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Step-by-Step Solution
Step 1: Identify how the numbers are distributed modulo 3
Consider a set of 3n consecutive natural numbers. When dividing numbers by 3, every set of three consecutive numbers contains:
Exactly one number that is divisible by 3 (remainder 0)
One number that leaves remainder 1
One number that leaves remainder 2
Hence, in 3n consecutive numbers, there will be:
n numbers with remainder 0 (multiples of 3)
n numbers with remainder 1
n numbers with remainder 2
Step 2: Calculate the total number of ways to choose 3 numbers
The total ways to choose any 3 numbers out of 3n is given by the binomial coefficient:
$ \binom{3n}{3} $.
Step 3: Count favourable outcomes (sum of chosen numbers divisible by 3)
The sum of three numbers is divisible by 3 in the following scenarios:
All three numbers come from the same remainder class.
They are all multiples of 3 (remainder 0), or all have remainder 1, or all have remainder 2.
The three numbers come from each of the three different remainder classes.
One number has remainder 0, another remainder 1, and another remainder 2.
Case A: All three from the same remainder class
Within each remainder class, there are n numbers. The number of ways to pick 3 from the same class is
$ \binom{n}{3} $. Since there are 3 such classes (0, 1, and 2 modulo 3), total ways are:
$ 3 \times \binom{n}{3} $.
Case B: One number from each remainder class
We select:
1 number from the n multiples of 3,
1 number from the n numbers giving remainder 1,
1 number from the n numbers giving remainder 2.
Thus, the number of ways is:
$ n \times n \times n = n^3 $.
Step 4: Combine the two cases
The total number of favourable outcomes is:
$ 3 \binom{n}{3} + n^3 $.
Step 5: Calculate the probability
The probability is the ratio of the favourable outcomes to the total number of outcomes:
$$
\text{Probability}
= \frac{3 \binom{n}{3} + n^3}{\binom{3n}{3}}.
$$
On simplification, this matches:
$$
\frac{\bigl(3n^2 - 3n + 2\bigr)}{\bigl(3n - 1\bigr)\bigl(3n - 2\bigr)}.
$$
