Consider the set of these 1008 consecutive odd numbers $S={3,5,7,\ldots,2017}$

Now I randomly choose 673 distinct elements of $S$ . What is the probability (in percent) that there exists 3 co-prime numbers among the chosen ones?

Give your answer to the nearest integer.

**
Bonus
**
: Generalize it!

The answer is 100.

**
This section requires Javascript.
**

You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.

Actually, I took inspiration from this problem.

Original problem:Randomly choose $n+1$ numbers from the firtst $2n$ positive integers.Prove that there are $2$ co-prime numbers in the choosen $n+1$Proof for the original problem:There must be at least $2$ consecutive numbers in the choosen numbers. These $2$ integers are co-prime (Q.E.D.)Simple, right?

We can solve

"253 days until 2017"similarly.Generalized problem: Randomly choose $2n+1$ numbers from the consecutive $3n$ odd numbers. Prove that there are 3 consecutive odd numbers in the choosen $2n+1$ .Proof: Suppose there are at most 2 consecutive, it is obvious that there are at most $2n$ choosen numbers(definitely false)With

$C$ : choosen number

$U$ : unselected number

The sequences below which maximize the choosen $2n+1$ form the $3n$ odd numbers:

$(CUC)$ $(CUC)$ $(CUC)$ $...$ $(CUC)$ (n groups)

$(CCU)$ $(CCU)$ $(CCU)$ $...$ $(CCU)$ (n groups)

$(UCC)$ $(UCC)$ $(UCC)$ $...$ $(UCC)$ (n groups)

Hence, there exists 3 consecutive odd numbers.(Q.E.D.). These 3 are also co-prime.

The answer is

100 %