To solve this problem, we need to find the probability of selecting three distinct numbers from the first 100 natural numbers that are divisible by both 2 and 3.

Total Possible Outcomes:
The total number of ways to select 3 distinct numbers from the first 100 natural numbers is given by the combination formula: C(100, 3) = 100! / (3! * (100 - 3)!) = 100 * 99 * 98 / (3 * 2 * 1) = 161,700.

Favorable Outcomes:
Out of the first 100 natural numbers, half of them are divisible by 2 (since every other number is divisible by 2), and one-third of them are divisible by 3 (since every third number is divisible by 3).

The numbers that are divisible by both 2 and 3 are the numbers that are divisible by their least common multiple, which is 6. We can determine the number of such numbers by dividing 100 by 6 and taking the floor value: ⌊100/6⌋ = 16.

Therefore, there are 16 numbers from 1 to 100 that are divisible by both 2 and 3.

Probability Calculation:
The probability of selecting three distinct numbers that are divisible by both 2 and 3 can be calculated as the ratio of the number of favorable outcomes to the total number of possible outcomes:

Probability = Number of Favorable Outcomes / Total Number of Possible Outcomes

Probability = 16 / 161,700

Simplifying this fraction, we get:

Probability = 4 / 40,425

Dividing both the numerator and denominator by 5, we get:

Probability = 4 / 8,085

Further simplifying by dividing both the numerator and denominator by 15, we get:

Probability = 4 / 1,355

Finally, dividing both the numerator and denominator by 5, we arrive at:

Probability = 4 / 1,155

Therefore, the probability that all three numbers selected are divisible by both 2 and 3 is 4/1,155, which corresponds to option D.