GingerAle

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

Edits: Error corrections

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

Heureka, your answer is correct if a box is restricted to a maximum of one piece of candy.

nCr(4, 1) +  nCr(3, 2) = 12

Reasoning:

There are nCr(4, 1)  =  4 ways to place the unique piece of candy in one of the four (4) boxes.  Then the 2 indistinguishable candies can be placed in the other 3 boxes nCr(3, 2)  = 8 ways.  There are (4+8) = 12 ways to distribute the candies.

If there are no restrictions on the maximum then:

\(\hspace{15pt} \left( \dfrac{4!}{2!}\right)* \left(2 \right) \;+\; \left( \dfrac{4!}{2!}\right) \;+\; \left(4\right)\\ \text {Reasoning:}\\ \text {All candy in 1 of 4 boxes (4 ways)}\\ \text {2 combinations (XX|Y & XY|X) distributed to boxes in (2)(4*3) = (2)(4!/2!) =(24) ways;}\\ \text {and 1 candy to each box distributed to boxes in (1)(4*3*2)/2 = 4!/2! = (12) ways.} \\ \text {Total 4+24+12 = 40 ways.} \)

Solution:

\(\text{Period } = \dfrac{2\pi}{c} = 1, \text{so } c = 2\pi\\ \text{Then } x = 20e^{(-bt)}* cos(2\pi*t)\\ \text{When t = 3, then x = 10}\\ \Rightarrow 10 = 20e^{(-3b)}cos(6\pi)\\ 0.5 = e^{(-3b)} * 1\\ Ln(0.5) = (-3b)\\ -3b = Ln(0.5)\\ -3b = -0.69314718\\ b= 0.23104906018\\ \text { }\\ 20e^{(-0.23104906018*3)}cos(6\pi*3)= 10 \text{ (check)}\\ a=20, \; b=0.23104906018, \; c=2\pi \)

_{Edit: Cosmetic}

GA

EP, the heat of fusion for 1g of water at 0 °C is approximately 334 joules or 79.7 calories. 

GA

To convert the standard deviations to percentage of AUC, use a Z-chart or calculator.

Using a Z-chart, find the 1.5 and 0.5 Sdvs area values, then subtract to find the area range.

The area from 0.5 to 1.5 Sdv is 0.9332 of the area (above 1.5) minus 0.6915 of the area (below 0.5) giving 24.2% of the total area.

Here are the percentages for the other deviations.

(A)  1.5 standard deviations above the mean or higher

       (1-0.9332) = 0.0668                          6.7%

(B)   0.5 to 1.5 standard deviations above the mean 

        (0.9332-0.6915) = 0.2417              24.2%

(C)   within 0.5 standard deviation of the mean

       (0.6925 - 0.3085) = 0.384              38.4%

(D)   0.5 to 1.5 standard deviations below the mean

       (0.3085 - 0.0668) = 0.2417           24.2%

(F)   1.5 standard deviations below the mean or lower

        (1-0.9332) = 0.0668                      6.7%

Note: when the standard deviations are below the mean the values are negative, use the “negative” or “left of mean” chart (or adjust by subtracting from 1).  

GA

Long ago, your ship sailed and sank. No one thought it worthy of salvage— then or now.  It needs warning buoys because of the toxic waste that it leaches.   

An arsonist thinks all fires start because of arson.

A thief thinks all missing items are stolen.

An embezzler thinks any missing funds are embezzled. 

A plagiarist thinks every posted solution is plagiarized.

 I wasn’t accusing you of plagiarism, Mr. BB. However, you didn’t cite the Mathematica program from which you copied your pasted SLOP.  The slipper fits because it belongs to you. _{You are what you are, Mr. BB.}

You are gifted Mr. BB, you are the only person I know that takes a brilliantly integrated mathematical program and finds a noxious use for it.  Another example of throwing slop on a diamond.

GA

I see you’ve thrown more slop on it –Surprise, Surprise!

By the way Mr. BB, n^3 is exactly 75; the (n =4.21725) should be approximate.

Mr. BB, your presentation is SLOP. Anyone who can figure out the solution from your slop does not need the slop in the first place. 

\(\log_{4n} 40\sqrt{3} = \log_{3n} 45\\ \text{Antilog form }\\ 4^x n^x=40\sqrt{3}\\ 3^x n^x=45\\ \left(\dfrac{4}{3}\right)^x=\dfrac {1}{9}*{8\sqrt{3}} = \dfrac{8}{3\sqrt 3}\rightarrow x=\dfrac {3}{2}\\ 8n^{\frac {3}{2}}=40\sqrt 3\\ n= 3^{\frac13} * 5^{(\frac23)} \rightarrow n^3= 3 * 5^2 =75 \)

_{Tell me, Mr. BB, do you actually find diamonds in SLOP, or do you find a diamond and throw SLOP on it? (Like, I don't already know.)}

GA