GingerAle

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 #2
avatar+2436 
+2

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.laugh

 

GA

 

Edits: Error corrections

Another solution and more comments here.   

 

.

Apr 5, 2018
 #10
avatar+2436 
+2

Mr. BB you are an enigma of chaotic stupidity!

 

First, you post this dumb blarney:

 

hectictar: You calculated the SA of a sphere. Wouldn't it be more accurate if you calculated it as SA of a circle, since it says "Assuming that the powdered sugar coating has negligible thickness and is distributed equally on all donut holes"??

At any rate, if you calculated them as circles, your numbers would be:

Niraek = 36pi

Theo   = 64pi

Akshaj= 40pi

Since the LCM of [36, 64, 40] =2,880, therefore:

Niraek will have finished = 2,880 / 36 =80 donut holes

Theo will have finished   = 2,880 /64  =45 donut holes

Akshaj will have finished= 2,880 /40  =72 donut holes.

Note: Somebody should check these numbers. hectictar: what do you think of this approach?

 

Then, after Hecticar answers, you replace the post with this:

 

This doesn't lead anywhere !!

 

Deleting your question after specifically asking Hectictar to answer, and receiving an answer, is just blòódy rude.  It looks like Hectictar is replying to empty air instead of a gasbag full of hot air.

 

Now you’ve replaced the post with this BS:

 

I think you should find the LCM of [6, 8, 10] =120, so you have:

120 / 6 =20 holes for N

120 / 8 =15 holes for T

120 / 10 =12 holes for A

 

(This is dumber blarney, because it’s not a square area)

 

Mr. BB, you are failing, fast! The toxic blarney circulating in your system is accelerating your dementia! You should check in to a neurological treatment center ASAP. They can’t save you, but maybe they will euthanize you and put you out of our misery.

 

 

GA

Mar 23, 2018