Anyone good at probability?

[Blzer]

Re: Anyone good at probability?

Not sure I agree with Question 2's reasoning because, and I think is the abstract math speaking, but I always get locked on this... because each member of the four person group has 55% chance of being Republican. This odds does not fluxate in regards to the next person in the group chosen to join the four...still just 55% chance it's also Republican....and so on. To calculate any other odds is adding up variables that have ceased to exist outside the individual chance.

I understand where you're coming from, but they aren't asking for probability per choice of selection. I'm not really sure how else to explain it, but I know for sure that this is a binomial distribution problem. The binomial distribution is used when there are exactly two mutually exclusive outcomes of a trial. These outcomes are appropriately labeled "success" and "failure", much like coin tosses as I explained earlier. On my sheet from the video, I showed by example that the binomial distribution is used to obtain the probability of observing x successes in n trials, with the probability of success on a single trial denoted by p. The binomial distribution assumes that p is fixed for all trials, just like you said and I won't wry away from that notion.

A problem like this is specifically different for this reason: we are determining a cumulative success issue. I worked around that by going the opposite way, doing 1 minus the probability of exact failure. So for instance, if they asked what the probability was of selecting 2 or less Republicans, you would have to do the probability of selecting exactly zero Republicans, plus the probability of selecting exactly one Republican, plus the probability of selecting exactly two Republicans. And that goes by the formula, which as we established is based on a common binomial distribution formula.

I actually have a calculator handy right now, so I'm just going to work out that math here. The probability would be:

(4)-Choose-(0) * (0.55)^0 * (0.45)^4
+ (4)-Choose-(1) * (0.55)^1 * (0.45)^3
+ (4)-Choose-(2) * (0.55)^2 * (0.45)^2

= 1 * 1 * 0.04100625
+ 4 * 0.55 * 0.091125
+ 6 * .3025 * .2025

= 0.04100625 + 0.200475 + .3675375

= .60901875 = 60.9%



You have to consider that each time you look at your selections, you must do the probability that you are selecting a Republican (success) times the probability that you aren't selecting a Republican (failure) for each of the four choices at a time, because you're looking at what the probability is that you're selecting exactly zero, or one, or two, or three, or four Republicans (which is based on the fact that you're also not selecting exactly four, or three, or two, or one, or zero Republicans at the same time). I never disputed what you said about probabilities not being the same, you must assume that each time a Republican has a 55% chance of being chosen, but this problem requires more calculations than that.

[GAMEC0CK2002]

Re: Anyone good at probability?

I understand where you're coming from, but they aren't asking for probability per choice of selection. I'm not really sure how else to explain it, but I know for sure that this is a binomial distribution problem. The binomial distribution is used when there are exactly two mutually exclusive outcomes of a trial. These outcomes are appropriately labeled "success" and "failure", much like coin tosses as I explained earlier. On my sheet from the video, I showed by example that the binomial distribution is used to obtain the probability of observing x successes in n trials, with the probability of success on a single trial denoted by p. The binomial distribution assumes that p is fixed for all trials, just like you said and I won't wry away from that notion.

A problem like this is specifically different for this reason: we are determining a cumulative success issue. I worked around that by going the opposite way, doing 1 minus the probability of exact failure. So for instance, if they asked what the probability was of selecting 2 or less Republicans, you would have to do the probability of selecting exactly zero Republicans, plus the probability of selecting exactly one Republican, plus the probability of selecting exactly two Republicans. And that goes by the formula, which as we established is based on a common binomial distribution formula.

I actually have a calculator handy right now, so I'm just going to work out that math here. The probability would be:

(4)-Choose-(0) * (0.55)^0 * (0.45)^4
+ (4)-Choose-(1) * (0.55)^1 * (0.45)^3
+ (4)-Choose-(2) * (0.55)^2 * (0.45)^2

= 1 * 1 * 0.04100625
+ 4 * 0.55 * 0.091125
+ 6 * .3025 * .2025

= 0.04100625 + 0.200475 + .3675375

= .60901875 = 60.9%



You have to consider that each time you look at your selections, you must do the probability that you are selecting a Republican (success) times the probability that you aren't selecting a Republican (failure) for each of the four choices at a time, because you're looking at what the probability is that you're selecting exactly zero, or one, or two, or three, or four Republicans (which is based on the fact that you're also not selecting exactly four, or three, or two, or one, or zero Republicans at the same time). I never disputed what you said about probabilities not being the same, you must assume that each time a Republican has a 55% chance of being chosen, but this problem requires more calculations than that.

Oh okay. So you're reasoning was good, but your estimation was off in your previous explanation. This answer sits well in line with my "common sense" and "no higher math skills" answer. Kudos to you. I'll stick with looking at cells and dissecting frogs.

[Blzer]

Re: Anyone good at probability?

Oh okay. So you're reasoning was good, but your estimation was off in your previous explanation. This answer sits well in line with my "common sense" and "no higher math skills" answer. Kudos to you. I'll stick with looking at cells and dissecting frogs.

What was my estimation again? Did I say higher than 90% I just didn't want to calculate it in my head is all, but 90% and 95.9% are pretty close to each other. All I was saying was that to figure that you'll get at least one Republican (or one heads on a coin toss) out of four trials should be pretty high in this probability case.

I'll wait on Kev to respond though, I need to hear about that last question again to make sure that he worded it correctly.


EDIT: Oh, you may have looked at my last post and thought I was answering his question. No, that was just an example if you were looking at "the probability of choosing at most two Republicans." My video when I wrote on paper is the correct answer to #2 of his homework.

[deaduck]

Re: Anyone good at probability?

Maybe I worded it akwardly Blzer...you're reasoning is perfectly sound from a math/probability stand point.

Just in the Abstract math class I took it ventured into a almost philosophical bent on so many probability questions...i.e. if you flipped a coin 99 times and all 99 times it was tails, what makes the odds any different than 50/50 on the hundredth toss just because you want to group the results?

[Skyboxer]

Re: Anyone good at probability?

↑ ↑ ↓ ↓ ← → ← → B A Start

My best guess...

[GAMEC0CK2002]

Re: Anyone good at probability?

↑ ↑ ↓ ↓ ← → ← → B A B A Start

My best guess...

Fixed.

[Blzer]

Re: Anyone good at probability?

Maybe I worded it akwardly Blzer...you're reasoning is perfectly sound from a math/probability stand point.

Just in the Abstract math class I took it ventured into a almost philosophical bent on so many probability questions...i.e. if you flipped a coin 99 times and all 99 times it was tails, what makes the odds any different than 50/50 on the hundredth toss just because you want to group the results?

Ah, I see what you are saying. But I think you are asking me a different question. You are asking me, once again, what the probability is that I will flip a tails at a certain time. That is, if I flip 99 tails out of 99 trials, what is the probability that I will flip a tails again? The probability, as you stated, is 50%. But that's not exactly what the question is asking. The question is asking something more like, "What is the probability that you will flip a coin 99 times and at least 80 of the flips will be tails?" That answer is not 50%, it is something much smaller. 50% would be a more accurate estimation if you were looking for at least 50 flips or something, though that wouldn't be the answer either. I guess in short, this isn't an abstract math class anyway so this isn't much of an issue.

But I fully well understand and am aware of what you are trying to say; that a probability has zero affect on individual outcomes, it is only there for best educated predicted outcomes. For example, if the odds of a weighted coin (however that works) gives one side a 51% probability of landing on it and the other is 49%, prediction says that there is absolutely no reason to ever go with the 49% probability, as it has a lesser chance of being landed on and all events are mutually exclusive (that is, there is no reason to believe that a balance in results must exist or lie in the way of the actual probability). Now people may go against the grain just by going against the grain, but that's why we are people and not computers.

But with all of this being said, the only way that you may have a point and that I may be wrong is if the answer actually relies on us to account for variable change; that is, that each event is not mutually exclusive and that if we choose a Republican on the first selection, that the probability of choosing a Republican the next time will be 10/19 instead of the same 11/20 from the beginning. This will have us say, "What is the probability of choosing a Republican given that we have already chosen a Republican?" This would be more like drawing cards out of a deck without replacement and less like a coin toss. That said, I can't confirm which way they want the question to be answered, but I simply answered it like this: "What is the opposite probability of selecting exactly zero Republicans?" That way, we get the opposite of what the question asks and grab the inverse probability (I know that term may not be semantically correct).

I wish Kev would pop back up in this thread though.

[GAMEC0CK2002]

Re: Anyone good at probability?

Ah, I see what you are saying. But I think you are asking me a different question. You are asking me, once again, what the probability is that I will flip a tails at a certain time. That is, if I flip 99 tails out of 99 trials, what is the probability that I will flip a tails again? The probability, as you stated, is 50%. But that's not exactly what the question is asking. The question is asking something more like, "What is the probability that you will flip a coin 99 times and at least 80 of the flips will be tails?" That answer is not 50%, it is something much smaller. 50% would be a more accurate estimation if you were looking for at least 50 flips or something, though that wouldn't be the answer either. I guess in short, this isn't an abstract math class anyway so this isn't much of an issue.

But I fully well understand and am aware of what you are trying to say; that a probability has zero affect on individual outcomes, it is only there for best educated predicted outcomes. For example, if the odds of a weighted coin (however that works) gives one side a 51% probability of landing on it and the other is 49%, prediction says that there is absolutely no reason to ever go with the 49% probability, as it has a lesser chance of being landed on and all events are mutually exclusive (that is, there is no reason to believe that a balance in results must exist or lie in the way of the actual probability). Now people may go against the grain just by going against the grain, but that's why we are people and not computers.

But with all of this being said, the only way that you may have a point and that I may be wrong is if the answer actually relies on us to account for variable change; that is, that each event is not mutually exclusive and that if we choose a Republican on the first selection, that the probability of choosing a Republican the next time will be 10/19 instead of the same 11/20 from the beginning. This will have us say, "What is the probability of choosing a Republican given that we have already chosen a Republican?" This would be more like drawing cards out of a deck without replacement and less like a coin toss. That said, I can't confirm which way they want the question to be answered, but I simply answered it like this: "What is the opposite probability of selecting exactly zero Republicans?" That way, we get the opposite of what the question asks and grab the inverse probability (I know that term may not be semantically correct).

I wish Kev would pop back up in this thread though.

That's what I was wondering. Do you go back to the 9 vs. 11 after each selection or is subsequent selections done without the previously chosen individual(s)

[Blzer]

Re: Anyone good at probability?

That's what I was wondering. Do you go back to the 9 vs. 11 after each selection or is subsequent selections done without the previously chosen individual(s)

I took a statistics class four years ago (or I guess it's five years ago now, hah!), and from what I learned there, my guess is that the probabilities are fixed. It would make the problem a hell of a lot easier that way, but I also imagine that's how it's being asked.

If it's being asked the other way, I can surely whip out another piece of paper and do another recording explaining how that's done (which is the same process, but with different numbers and instead burdens a cumulative effect).

[Tomahawk]

Re: Anyone good at probability?

#1 -
Probability of a student taking English - 60%
Probability of taking both English and Math - 38%

Probability of someone taking English but not Math of the entire student population.
60 - 38 = 22%

#2 - Probability of getting one Republican in 4 draws if 11/20 in the pool are Republicans.

First draw - 11/20
Democrat on the 1st draw, 2nd draw - 11/19
Democrat on the 1st two draws, 3rd draw - 11/18
Democrat on the 1st three draws, 4th draw - 11/17

Chance you draw a Republican in at least once in four pulls 1 - (11/20 x 11/19 x 11/18 x 11/17) = 87.4%

#3 - Females don't matter here. 58 males on campus divided by 131 total males = 44.3 %.

[Blzer]

Re: Anyone good at probability?

#2 - Probability of getting one Republican in 4 draws if 11/20 in the pool are Republicans.

First draw - 11/20
Democrat on the 1st draw, 2nd draw - 11/19
Democrat on the 1st two draws, 3rd draw - 11/18
Democrat on the 1st three draws, 4th draw - 11/17

Chance you draw a Republican in at least once in four pulls 1 - (11/20 x 11/19 x 11/18 x 11/17) = 87.4%

This logic is flawed. What you just calculated was the probability that you will draw at most three Republicans, given the events are not mutually exclusive (you did 1 minus the probability that the first, second, third, and fourth selections are all Republicans... again, without any sort of mutual exclusion, which is still this thread's hot topic debate).

[mudman]

Re: Anyone good at probability?

#1 -
Probability of a student taking English - 60%
Probability of taking both English and Math - 38%

Probability of someone taking English but not Math of the entire student population.
60 - 38 = 22%

#2 - Probability of getting one Republican in 4 draws if 11/20 in the pool are Republicans.

First draw - 11/20
Democrat on the 1st draw, 2nd draw - 11/19
Democrat on the 1st two draws, 3rd draw - 11/18
Democrat on the 1st three draws, 4th draw - 11/17

Chance you draw a Republican in at least once in four pulls 1 - (11/20 x 11/19 x 11/18 x 11/17) = 87.4%

#3 - Females don't matter here. 58 males on campus divided by 131 total males = 44.3 %.

#1 is 22% and can be seen easily with a venn diagram. Just draw two circles that overlap. One being M and the other being E. The overlap is .38 so the rest of M is .12 and the rest of E is .22. The .22 is the part that is English and not Math.

#2 No, as Blzr said, this is wrong.

The best way to think about this is to realize that the only way to NOT have 1 republican is to have all 4 democrats. P(all Dem) = (9/20)*(8/19)*(7/18)*(6/17) = .026006.

So P(not all Dem) = P(at least 1 repub) = 1 - .026006 = .973994 or 97.3994%

A tree diagram will also get you this result, although it's a bit tedious.

#3 is correct as the female data is irrelevant. The problem tells you it is a male that is drawn at random making it a conditional probability.

[deaduck]

Re: Anyone good at probability?

But with all of this being said, the only way that you may have a point and that I may be wrong is if the answer actually relies on us to account for variable change; that is, that each event is not mutually exclusive and that if we choose a Republican on the first selection, that the probability of choosing a Republican the next time will be 10/19 instead of the same 11/20 from the beginning. This will have us say, "What is the probability of choosing a Republican given that we have already chosen a Republican?" This would be more like drawing cards out of a deck without replacement and less like a coin toss. That said, I can't confirm which way they want the question to be answered, but I simply answered it like this: "What is the opposite probability of selecting exactly zero Republicans?" That way, we get the opposite of what the question asks and grab the inverse probability (I know that term may not be semantically correct).

See now you're starting to think like I am...the question is flawed.

If each of the pick is consider a single possibility then the result of that pick has to change the probability of the next pick...but if picked a unit then the individual probaility of each remains unchanged.

To explain it to my wife I had to simplify it to three balls...two red and one blue and the question is what are the chances of picking both red balls?

Well, the chance of picking any of the balls is a 33.333% chance. So if you reach in the bag holding them with both hands at the same time...each hand has a 33.333% chance of draw a blue ball and a 66.666% chance of drawing red...but if you pull your left hand out first and it's holding a red ball. Then you'd think logically your odds in the right hand suddenly switch to 50/50 but that's a flawed logic as the action occured in isolation of the results of the left hand.

Math is not the solid thing it's taught to be.

[mudman]

Re: Anyone good at probability?

See now you're starting to think like I am...the question is flawed.

If each of the pick is consider a single possibility then the result of that pick has to change the probability of the next pick...but if picked a unit then the individual probaility of each remains unchanged.

To explain it to my wife I had to simplify it to three balls...two red and one blue and the question is what are the chances of picking both red balls?

Well, the chance of picking any of the balls is a 33.333% chance. So if you reach in the bag holding them with both hands at the same time...each hand has a 33.333% chance of draw a blue ball and a 66.666% chance of drawing red...but if you pull your left hand out first and it's holding a red ball. Then you'd think logically your odds in the right hand suddenly switch to 50/50 but that's a flawed logic as the action occured in isolation of the results of the left hand.

Math is not the solid thing it's taught to be.

Oh yes it most certainly is.

There is absolutely zero ambiguity in this question as it's crystal clear. People's inability in this thread to understand the question doesn't mean it isn't clear.

When choosing people, w/o replacement is assumed, always has been and always will be as choosing the same person twice doesn't make sense. It specifically says you are choosing a group of 4. It's impossible to guarantee a group of 4 is chosen if you assume replacement.

The question is clear. Mathematics is based on things that are proven to be true, it doesn't get any more solid than that.

Also, your blue ball red ball problem is not related in any way to this dem/repub question. It may sound similar but it's not.

[Blzer]

Re: Anyone good at probability?

See now you're starting to think like I am...the question is flawed.

I wouldn't really say that it's flawed, but more like it's ambiguous. Kev never told us what kind of statistics they're learning and if they are at all getting into mutual exclusion, fixed probabilities, and binomial distributions. If they aren't learning that, then I suppose this could be a dependent probability problem, like we have been discussing.

To explain it to my wife I had to simplify it to three balls...two red and one blue and the question is what are the chances of picking both red balls?

Well, the chance of picking any of the balls is a 33.333% chance. So if you reach in the bag holding them with both hands at the same time...each hand has a 33.333% chance of draw a blue ball and a 66.666% chance of drawing red...but if you pull your left hand out first and it's holding a red ball. Then you'd think logically your odds in the right hand suddenly switch to 50/50 but that's a flawed logic as the action occured in isolation of the results of the left hand.

Exactly, but I think that your example has some holes. As a better explanation, think of the Game Show Host problem, which seems to constantly be disputed by the masses.

This is a problem where you have three doors which behind them sits hidden objects. One of the doors contains a valuable prize and the other do not (normally I hear this problem involve a car and two goats). The game show host will ask you, the contestant, to choose one of the three doors without zero penalty. The door options are #1, #2, and #3. Let's say that you choose door #2. The game show host, who know what is behind each of the doors, will then show you one door which he knows has a goat behind it, so let's say that he shows you door #1. Here is the kicker: the game show host will now ask you whether you want to switch your choice.

A lot of people will be instinctive, paranoid, or superstitious, and say that because it's a 50/50 chance, it might as well be in your best interest to stick with your gut choice. But of course, since this is simply a probability problem, this is where you are tricked into thinking that you are really looking at a 50/50 probability. What we all have to remember is that the game show host isn't just showing us where one goat is, he is showing us where the car isn't, and is asking us whether we want to switch our choice to one other door or not. When we first chose a door, our probability was 33.3% of being correct. When he showed us that a car was not in the other door, that now increases our chances of the car being in the other door by another 33.3%, as our choice, no matter how you look at it, was always and will always be 33.3%. That's basically how you have to look at it.

To better get a grasp of this example, let's make it 100 doors with 99 goats and 1 car. If we choose a door, the probability we landed on the car is 1%. Now let's say that the game show host shows us 98 other doors revealing goats, and we're left with the door that we chose and one other door. Is it in our best interest to switch? Hell yeah it is! Now see, does that make sense that our probability is now 50/50 since we're reduced to two doors? No way! He simply showed us where the car isn't, and he's asking us whether we want to switch from our 1% probability to our 99% probability.

Always account for variable change. But the question is... does this problem want us to do so as well? If so, mudman provided the correct answer, and my answer would be for a problem given that the probability is fixed.