91Ó°ÊÓ

The total number of ways to select 3 distinct integers from the set {1,2,3, ...., 1000} is given by the binomial coefficient, also known as a combination. The notation is C(n, k) where n is the total number of items (in our case, 1000), and k is the number of items to choose (in our case, 3). Using the formula for combinations, we find C(1000, 3) = \(\frac{1000!}{3!(1000-3)!}\)

When a number is divisible by 3, it means that the sum of its digits is also divisible by 3. We need to find the number of ways to select 3 distinct numbers so that their sum is divisible by 3. That happens when the three numbers selected are either all divisible by 3 or two of them give a remainder 1 and the other gives a remainder 2 when divided by 3. We can find these two probabilities separately and then add them together since they are disjoint events. For the case where all three numbers are divisible by 3, there are 334 such numbers between 1 and 1000, hence, the number of sets is C(334, 3). For the case where two numbers give a remainder of 1 and one number gives a remainder of 2, there are 333 numbers between 1 and 1000 that give a remainder of 1 when divided by 3 and similarly, there are 333 numbers that give a remainder of 2. Hence, the number of sets is C(333, 2) * 333 = 333 * C(333, 2).

Now that we have the total number of sets and the number of sets whose sums are divisible by 3, we can find the probability. Probability is defined as the number of desired outcomes (sets whose sum is divisible by 3) over the total number of outcomes (total number of sets). Hence, the probability is \(\frac{C(334, 3) + 333 \cdot C(333, 2)}{C(1000, 3)}\)