Ok, so I'll use GingerAle's percception of the question, although I did get a different answer

MIRB16 I also think you are in my AoPS class???

[b]The numbers $1,2,3,4,5,6$ are written down and some of them are colored blue. A coloring is called factorific if there is at least one blue number, and for each blue number, all of its divisors are also blue.

If Grogg randomly colors some, all, or none of the numbers from 1 to 6 blue, what is the probability that his coloring is factorific?[/b]

So, to find the probability, we have to first find the total number of possibilities.

If Grogg colors none of them, that is $\binom{6}{0}$ ways he can color

If Grogg colors 1 number, that is $\binom{6}{1}$ ways he can color

If Grogg colors 2 numbers, that is $\binom{6}{2}$ ways he can color

If Grogg colors 3 numbers, that is $\binom{6}{3}$ ways he can color

If Grogg colors 4 numbers, that is $\binom{6}{4}$ ways he can color

If Grogg colors 5 numbers, that is $\binom{6}{5}$ ways he can color

If Grogg colors 6 numbers, that is $\binom{6}{6}$ ways he can color

$$\binom{6}{0} + \binom{6}{1} + \binom{6}{2} + \binom{6}{3} + \binom{6}{4} + \binom{6}{5} + \binom{6}{6}$$

$$1 + 6 + 15 + 20 + 15 + 6 + 1$$

$$64$$

There are 64 ways to color numbers.

Now we have to set some rules for our casework.

For this casework, let's include the factorifics that have more than one main "blue" number, so 5 and 6 could be colored as main numbers (just an example).

PS: I will also can the numbers colored "sets"

$\bf\binom{6}{0}$

For this option, this cant work because 0 has no divisors! (You cannot divide by 0)

c

Since we're focusing on numbers with only 1 divisor, that would be ... only 1. The reason is because if you only have to have 1 divisor, then the divisor HAS to be the number itself - and hat only applies for one because when you divide 1 by 1, it is 1.

$\bf\binom{6}{2}$

Theses numbers can only have 2 divisors, and that means it can't be composite numbers, because the minimum amount of divisors a composite number can have are 3 divisors (Perfect Square Numbers). That means that the numbers are all [b]prime[/b]. The only numbers that would work are prime numbers less than 6, and those are 2, 3, and 5. So the numbers colored are

$1,2$

$1,3$

$1,5$

$\bf\binom{6}{3}$

Now here is where it gets a little bit confusing, but it is manageable, same for the rest.

So, we can first list out the perfect square numbers, (they have 3 divisors, remember) and the only one less than 6 is 4.

$1, 2, 4$

Now, the reason why the rest are a bit more confusing is because we are combining sets from the previous case to make our present case, so that means cases like, say, $1, 2, 3$ will work because $1, 3$ is one and $1, 2$ is the next. The next set of colored numbers that work are $1, 2, 5$ because again, they include two sets from the previous. The next set is $1, 3, 5$ because there are two different cases (again) 3, and 5. And if we try the next set... we see that there are no more sets left that meet the specifications.

$1, 2, 3$

$1, 2, 4$

$1, 2, 5$

$1, 3, 5$

$\bf\binom{6}{4}$

Again, it's a bit more complicated but we can handle it

Ok, so the actual composition numbers besides perfect square numbers under or equal to 6 are only...6. 6 has the divisors, or factors, $1, 2, 3, 6$

$1, 2, 3, 6$

The next ones we will combine from the previous case.

The first set should be $1, 2, 3, 5$ because all of the sets apply here, too. The next set should be $1, 2, 3, 4$ because that applies too - the perfect square 4 and the prime number 3 to provide this set. and lastly, we have the set $1, 2, 4, 5$ and we can apply the same concept from the set $1, 2, 3, 4$, 5 is also a prime number.

$1, 2, 3, 4$

$1, 2, 3, 5$

$1, 2, 3, 6$

$1, 2, 4, 5$

$\bf\binom{6}{5}$

Hang in there, we're almost done!

So, this, we clearly don't have a number with 5 divisors that is below or equal to 6, because 6 had 4 divisors and it was the largest number possible, so now all we can do is to combine sets (again)

The first set we can try is by combining the two sets $1, 2, 3, 5$ and the set $1, 2, 4$, and what results is $1, 2, 3, 4, 5$, which works! The next possible set can be $1, 2, 3, 4, 6$, but this set is composed of the sets $1, 2, 3, 6$ and $1, 2, 4$. The last, one we can try is $1, 2, 3, 5, 6$ composed of the sets $1, 2, 3, 5$ and the set $1, 2, 3, 6$.

$1,2,3,4,5$

$1,2,3,4,6$

$1,2,3,5,6$

$\bf\binom{6}{6}$

And finally, the last possibility where Grogg colors all of the numbers. How delightful :)

This case has only one set $1, 2, 3, 4, 5, 6$, and all we have to do is to check this if it works. we can obviously do this by combining the two sets from before $1,2,3,4,5$ and $1,2,3,5,6$. It does work! Now we can put everything together.

$\bf\binom{6}{0}$ has 0 cases

$\bf\binom{6}{1}$ has 1 case

$\bf\binom{6}{2}$ has 3 cases

$\bf\binom{6}{3}$ has 4 cases

$\bf\binom{6}{4}$ has 4 cases

$\bf\binom{6}{5}$ has 3 cases

$\bf\binom{6}{6}$ has 1 case

$\frac{1+3+4+4+3+1}{64} = \frac{16}{64} = \frac{1}{4}$

The probability that Grogg has colored in a factorfic is $\bf\frac{1}{4}$.

Hopefully that helped?