I cant hack it..

#21
i already gave you the number. and my professor was very clear "u dont need to use binomial distribution for that task"!
he the author of the book: https://digikogu.taltech.ee/et/Download/32c8edd9-f716-409a-8bce-76c189bea0cf

task 2.24, page 143.. answer given at page 664.
0,7^12 + 0,7^13 + 0,7^14 + 0,7^15
0.013841 +0.009689 +0.006782 +0.004748 = 0.035

all i needed was a verification.. one that you are clearly incapable of providing. but i see you point, and i was having the same doubts about his solution. i think he maybe just wrong.
like, the probabilities are given for single votes, not the total outcome. so why doesn't he include the combinations into the sample space?
also, by his algorithm receiving at least 3 votes has a probability of 113%,
also, wouldn't 12 votes have the same probability as 12 votes, including 3 denials. cause how else would we apply the same algorithm if expected outcome would be 4 greens, 5 yellows and 6 reds.

what vexes me most, is that later in his book, he has a chapter about binomial distribution,
 
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#22
A loan application is being processed by a committee of 15 members. Successful application has to be approved by at least 12 members. The probability of a single approval is .7 and it doesn't depend on other members. How big is the probability of loan approval?
The probability of twelve members approving the loan, where each member has a 70% chance to do so is calculated by multiplying the individual probabilities together. In this case they are all the same, so we can represent it by

P(loan approved) = P(individual approval)^12 = 0.7^12 = 1.38%
 
#23
The probability of twelve members approving the loan, where each member has a 70% chance to do so is calculated by multiplying the individual probabilities together. In this case they are all the same, so we can represent it by

P(loan approved) = P(individual approval)^12 = 0.7^12 = 1.38%
well, yeah.. that's what the professor is claiming. but then

1. how come the probability of 3 or more votes is 113% ?
2. how would you then calculate if there would be more then one option to vote?
 
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#24
I see the shrinking value of multiplying together the probabilities isn't quite making sense to you. So here's another way to look at it:

What's the chances of rolling a 6 with 1d6? There's only one outcome out of six, 1/6
What about getting a 12 with 2d6? Now there's 36 outcomes, and only 1 of them is a 12, 1/36
What bout getting an 18 with 3d6? Now there's 216 outcomes, and only 1 of them is an 18, 1/216

Or again with flipping three coins:
coinflip.png

In each case we're taking a fraction of a fraction of a fraction and so it get's smaller with each additional probability.
1. how come the probability of 3 or more votes is 113% ?
2. how would you then calculate if there would be more then one option to vote?
1). 0.7^3 = 34.3% for three votes at 70% chance of success each
2). This question isn't clear to me; could you rephrase it?
 
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#25
P(min 12 votes) = P(12) + P(13) + P(14) + P(15) .. That's the professor's algorithm. And yours also. No?
P(min 3 votes) = P(3) + ... + P(15) = 1.13
P(min 2 votes) = 1.62
P(min 1 vote) = 2.32

correct me if i'm wrong, but probability cant be more than 100%
 
#26
Ordinarily I would say no, probabilities are usually multiplied together, not added. But we may be talking about different problems. I'm downloading the book you linked to read up on the details. I'll post again afterward
 
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#27
Problem 4.24, translated from Estonian on pp143
The Supervisory Board of the Company has 15 members. For reorganization the consent of at least 80% of the members of the supervisory board is required. The probability that each individual member of the Supervisory Board votes in favor is 0.7, regardless of how the other members vote. How likely it is reorganization proposal in the Council?
Answer given on pp664 to problem 4.24: 0.0351
80% of 15 members is 12 members to consent, out of a set of 15.
Given 15 tries, what are the chances of at least 12 successes, where P(success)=0.7?
Or equivalently what are the chances of getting less than 4 failures, where P(failure)=0.3?

Binomial Distributions
 
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#31
I am not sure what the point of this thread is anymore.
my point is, that i'm waiting for the answer.
the point of a forum is to discuss statistics and probability

additional point is that you cant seem to come to an agreement of what would be the correct algorithm for the solution.
e.g. @AngleWyrm and @Dason have different results.

question remains: is there a single right answer to that problem?
 
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#32
Let me rephrase that. Lets say there is a bucket of skittles, containing 7 blue, 2 red and 1 green.
A candy is taken from the bucket, and then put back, 15 times.
What is the probability of drawing at least 3 blue candies?

@AngleWyrm using your (or the professors) algorithm, it would be 113% which is clearly false, cause there exists a probability to draw all red or even all green, even tho its a very small number.

@Dason in that case you wouldnt even be able to use binomial distribution, as there is more than 2 options.
 

Dason

Ambassador to the humans
#33
@Dason in that case you wouldnt even be able to use binomial distribution, as there is more than 2 options.
Once again... incorrect. There is "blue" and "not blue". Which is all that matters for this problem.

But hey maybe one time just for giggles you could actually show your full work and what you're doing. You know - the thing I was asking you to do before.
 

Dason

Ambassador to the humans
#37
You may not have formally learned about the binomial distribution but you have the foundation to calculate these probabilities. But it will require using combinations. So how many ways are there for exactly 3 of the 15 people to vote yes.
 
#40
I just feels good proving the leading professor in my country is wrong.. :) and i'm not even a scientist.
we had 30 min phone call, arguing about the validity of the solution provided earlier, also by @AngleWyrm
its quite sad actually the way the fool students..