To solve the problem, we need to find the probability that all three distinct numbers selected from the first 100 natural numbers are divisible by both 2 and 3.
### Step-by-Step Solution:
1. **Understanding Total Outcomes**:
The total number of ways to choose 3 distinct numbers from the first 100 natural numbers is given by the combination formula \( \binom{n}{r} \), where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose. Here, \( n = 100 \) and \( r = 3 \).
\[
\text{Total Outcomes} = \binom{100}{3} = \frac{100 \times 99 \times 98}{3 \times 2 \times 1} = \frac{970200}{6} = 161700
\]
2. **Finding Favorable Outcomes**:
We need to find the numbers that are divisible by both 2 and 3. The least common multiple (LCM) of 2 and 3 is 6. Therefore, we need to find the multiples of 6 within the first 100 natural numbers.
The multiples of 6 from 6 to 96 are:
\[
6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96
\]
To find how many multiples of 6 are there, we can use the formula for the \( n \)-th term of an arithmetic progression (AP):
\[
a_n = a + (n-1)d
\]
where \( a = 6 \) (the first term), \( d = 6 \) (the common difference), and \( a_n = 96 \) (the last term).
Setting up the equation:
\[
96 = 6 + (n-1) \cdot 6
\]
\[
90 = (n-1) \cdot 6
\]
\[
n - 1 = 15 \implies n = 16
\]
Thus, there are 16 numbers that are divisible by both 2 and 3.
3. **Calculating Favorable Outcomes**:
Now, we need to choose 3 distinct numbers from these 16 favorable outcomes:
\[
\text{Favorable Outcomes} = \binom{16}{3} = \frac{16 \times 15 \times 14}{3 \times 2 \times 1} = \frac{3360}{6} = 560
\]
4. **Calculating the Probability**:
The probability \( P \) that all three selected numbers are divisible by both 2 and 3 is given by the ratio of favorable outcomes to total outcomes:
\[
P = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{560}{161700}
\]
To simplify this fraction:
\[
P = \frac{560 \div 280}{161700 \div 280} = \frac{2}{577.5} \approx \frac{4}{1155}
\]
### Final Answer:
The probability that all three numbers selected are divisible by both 2 and 3 is:
\[
\frac{4}{1155}
\]
