 6 (\text{three-digit}) = 9$

Step 3: Probability

The probability that a randomly chosen number is divisible by 3 is: $P = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{9}{15} = \frac{3}{5}$

Final Answer:

The probability is $\frac{3}{5}$.

Would you like further clarification or step-by-step details? Here are 5 related questions you might explore:

1. How is divisibility by 3 determined mathematically?
2. What is the general formula for permutations with restrictions?
3. How does this change if repetition of digits is allowed?
4. Can we apply the same logic for other divisors like 2 or 5?
5. How do probabilities change if the number length is fixed (e.g., only three-digit numbers)?

Tip: Always check divisibility rules carefully when working with probabilities involving number properties.