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"?

Question

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?

Hi! I'm a student in middle school, so simpler terms would be very appreciated.

I was wondering if there was a way to solve this problem without solely relying on casework. My first approach was to count all the ways by seeing which lists worked and which didn't. For example, (1, 2, 3) works, while (1, 4, 6) doesn't. This was very tedious, and I sometimes wondered if I had missed any cases.

I ended up with 1/4, which I believe is correct. Is this the correct answer? If so, is there any way to get this probability without solely using casework?

Answer 1

Obviously, $1$ must be colored, because it divides everything; anything that doesn't color $1$ fails.

Since $2$, $3$, and $5$ are prime, we don't have to worry about those, mostly: only $1$ divide these, so since we need $1$ anyway, we're free to have or not have these and it affects nothing else. However, when we deal with other numbers we might need to consider these.

This leaves only $4$ and $6$. Time for some case work.

• Case 1: neither $4$ nor $6$ are colored. In this case, $2$, $3$, and $5$ are all free, so we get $2^3=8$ successful cases.
• Case 2: $4$ is colored but $6$ is not. In this case, $2$ needs to be colored, but $3$ and $5$ are free. This is $2^2=4$ successful cases.
• Case 3: $6$ is colored. Both $2$ and $3$ must be colored; since all of $4$'s divisors are colored, that one is free, along with $5$. This is $2^2=4$ successful cases.

So all told, we have $2^6=64$ total cases, and $8+4+4=16$ of them are successful, the answer is indeed $16/64 = 1/4$.