GingerAle

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Nombre de usuarioGingerAle
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Preguntas 2
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 #9
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+2

Solution:

 

\(\text {The series ends after a team has a fourth win.}\\ \text{Here are the four scenarios with derived probabilities }\\ \text{where the Cubs win the series. }\\ \text{1) The Cubs win the first four games.}\ \text{ }\\ \hspace{35 mm} \rho(\text {Cbs win on } 4^{th} \text{game )= }\ \text{ }\\ \hspace{35 mm} (3/5)^4 = (81/625) = 12.96\%\\ \text{ }\\ \text{2) Cubs win series in game five. }\\ \text{For the Cubs to win the series in game five,}\\ \text{ they need to win three games in four trials and then win the fifth game. }\\ \text{ } \hspace{34 mm} \rho \text {(Cbs win 3 in 4})*\rho \text{(Cbs win 5th game}) =\\ \text{ } \hspace{35 mm} \ \binom{4}{3}(3/5)^3*(2/5) *(3/5) = (648/3125) \approx 20.74\% \\ \text{ }\\ \text{3) Cubs win series in game six. }\\ \text{For the Cubs to win the series in game six, }\\ \text{they need to win three games in five trials and then win the sixth game. }\\ \text{ } \hspace{34 mm} \rho \text {(Cbs win 3 in 5})* \rho\text{(Cbs win 6th game}) =\\ \text{ } \hspace{34 mm} \binom{5}{3}(3/5)^3*(2/5)^2 *(3/5) = (648/3125) \approx 20.74\% \\ \text{ }\\ \text{4) Cubs win series in game seven. }\\ \text{For the Cubs to win the series in game seven, }\\ \text{they need to win three games in six trials and then win the seventh game. }\\ \text{ } \hspace{34 mm} \rho \text {(Cbs win 3 in 6})* \rho \text{(Cbs win 7th game}) =\\ \text{ } \hspace{34 mm} \binom{6}{3}(3/5)^3*(2/5)^3 *(3/5) \approx 16.59\% \\ \text{ }\\ \text{ The sum of the individual probabilities gives }\\ \text{ the overall probability of the Cubs winning the series.}\\ \text {Sum of individual probabilities: } \\ (20.74+20.74+16.59+12.96) \bf \approx 71.0\% \)

 

 

GA

GingerAle 14-abr-2018
 #11
avatar+965 
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 13-abr-2018
 #9
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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???

 

WTF! 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 Sweet Polly pure piss 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.indecision

 

 I'm sorry if it is not pleasing to you

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

 

GA

Edit Corrected typo.

GingerAle 11-abr-2018
 #7
avatar+965 
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 09-abr-2018
 #2
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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.laugh

 

GA

 

Edit: Corrected error

GingerAle 05-abr-2018
 #10
avatar+965 
+1

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

GingerAle 23-mar-2018