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?

Thanks! Can I get a thumbs up?

MIRB16
Apr 4, 2018

#1**0 **

See the answer here. If not correct, let us know:

**https://web2.0calc.com/questions/help_47515**

Guest Apr 4, 2018

#2**0 **

The solution to this question requires analysis of mutually exclusive (conditional) sets. Each set has an individual probability. The sum of these probabilities times the probability of choosing one of the sets determines the overall probability of a “factorific” coloring.

From the question: The condition requires all divisors of a blue number to be blue. Note a number can be blue __without__ it being a divisor of all greater numbers __as long as its divisors are blue__.

*Grogg chooses to color **none** **(0)** of the numbers.

__Probability of factorific__: Zero (0)

*Grogg chooses to color **one (1) **of the six (6) numbers.

__Probability of colorific__: **(1/6).**

__Explanation:__ Number one (1) is the only positive integer that has one divisor.

*Grogg chooses to color **two** **(2)** of the six (6) numbers.

__Probability of factorific__: = **(1/5)**

__Explanation:__ Number of ways to choose two numbers from 6 numbers = (nCr(6, 2)) = 15.

Number one (1) must be in the set because it’s a divisor of all integers. The other number has to be a prime number (2, 3, or 5)

{1, 2}

{1, 3}

{1, 5}

Number of sets that have a one (1) and a prime: Three (3).

Probability (3/15) = (1/5)

*Grogg chooses to color **three (3)** of the six (6) numbers.

__Probability of factorific__: (5/20) = **(1/5)**

__Explanation:__ Number of ways to choose 3 numbers from 6 numbers = (nCr(6, 3)) = 20

Number one (1) must be in the set because it’s a divisor of all integers.

These 4 sets meet the conditions:

{1, 2, 3} Two (2) is blue here, it’s not a divisor of three, but one (1) is its divisor, so this is valid.

{1, 2, 4} all divisors are blue for all blue numbers.

{1, 2, 5} ‘’

{1, 3, 5} ‘’

~~{1, 5, 6} ‘’~~

Five of 20 sets meet the conditions: (1/5)

*Grogg chooses to color **four (4)** of the 6 numbers.

__Probability of factorific__: **(4/15)**

__Explanation:__ Number of ways to choose 4 numbers from 6 numbers = (nCr(6, 4)) = 15

Number one (1) must be in the set because it’s a divisor of all integers.

These 3 sets meet the conditions

{1, 2, 3, 4} all divisors are blue for all blue numbers.

{1, 2, 3, 5} ‘’

{1, 2, 3, 6} ‘’

{1, 2, 4, 5} ‘’

Four of 15 sets meet the conditions: (4/15)

*Grogg chooses to color **five** **(5)** of the six (6) numbers.

__Probability of factorific__: **(1/2)**

__Explanation:__ Number of ways to choose 5 numbers from 6 numbers = (nCr(6, 5)) = 6

Number one (1) must be in the set because it’s a divisor of all integers. Number of sets of five (5) that have one (1) as an element = (nCr(5, 4)) = 5.

Three sets meet the conditions

{1,2,3,4,5} all divisors are blue for all blue numbers.

{1,2,3,4,6} ‘’

{1,2,3,5,6} ‘’

Three sets of 6 sets meet the conditions (3/6) = (1/2)

*Grogg chooses to color six of the six (6) numbers.

__Probability of factorific__: **(1) or 100%**

Explanation:All numbers are blue and all numbers have all their divisors colored blue.

---------

Sum of individual (mutually exclusive) probabilities: ((0)+(1/6)+(1/5)+(1/4)+(4/15)+(1/2)+(1))

Grogg has a (1/7) probability of picking one of the seven sets (including the empty set).

~~(1/7)*((0)+(1/6)+(1/5)+(1/4)+(4/15)+(1/2)+(1)) = (143/420) ~~**≈ **34.05**%**

(1/7)*((0)+(1/6)+(1/5)+(1/5)+(4/15)+(1/2)+(1)) = (143/420) ≈ 33.33%

The overall probability that Grogg’s coloring is factorific is **≈ 33.33%**

**Sources: A genetically enhanced chimp brain, and comprehensive programming of basic set theory from Lancelot Link.**

GA

Edit: Corrected error

GingerAle
Apr 5, 2018

#3**0 **

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?

jollyjellyjojo
Apr 6, 2018

#4

#6**-1 **

I think I need powerful anti-nausea meds to wade through this irrelevant, wordy, neurotic swill. Jesus Christ! __It’s amazing that someone far surpassed Mr. BB in pasting mathematically infused slop on this forum.__ It’s interesting that someone can write reasonably tight LaTex code but doesn’t know how to implement its execution here; probably because the math part doesn’t belong to the poster, but I’m sure the neurotic commentary does.

After pumping the sweet swill down the sewer, the math part seems fine. It’s not the correct answer though. Jojo didn’t correctly perceive my “percception,” else the answer would be the same.

GA

GingerAle
Apr 9, 2018

#5**0 **

**I have made an ****assumtion**** here. The likelihood that any individual number is blue is 50%**

For this to work number 1 must be blue because it is a factor of all the others.

Since 1is blue, it really doesn't matter whether 2,3 or 5 are blue as their only factor other then themselves is 1.

6 blue , 3blue, 2 blue, 1 blue others don't matter. Prob = 0.5^4 = 1/16

6 not blue, 4 blue, 2 blue, 1 blue others don't matter Prob = 1/16

6 not blue, 4 not blue, 1 blue and others don't matter Prob = 1/8

1/16 + 1/16 + 1/8 = 4/16 =** 1/4 **

I think that the chance that his colouring is factoric is ** 1/4**

**Melody Mar 18, 2018**

**edited by Melody Mar 18, 2018**

Guest Apr 6, 2018

#7**0 **

Continued from above.

Because of the buckets of slop, it’s not easy to tell that Jojo’s answer is the same as Melody’s result; it’s still the wrong answer. The reason is the technique used for the solution does not represent the method Grogg uses to select which numbers are colored.

My answer is spot on. I can “prove” my answer is correct, though not with precise mathematics, because I lack the proficiency. (Lancelot could explain why in a blink.) My solution method follows the selection schemes (0-6) Grogg uses to color the numbers. Choose 0 through 6 numbers (for choices of 1-5 the numbers are selected randomly), color them, then test if it’s factorific. Each scheme has an individual probability of producing a factorific coloring.

Taking the sum of probabilities for each individual scheme and multiplying by a scheme’s selection probability (1/7), gives the true probability of selecting a factorific coloring. This is different from assuming a number has a 50% probability of being blue. The net change is 75% greater than the ratio of the number of success to the total sample space (1/3) vs, (1/4).

This question and others like it is interesting to me because a random selection process produces a success rate (factorific coloring). This question has minimal interest. There is no comment from the person who asked it. There are only 17 additional views since I posted my answer. Melody has not commented on it –she may not have seen it.

When Lancelot Link was teaching me combinatorics and statistics, he explained that there were many statistics questions where the “natural” solution was wrong. The most famous of these is the “Monty Hall” problem. The correct solution to this problem eluded many highly educated and experienced statisticians. Many disagreed with the correct solution after its presentation. Lancelot said though this is quantifiable by using Bayesian statistics, it was easy enough for any dumb dumb to construct a Monte Carlo simulation to demonstrate the correct solution.

This question does not require Bayesian statistics to solve –a simple weighting of the expectations solves this. To be sure that I’m not full of blarney and BS (which is becoming epidemic on here), I will write a Monte Carlo simulation. I’m certain the results will hover very close to the calculated (1/3) probability. Further, analyzing the data will show a mostly flat distribution of the numbers—in both the selected sets (0-6) and selected numbers (1-6).

GA

GingerAle
Apr 9, 2018

#8**0 **

I guess I submitted my answer and got it right???? It is a bit wordy

although GingerAle I used the fact how you used your casework and how you organized them.

And there's no need to be rude? Chill? pls?

Uh, This work did actually belong to me, although like I said, is wordy and not the most efficient way, :/

"You correctly used constructive counting within casework to count the number of factorific colorings and total number of colorings, then used these values to calculate the desired probability.

Your casework to count the successes did work, but isn't the most efficient such approach as you noted. Plus, it is hard to check some of your cases to ensure you found all of the successful colorings and did not miss any, although your method of "combining" smaller sets could have helped you do this. Can you think of a faster and simpler approach here? You may still need some form of casework, although not necessarily based on the number of blue numbers. There are other approaches that have fewer cases and less complex ways to count the successes within each case.

Your second case when counting the successes was not labeled correctly. Also, it is a bit confusing to label the cases with bolded combinations, since some of those technically have the same value. Instead, you could mark them with a short description like "Exactly numbers colored:" and so on."

That was from AoPS???

"My answer is spot on. I can “prove” my answer is correct,"

-???

"I think I need powerful anti-nausea meds to wade through this irrelevant, wordy, neurotic swill. Jesus Christ! It’s amazing that someone far surpassed Mr. BB in pasting mathematically infused slop on this forum. It’s interesting that someone can write reasonably tight LaTex code but doesn’t know how to implement its execution here; "

-Uh...Ok??? I'm sorry if it is not pleasing to you, but ¯\_(ツ)_/¯

jollyjellyjojo
Apr 11, 2018

#9**0 **

^{I guess I submitted my answer and got it right???? }

You guess you submitted your answer? Either you did or didn’t.

^{It is a bit wordy}

No kidding; who’d ever notice?

^{although GingerAle I used the fact how you used your casework and how you organized them. }

Then you submitted my answer, not yours. Both of our answers cannot be correct.

^{ “...}

^{Your casework to count the successes did work, but isn't the most efficient such approach as you noted.}

I made no comment about efficiency

^{Your second case when counting the successes was not labeled correctly. }

That was the fourth case, and it was corrected. This was an easy catch, if you had reviewed the work.

^{Also, it is a bit confusing to label the cases with bolded combinations, since some of those technically have the same value. Instead, you could mark them with a short description like "Exactly numbers colored:" and so on." That was from AoPS???}

W__T__F! My presentation is a concept based teaching post; it’s not something I would submit to an __AoPS__ teacher/professor, or any teacher/professor. If you wanted to use my solution method, you should have reworded it, but not with your neurotic __S__weet __P__olly __p__ure __p__iss encouragement preceding every number.

^{"My answer is spot on. I can “prove” my answer is correct,"}

^{-???}

I completed the Monte Carlo simulations. As expected, it hovers near 1/3 using the “Grogg” selection method. Randomly choosing to color or not to color numbers 1-6, gives a slight variation around a 25% probability for a factorific coloring, and the counts of colored numbers, from 0 to 6 tends toward a binomial distribution. There are other interesting statistics based on various restrictions.

^{And there's no need to be rude? Chill? pls?}

Yes there is. You assaulted the senses of everyone who read your post. At least, those who are not on thorazine or have a natural version of it.

^{ I'm sorry if it is not pleasing to you}

You’re not, and that’s an understatement!

GA

Edit Corrected typo.

GingerAle
Apr 11, 2018

#10**0 **

Ok so I think you misunderstood my statement, and I did submit my answer. I SUBMITTED MY ANSWER.

"You correctly used constructive counting within casework to count the number of factorific colorings and total number of colorings, then used these values to calculate the desired probability.

Your casework to count the successes did work, but isn't the most efficient such approach as you noted. Plus, it is hard to check some of your cases to ensure you found all of the successful colorings and did not miss any, although your method of "combining" smaller sets could have helped you do this. Can you think of a faster and simpler approach here? You may still need some form of casework, although not necessarily based on the number of blue numbers. There are other approaches that have fewer cases and less complex ways to count the successes within each case.

Your second case when counting the successes was not labeled correctly. Also, it is a bit confusing to label the cases with bolded combinations, since some of those technically have the same value. Instead, you could mark them with a short description like "Exactly numbers colored:" and so on."

This comment was from a reviewer on AoPS

And I did not submit your answer, I submitted mine.

And If you do want to see the answer OFFICIALLY posted by AoPS, here it is

We will compute this asThe total number of colorings is just because there are two choices for each number, it can be either blue or not blue.

The number of factorific colorings is a bit more difficult to compute. First, we notice that every factorific coloring has 1 colored blue.

So, we only have to deal with the numbers 2 through 6. The only overlap in divisors comes from 2 being a factor of both 4 and 6, so we count by casework on whether 4 and 6 are colored blue.

Case 1: 6 is colored blue and 4 is not.

In this case, we must color 2 and 3 blue, because they are divisors of 6.

We know 4 is not colored blue, and 5 can go either way. So, there are 2 such colorings.

Case 2: 6 and 4 are both blue.

In this case, we must color 2 and 3 blue, because they are divisors of 6.Again, 5 can go either way. So, there are again 2 colorings in this case.

Case 3: 4 is blue and 6 is not.

In this case, we know that 2 is colored blue and 6 is not.There are two options for both 3 and 5. So there are such colorings in this case.

Case 4: Neither 4 nor 6 is blue.

In this case, we can simply color 2, 3, and 5 however we want. There are ways to do that.

Combining our work, we find that there are factorific colorings. So, the probability that Grogg's coloring is factorific is 1/4.

1/4. YES MY ANSWER IS CORRECT.

And it is not nice to go around media like this with an attitude like that.

jollyjellyjojo
Apr 11, 2018

#11**0 **

You submitted your solution to AoPS and the reviewer said your solution (1/4) is correct. Well, assuming you are not telling us a big fib (that means lying), the reviewer is wrong. I’m sure the official answer/solution goes well into the hierarchy, where a Ph.D. approved it, without a proper review. The wrong answers that often appear in the back of math books also receive approval. Once a solution is believed correct, the analysis ends, and that’s the end of it.

This question has some powerful juju, Jojo. (Juju means magical karma.) Despite the impeccable credentials of AoPS Professors, teachers, and reviewers, (1/4) is the WRONG answer. The answer is wrong is because the solution method is wrong for this question. It should be obvious that Grogg does not arbitrarily choose whether to color a number blue; if he did then that would be the correct solution.

Grogg randomly chooses a set of numbers from zero to six, then he randomly colors numbers blue to match the count of the set. If Grogg chooses three numbers then he randomly colors three of the six numbers blue. If he chooses zero numbers the he colors none, if he chooses 6 then he colors all six. Sets of zero, one, two, three, etc, all have an equal (1/7) probability of selection. Any of the six numbers also have an equal probability of selection in each set.

However, if Grogg considers each of the six numbers individually, and randomly chooses to color it or not, then the “set counts” become a binomial distribution: 1,6,15,20,15,6,1.

The sum of this binomial distribution is 64 and the probabilities of Grogg selecting each set are 1.5625%, 9.375%, 23.4375%, 31.25%, 23.4375%, 9.375%, and 1.5625%.

As you can see this distribution is very different from the uniform probability of 14.285% for each set, where Grogg chooses each set independently.

In any case, good for you, for getting the correct answer. One lesson I never learned very well in the lower grades is, “the ‘correct’ answer is the one the teacher says it is.” This is good advice because the teacher usually is correct. Personally, though, if there is a conflict, I’d rather have a lower grade and the correct solution. I definitely got the lower grade, but I was less sure about the correct solution.

In this case, I’m very sure of the correct solution.

^{And it is not nice to go around media like this with an attitude like that.}

Really? What kind of media would make it nice?

GA

GingerAle
Apr 13, 2018