Are both of these the same thing? (logical NOT and the complement rule)

[Rev. Cheeseman]

Hi everyone,

This is an example of binary variable called as logical NOT https://www.fico.com/fico-xpress-op.../mipform/dhtml/chap2s1.html?scroll=sseclognot

...and this is the complement rule of probability https://brilliant.org/wiki/probabil...tal to 1.&text=Equivalently, P ( A c ),1−P(A)

Are both of these things the same?

 

[PeroK]

The logical NOT is the same concept as the complement in set theory.

 

[Rev. Cheeseman]

The logical NOT is the same concept as the complement in set theory.

Thank you.

 

[FactChecker]

A logical variable is a True/False or a 1/0 variable. Probabilities can have any real value from 0 to 1, including 0 and 1. In other words, it can express True/False/Maybe. In a coin toss, the event of Tails is the complement of the event of Heads. Both have a probability of 1/2.

 

[Rev. Cheeseman]

Thank you for these responses. I wonder if we can use probability formulas instead of crime per capita (for example in this context, rape) in order to see the likelihood of which of these two US states are more likely to commit rape against someone. For instance, Alaska has a rape rate of 134.0 per 100k and Tennessee has a rape rate of 38.2 per 100k according to https://en.m.wikipedia.org/wiki/List_of_U.S._states_and_territories_by_violent_crime_rate. How do we formulate a probability formula question using these datas?

 

[FactChecker]

For instance, Alaska has a rape rate of 134.0 per 100k and Tennessee has a rape rate of 38.2 per 100k according to https://en.m.wikipedia.org/wiki/List_of_U.S._states_and_territories_by_violent_crime_rate. How do we formulate a probability formula question using these datas?

Suppose R is the event of a particular person being raped, A is the event that the person is in Alaska, and T is the event that the person is in Tennessee. Then the probability of a particular person being raped, given he is in Alaska is denoted P(R|A) = 134/100,000 = 0.00134. The probability of a particular person being raped, given he is in Tennessee is denoted P(R|T) = 38.2/100,000 = 0.000382.

 

[Rev. Cheeseman]

Suppose R is the event of a particular person being raped, A is the event that the person is in Alaska, and T is the event that the person is in Tennessee. Then the probability of a particular person being raped, given he is in Alaska is denoted P(R|A) = 134/100,000 = 0.00134. The probability of a particular person being raped, given he is in Tennessee is denoted P(R|T) = 38.2/100,000 = 0.000382.

Thank you so much. But in the real world, per capita tells us more details than using probability formula. Is that accurate? Sorry.

 

[FactChecker]

Thank you so much. But in the real world, per capita tells us more details than using probability formula. Is that accurate? Sorry.

I can't think of anything else it says unless some more facts are given.

 

[Rev. Cheeseman]

I can't think of anything else it says unless some more facts are given.

Alright. Let's see if I can formulate a probability question by myself correctly.

Say, the numbers of crime in population A which is around 15-16 millions (let's say 15 millions) are 889 crimes vs the numbers of crime in population B which is around 106-107 millions (106 millions) are 3333. As we understand, probability formula is favorable event divided by numbers of possible events. 889 divided by 15 millions is 0.00005 and 3333 divided by 106 millions is 0.00003.

But because there are too many zeroes in there, can we do this instead? What I mean is, for example for population A we just divide 889 by 1500 so we will get 0.5 instead of 0.00005 when we divide 889 by 15 millions. It is still centered around 5 but with less zeroes. Also, the probability range is from 0 to 1.

 

[FactChecker]

Say, the numbers of crime in population A which is around 15-16 millions (let's say 15 millions) are 889 crimes vs the numbers of crime in population B which is around 106-107 millions (106 millions) are 3333. As we understand, probability formula is favorable event divided by numbers of possible events. 889 divided by 15 millions is 0.00005 and 3333 divided by 106 millions is 0.00003.

But because there are too many zeroes in there, can we do this instead? What I mean is, for example for population A we just divide 889 by 1500 so we will get 0.5 instead of 0.00005 when we divide 889 by 15 millions. It is still centered around 5 but with less zeroes. Also, the probability range is from 0 to 1.

I see your point that the numbers are very small for this example. Probabilities are just another way of representing the same facts as the rate per 100,000. But it is valid to compare probabilities (or rates). You can say that the probability of being raped in Alaska is 3.8 times higher than in Tennessee. (0.00134/0.000362 = 134/38.2 = 3.70)

 

[Rev. Cheeseman]

I see your point that the numbers are very small for this example. Probabilities are just another way of representing the same facts as the rate per 100,000. But it is valid to compare probabilities (or rates). You can say that the probability of being raped in Alaska is 3.8 times higher than in Tennessee. (0.00134/0.000362 = 134/38.2 = 3.70)

Yes, even from your previous calculations for Alaska it is 0.00134 and for Tennessee it is 0.000382. The probability range for Alaska is higher than Tennessee, therefore the probability of being raped is higher in Alaska.

But for my own formulated probability question, is it appropriate if I change the population from 15 millions to 1500 to get appropriate numbers so we can use the probability range which is from 0 to 1? By the way, in my formulated probability question instead of using rate I was using the numbers of crime or criminals.

 

[FactChecker]

Alright. Let's see if I can formulate a probability question by myself correctly.

Say, the numbers of crime in population A which is around 15-16 millions (let's say 15 millions) are 889 crimes vs the numbers of crime in population B which is around 106-107 millions (106 millions) are 3333. As we understand, probability formula is favorable event divided by numbers of possible events.

That is valid if all the events have an equal probability. Otherwise, it is not valid.

889 divided by 15 millions is 0.00005 and 3333 divided by 106 millions is 0.00003.

But because there are too many zeroes in there, can we do this instead? What I mean is, for example for population A we just divide 889 by 1500 so we will get 0.5 instead of 0.00005 when we divide 889 by 15 millions. It is still centered around 5 but with less zeroes. Also, the probability range is from 0 to 1.

It does not represent a probability if you divide by the wrong number. That is important. You do not want to tell a person that he has a 50% probability of being a victim of crime when the reality is that he has a 0.005% probability of being a victim.

 

[Rev. Cheeseman]

That is valid if all the events have an equal probability. Otherwise, it is not valid.

It does not represent a probability if you divide by the wrong number. That is important. You do not want to tell a person that he has a 50% probability of being a victim of crime when the reality is that he has a 0.005% probability of being a victim.

Ok. Thanks for the important informations. For population A which is 889 crimes or criminals divided by 15 millions is 0.00005 and for population B which is 3333 crimes or criminals divided by 106 millions is 0.00003.

Therefore, from probability range 0 to 1, the probability of someone getting assaulted by a criminal in population A is 0.00005 and the probability of someone getting assaulted by criminal in population B is 0.00003. So, someone getting assaulted by a criminal is more likely in population A than population B.

Is this correct?

 

[hutchphd]

probability of someone getting assaulted

The answer is yes you are correct . Of course the semantics matter very much. This is the probabiliity of assault on some particular person (someone). The probability of "anyone" being assaulted will depend explicitly upon the population. Most fallacious inferences (I like to call them call them "Trumpisms") depend upon this type of error.

 

[Rev. Cheeseman]

The answer is yes you are correct . Of course the semantics matter very much. This is the probabiliity of assault on some particular person (someone). The probability of "anyone" being assaulted will depend explicitly upon the population. Most fallacious inferences (I like to call them call them "Trumpisms") depend upon this type of error.

Thank you. So, the favorable event can be rate, numbers of everything, anything, etc.?

 

[hutchphd]

So long as they are correctly used (usually this is a question of "normalizing " the compared ratios in a fashion that makes sense.) $\frac 3 5 \neq \frac 3 8$ even though $3=3$

 

[Rev. Cheeseman]

So long as they are correctly used (usually this is a question of "normalizing " the compared ratios in a fashion that makes sense.) $\frac 3 5 \neq \frac 3 8$ even though $3=3$

Sorry for not telling the details. I mean both the favorable event or outcome and the total events or outcomes can be expressed as the numbers of crimes committed or numbers of criminals (or even rate per 100k, and so on) as the favorable outcome and the total population of something as the total outcome, isn't it?

Just like, for example, 10 red marbles as the favorable outcomes divided by 20 marbles as the total outcomes.

I keep forgetting about probability formula because sometimes it messes up my thinking trying to make sense both the question and the formula.

 

[Agent Smith]

Non liquet whether I understood the question, but if you're worried about "too many 0s", go the percentage way and scale up the values. If crime rate is 300 per 3,000,000, we get a crime rate = 0.0001. We could then scale this value to per 10000 like so: $0.0001 \times 10000 = 1$ per ten thousand.

 

[Agent Smith]

Ok. Thanks for the important informations. For population A which is 889 crimes or criminals divided by 15 millions is 0.00005 and for population B which is 3333 crimes or criminals divided by 106 millions is 0.00003.

Therefore, from probability range 0 to 1, the probability of someone getting assaulted by a criminal in population A is 0.00005 and the probability of someone getting assaulted by criminal in population B is 0.00003. So, someone getting assaulted by a criminal is more likely in population A than population B.

Is this correct?

Si, this is correct! Remember the denominator for both 0.00005 and 0.00003 is 1. Once you have a common denominator, we may compare fractions with each other as being greater/lesser/equal. Someone in population A is more likely to be a victim of a crime than someone in population B.

 

[jbriggs444]

Si, this is correct! Remember the denominator for both 0.00005 and 0.00003 is 1. Once you have a common denominator, we may compare fractions with each other as being greater/lesser/equal. Someone in population A is more likely to be a victim of a crime than someone in population B.

There is at least one unstated assumption here. It seems to be assumed that the probability of being vicimized is uniform across the population and is independent of any prior victimization. However, it would be consistent with the data if 889 crimes in a population of 15 million were all committed against one individual. In that case the probability that a randomly selected individual from the population turns out to have been a crime victim would be one in 15 million rather than 889 in 15 million.

 

[Hornbein]

Thank you so much. But in the real world, per capita tells us more details than using probability formula. Is that accurate? Sorry.

I would say the state per capita statistic and the state probability are pretty much the same thing.

 

[Hornbein]

I would say the state per capita statistic and the state probability are pretty much the same thing.

I guess these men are right though, probability somewhat implies that everyone has the same chance. (I don't notice that because this condition seldom holds in the real world outside of gambling games.)

 

[Agent Smith]

There is at least one unstated assumption here. It seems to be assumed that the probability of being vicimized is uniform across the population and is independent of any prior victimization. However, it would be consistent with the data if 889 crimes in a population of 15 million were all committed against one individual. In that case the probability that a randomly selected individual from the population turns out to have been a crime victim would be one in 15 million rather than 889 in 15 million.

I didn't know that. Gracias. What's the probability of "winning the lottery" 889 times in a row? Is this hapless person always getting chosen despite true randomness in criminal activity? I was taught that for true randomness, choosing has to be done with replacement.

 

[jbriggs444]

What's the probability of "winning the lottery" 889 times in a row? Is this hapless person always getting chosen despite true randomness in criminal activity? I was taught that for true randomness, choosing has to be done with replacement.

We are not told that the scenario is "truly random", whatever that means. We are told very little at all.

In the real world, the probability distribution for crime across potential victims is not uniform. Nor is it clear that the victim selection for the next crime will be independent of the victim selection for the current one.

Yes, I agree that 889 times in a row for one victim would be remarkable and unrealistic. But we are in a math forum. We are allowed to consider situations that are unrealistic.

 

[Agent Smith]

We are not told that the scenario is "truly random", whatever that means. We are told very little at all.

In the real world, the probability distribution for crime across potential victims is not uniform. Nor is it clear that the victim selection for the next crime will be independent of the victim selection for the current one.

Yes, I agree that 889 times in a row for one victim would be remarkable and unrealistic. But we are in a math forum. We are allowed to consider situations that are unrealistic.

As far as I know, a truly random sample, taken from a population, is one such that the probability of being in the sample is uniform i.e. doesn't change. Does that mean the trials (each chosen data point) are independent? I should think so. In my book that means a person can be chosen more than once ("balls must be picked with replacement"). We ignore the other times the person shows up in the sample though, right? We don't double count that is.

Can we do something like the following though?
Say we have 6 balls, 4 are green and 2 are red. We pick a random ball. P(green) = 4/6 = 2/3. Now, to keep the probability constant at 2/3, the recommended procedure is to replace the ball back into the pool/population. But, we could also do this. Don't put the ball back. Instead do this:
1. If we picked a green ball. We have 3 green and 2 red balls. Remove 1 green ball and 1 red ball and we're left with 2 green balls and 1 red ball. P(green) = 2/3. No change in the probability.
2. If we picked a red ball. We have 4 green balls and 1 red ball. Remove 2 green balls and the result is 2 green balls and 1 red ball and P(green) = 2/3. The probability remains constant.