Talk:Conditional probability

message: I've struck out the sentence about decision trees. There is  certainly no sense in which conditional probability calculations are generally easier with decision trees. Decision trees can indeed be interpreted as conditional probability models (or not), but in any event, they are a very, very small part of the world of conditional probability, and making an unwarranted assertion about a minor topic is out of place. Wile E. Heresiarch 17:13, 1 Feb 2004 (UTC)
thread_title: Decision trees

message: In the Other Considerations section, the statement If P(B) = 0, then P(A \mid B) is left undefined. seems incorrect. Is it not more correct to say that P(A \mid B) is indeterminate?
If P(B) = 0, then P(A \cap B) = 0 regardless of P(A) or P(A \mid B).
Bob Badour 04:36, 11 June 2006 (UTC)
replies: text: It's undefined. If you think it's not undefined, then what do you think its definition is? --Zundark 08:54, 11 June 2006 (UTC)
replies: text: Indeterminate as I said, the definition of which one would paraphrase to incalcuable or unknown. However, an indeterminate form can be undefined, and the consensus in the literature is to call the conditional undefined in the abovementioned case. There are probably reasons for treating it as undefined that I am unaware of, and changing the text in the article would be OR. Thank you for your comments, and I apologize for taking your time. -- Bob Badour 00:07, 12 June 2006 (UTC)

text: I think you are absolutely right. However, the theory needed to obtain this is a lot more complicated than the theory needed to understand conditional probability as such. Should the article state clearly from the start that we are deling with discrete distributions only, and then perhaps have a last section dealing with generalization to continuous distributions?--Niels Ø (noe) 19:33, 22 February 2007 (UTC)
replies: text: It's already valid for continuous distributions, at least now that I've cleaned up the definition section. It's not usual to define P(A|B) when P(B) = 0, but if someone can find a decent reference for this then it might be worth adding to the article. --Zundark 11:25, 24 February 2007 (UTC)
replies: text: I was not able to find any source defining P(A\mid B) when P(B)=0. I posted in a math forum, and after an interesting discussion someone gave a nice argument (a bit long to be copied here) justifying why it does not make sense. I consider the point clarified. --zeycus 15:30, 28 February 2007 (UTC)
replies: text:  I'm afraid these comments are almost entirely wrong.  It is perfectly possible to condition on events of probability zero, and this is in fact common.  Consider tossing a coin.  If one does not know if the coin is fair or not, in the Bayesian world one assigns a probability distribution to the parameter p representing the probability of getting a head.  This distribution reflects the degree of belief one has in the fairness of the coin.  In the event that this distribution is continuous it is perfectly reasonable to condition on the event that p = 1/2, even though this event has probability zero.  To define these conditional probabilities rigourously requires measure theory, and this approach agrees with the naive interpretation given in first level courses.  A good reference is Basic Stochastic Processes by Zastawniak and Brzezniak.  PoochieR 21:45, 6 November 2007 (UTC)
replies: text: To repeat myself from above: Should the article state clearly from the start that we are deling with discrete distributions only, and then perhaps have a last section dealing with generalization to continuous distributions?--Niels Ø (noe) 12:31, 7 November 2007 (UTC)
replies: text: No, because we aren't dealing only with discrete distributions. --Zundark 12:39, 7 November 2007 (UTC)
replies: text: Some of the most important modern uses of conditional probability are in Martingale theory, with direct practical applications in all areas of mathematical finance.  It is simply impossible to deal with these without conditioning on events of probability zero, so I think it's important that you should include these.  A way round would be to make it clear that the definition you have given is a naive definition, which only works for conditioning on events with probability > 0; however to give the definition which works for conditioning on any event requires the use of measure theory.  The measure theoretic definition agrees with the naive definition where that is applicable.  The natural way to express the measure theoretic formulation is in terms of conditional expectations, conditional on sigma-algebras of events; in this formulation P(A|B) = E(I(A)|\sigma (I(B))), where I(A) is the indicator random variable of event A.  A better reference than the one I gave before is: Probability with Martingales, David Williams, Ch.9. PoochieR 09:41, 8 November 2007 (UTC)

text: I am quite happy to edit your definition with respect to P(A|B) when P(B)=0 but you cannot leave it as it is.  The correct definition to make the discrete case correspond with the more general case is to define P(A|B)=0 when P(B)=0.  There are no problems then with the naive interpretation, and the benefit of agreeing with the more sophisticated approach.  In many ways it is similar to the debates that used to go on regarding 0!.PoochieR 18:16, 15 November 2007 (UTC)
replies: text:  Not to be too rude, PoochieR, but P(A|B)=0 whenever P(B)=0 is missing an obvious case: we clearly want P(B|B)=1!  Don't forget that the marginal distribution is, nonetheless, intended to be a distribution. ub3rm4th (talk) 10:10, 14 December 2008 (UTC)

text: About conditioning on zero-probability condition: see also Conditioning (probability). Boris Tsirelson (talk) 16:34, 14 December 2008 (UTC)
thread_title:  Undefined or Indeterminate? 

message: Are the modulus signs in the "Definition" section intended to refer to the cardinality of the respective sets? It's not clear from the current content of the page. I think the set theory background to probability is a little tricky, so perhaps more explanation could go into this section?
:I absolutely agree.--Niels Ø (noe) 14:13, 29 January 2007 (UTC)
replies: text: The text talks about elements randomly chosen from a set. The author's intent clearly is that this implies symmetry.--Niels Ø (noe) 08:29, 24 February 2007 (UTC)
replies: text: Yes, you are absolutely right. But then, why defining conditional probability only in that particular case, when it makes sense and is usually defined for any probabilistic space with the same formula P(A\mid  B) = \frac{P(A\cap B)}{P(B)}. --zeycus 8:43, 24 February 2007 (UTC)
replies: text: I agree this is a weak section in the article; one should not have to guess about the author's intentions. Anyway, I think the idea is to gereralize from the fairly obvious situation with symmetry to the general formulae. Of course, that kind of reasoning does not really belong under the heading "Definition". Go ahead; try your hand at it!--Niels Ø (noe) 10:07, 24 February 2007 (UTC)
replies: text: I've restored the proper definition. --Zundark 11:06, 24 February 2007 (UTC)
thread_title:  Use of 'modulus signs' and set theory 

message: Someone ought to expand the discussion of independence to n terms.  For instance, three events E, F, G are independent iff:
P(EFG)=P(E)P(F)P(G),
P(EF)=P(E)P(F)
P(EG)=P(E)P(G)
P(GF)=P(G)P(F)
And so on.  Verbally, every combination k (as in n choose k) of the n events (k=2,3,...,n), must be independent for ALL of them to be independent of each other.  Most textbooks I've seen include independence definitions for more than two events.
—The preceding unsigned comment was added by 171.66.41.25 (talk • contribs).
:This article is only concerned with independence as far as it relates to conditional probability. The general case is covered in the article on statistical independence. --Zundark 21:27, 20 July 2007 (UTC)
thread_title: Improving the Independence Section

message: This article was automatically assessed because at least one WikiProject had rated the article as start, and the rating on other projects was brought up to start class. BetacommandBot 03:52, 10 November 2007 (UTC)
thread_title:  WikiProject class rating

message: The Marginal distribution article does not present enough information to stand on its own.  It should be merged into this article.
Neelix (talk) 14:52, 13 April 2008 (UTC)
replies: text: I disagree.  Marginal distribution should be expanded. Michael Hardy (talk) 15:47, 13 April 2008 (UTC)

text: Oppose, per Michael Hardy. Marginal distribution is a sufficiently important and distinctive idea that it deserves its own article.  Plus it's a rather different thing from a conditional distribution.  Jheald (talk) 16:06, 13 April 2008 (UTC)

text: That makes does make sense.  I will remove the merge suggestion.  The marginal distribution article, however, still requires expansion.  I will place a proper notice on that article. Neelix (talk) 18:10, 13 April 2008 (UTC)
thread_title:  Merge with [[Marginal distribution]] 

message: The first sentence of the lead says:
:This article defines some terms which characterize probability distributions of two or more variables.
I think this sentence is superfluous, and makes things harder than they need be. One can teach good parts of probability theory, including conditional probability, without ever mentioning distributions or variables. E.g., rolling a die, you have a space U={a,b,c,d,e,f} (representing 1,2,3,4,5,6, but I use letters to emphasize that I'm NOT introducing a random variable taking numerical values), a simple probability function P(i)=1/6 for all i in U, events like A and B being subsets of U, an extended probability function P(A) (in this symmetrical case, P(A) = n(A)/n(U) where n counts elements in a set). With A = {even} = {b,d,f} and B = {more than four eyes} = {e,f}, you can find P(B|A), say.--Noe (talk) 08:55, 18 September 2008 (UTC)
thread_title:  Distributions and variables? 

message: What does it mean to condition on a continuous random variable? The definition given here does not seem to extend to such cases. 58.88.53.246 (talk) 16:22, 27 September 2008 (UTC)
replies: text: See also Conditioning (probability). Boris Tsirelson (talk) 16:45, 14 December 2008 (UTC)
thread_title:  Conditioning on a random variable 

message: I am removing the "other considerations" section. The full text of the section was as follows:
* If B is an event and P(B) > 0, then the function Q defined by Q(A) = P(A|B) for all events A is a probability measure.
* Many models in data mining can calculate conditional probabilities, including decision trees and Bayesian networks.
These remarks are stated completely out of context and I feel they do not belong here, at least without furhter explanation. Instead, I have moved the remark about P(A|B) being a probability measure to the article on probability spaces, where it fits more naturally. I have simply removed the remark about data mining entirely; perhaps a section on "applications of conditional proability" (or bayesan theory) could use it, but as an isolated remark it seems irrelvant. Birkett (talk) 08:04, 16 December 2008 (UTC)
thread_title:  Other Considerations Section 

message: I think it would be much cleaner to keep condition as C. I know letters don't matter in math and it is contextual but nevertheless ... 
It made me little confused when I read it first time. 86.61.232.26 (talk) 15:54, 1 January 2009 (UTC)
thread_title:  condition should be called C  not an A (in which case non condition is called C) 

message: The related Dutch article about conditional probability states:
In probability theory we use the term 'conditional probability' if we know that an event, say B, has happened, by which the possible outcomes are reduced to B.
However, on this page it is defined as the probability of some event A, given the occurrence of some other event B, without saying anything about statistical dependency. Further on it says that if A and B have no impact on the probability of each other, events A and B are independent, which means that P(A|B) = P(A).
This doesn't say anything about the situation being conditional or not; it only says that probabilities are in both perspectives the same. "given the occurrence of some other event B" is a perspective rather than a (limited) situation.
Furthermore, there is a fundamental difference between reducing the sample space and affecting probabilities. Throwing two dice don't affect each other's probabilities, but the total sample space is reduced by knowing the outcome of one die.
What is your opinion about the given definition on the Dutch page? Heptalogos (talk) 10:52, 13 February 2009 (UTC)
thread_title:  Reducing sample space 

message: There are quite a few articles that attempt to to deal with conditional probability, with varing levels of quality:
* conditional probability
* conditioning (probability)
* conditional probability distribution
* conditional expectation
* regular conditional probability
Perhaps some of these should be combined? —3mta3 (talk) 09:53, 27 June 2009 (UTC)
replies: text: I totally agree. Now it is quite arbitrary which contains what. I think reorganizing, unifying, and probably merge somewhat into fewer articles should be necessary. 80.98.239.192 (talk) 12:37, 3 November 2013 (UTC)
thread_title:  Other articles 

message: Given P(B|A) = 1, and P(B|~A)=p, can I find P(A|B) without knowing anything about P(A) or P(B)?  
I have an algorithm, which if correct, gives the correct result all the time. If incorrect, it gives the correct result 10% of the time.  It seems that if I get a correct result, I should be able to have some confidence that the algorithm is correct.  Intuitively, I think it should be possible, but I can't seem to formulate it right.  —Preceding unsigned comment added by 71.193.61.249 (talk) 17:21, 10 July 2009 (UTC)
replies: text: Short answer: no, it depends on P(A). Longer answer: See Bayes theorem or ask at Wikipedia:Reference desk/Mathematics. —3mta3 (talk) 17:47, 10 July 2009 (UTC)
thread_title:  Question 

message: This article would be very confusing to someone who is not mathematically minded (e.g. me). When reading the article I was confronted with a large wall of formulae with no clear explanation of what they meant. There needs to be a section in this article with a clear explanation with examples of what conditional probability is and how it can be used by a non mathematically minded person without the use of complex formulae.  —Preceding unsigned comment added by 79.74.192.244 (talk) 10:38, 3 August 2009 (UTC)
replies: text: I totally agree! The terminology and pictography is initially unfathomable :( We're not here to show off guys, we here to share information in a digestible format.Dickmojo (talk)  —Preceding undated comment added 11:40, 21 February 2012 (UTC)
thread_title:  Clarity 

message: The proposed link is 
http://www.opentradingsystem.com/quantNotes/Conditional_probability_.html
It provides examples of calculations with Conditional probability.
Does anyone object?  —Preceding unsigned comment added by Kaslanidi (talk • contribs) 20:22, 30 December 2009 (UTC)
replies: text: I second the objection. OhNoitsJamie Talk 20:41, 30 December 2009 (UTC)
thread_title:  I propose adding a link in external references 

message: I hope the following is true and clear:
Statement: conditional method is useful solving the probability of a cause (C) given the effect (E), when the effect is unequally distributed over the sample space.
I will use three somewhat similar examples to demonstrate. The first one also explains the conditional method. The similar part of all three:
:There are two baskets with fruit. Basket 1 has two apples and an orange. Basket 2 has an apple, an orange and a peach. A basket is picked randomly, from which an item is picked randomly.
*Example 1, calculating a cause given the effect.
Given that an apple is picked (event E), what is the chance that it comes from basket 1 (event C)?
:1. The probability to pick basket 1 is 1/2.
:2. The probability to pick an apple from it is 2/3.
:3. The probability to pick both basket 1 and the apple is 1/2 * 2/3 = 1/3 = 2/6.
:4. The probability to pick basket 2 is 1/2.
:5. The probability to pick an apple from it is 1/3.
:6. The probability to pick both basket 2 and the apple is 1/2 * 1/3 = 1/6.
Both probabilities to eventually pick an apple are 2/6 for basket 1 and 1/6 for basket 2, together 3/6. So, the probability that the apple was picked from basket 1 is 2/6 / 3/6 = 2/3. In words: the probability that basket 1 provided the picked apple is the probability of both events together (3) divided by the overall probability to pick an apple (3+6).
In formula: P(C|E) = P(C and E) / P(E).
*Example 2, calculating an effect given the cause.
Given that basket 1 is picked (event C), what is the chance to pick an apple (event E)?
P(E|C) = P(C and E) / P(C) = 2/6 / 1/2 = 2/3.
Since P(C) = 1 (the cause is given) and thus P(C and E) = P(E), P(C and E) / P(C) = P(E)/1 = P(E). Therefore P(E|C) = P(E).
The probability to pick an apple from it is 2/3, as in statement 2, which is unconditional. There's no need for conditional method.
*Example 3, calculating a cause given the effect, when the effect is equally distributed over the sample space.
Given that an orange is picked (event E), what is the chance that it comes from basket 1 (event C)?
P(C|E) = P(C and E) / P(E) = (1/6) / (1/6 + 1/6) = 1/2.
The orange has the same probability to be picked from every basket, because it's equally distributed over all baskets. This makes the probability the same as the probability of the cause, as in statement 1. Therefore P(C|E) = P(C). There's no need for conditional method. [[Us
:You're missing the point: there is no such as a conditional method. In every case you mention, a conditional probability has to be calculated. What you seem to demonstrate is that in some cases this conditional probability is easily determined without using Bayes' formula. Nothing new or surprising. Nijdam (talk) 17:20, 18 January 2010 (UTC)
replies: text: At least you're not missing the point; thanks for using the right terms. Good to know that this is not surprising to you. Now let's bother about the rest of our customers. First of all I tried to provide an example (1) that would be easy to understand and at the same time explaining the formula. Secondly I try to address the usefulness of what you call Bayes' formula, which usefulness doesn't seem to be a maths thing, but would IMO be so nice on Wikipedia. I've read Bayes' theorem in which the formula P(C|E)=P(C and E)/P(E) is only mentioned somewhere halfway as the definition of conditional probability. I do miss an introduction/ definition/ description of this formula. Heptalogos (talk) 20:16, 18 January 2010 (UTC)
thread_title:  The usefulness of conditional solution 

message: What is the meaning of the following section from the article?
In general it does not make much sense to ask after observation of a remarkable series of events, "What is the probability of this?"; which is a conditional probability based upon observation. The distinction between conditional and unconditional probabilities can be intricate if the observer who asks "What is the probability?" is himself/herself an outcome of a random selection. The name "Wyatt Earp effect" was coined in an article "Der Wyatt Earp Effekt" (in German) showing through several examples its subtlety and impact in various scientific domains.
Nijdam (talk) 13:48, 31 August 2010 (UTC)
thread_title:  Meaning 

message: The graph on this page implies the area is important, but is not. I'd suggest a 2d line. 2010-11-29  —Preceding unsigned comment added by 66.35.36.132 (talk) 04:31, 30 November 2010 (UTC)
thread_title: Graph 2004

message: The conditional probability is written P(A|B), or perhaps p(A|B) or Pr(A|B). Why not omit the P and write (A|B) instead of P(A|B) for the conditional probability, and  (A| ) instead of P(A) for the unconditional probability? Like the bra-ket notation of quantum mechanics.
The formula
:(A∩B| ) = (A|B)(B| ) = (B|A)(A| ) 
is easier to read than  
:P(A∩B) = P(A|B)P(B) = P(B|A)P(A)
Bo Jacoby (talk) 15:03, 6 July 2011 (UTC)
replies: text: We have to follow established notation.  Anything else would clearly be WP:OR.  As for the wider question of whether the P actually serves anything, I think maybe it does -- for example, avoiding any confusion with certain inner product notations.  Jheald (talk) 11:14, 7 July 2011 (UTC)
thread_title:  Write (A|B) instead of P(A|B) 

message: In my view, the section Terminology is unnecessary. Although I can see the logic in that joint and marginal probabilities are part of the definition of conditional probability, this article should not be for explaining these concepts - it should be sufficient to link to the relevant articles. This material stems from very early in the article history when it was stated "This article defines some terms which characterize probability distributions of two or more variables." I deleted the section, however it was reverted, with the reason that its existence was "to avoid entries in see also section". I am of the opposite opinion - if anything, it is more appropriate and concise to link from See Also to the relevant articles. Thoughts? Gnathan87 (talk) 13:58, 4 August 2011 (UTC)
replies: text: The purpose of the "See also" section is plainly set out at WP:SEEALSO. There needn't be a "terminology" section, but the article needs to say enough about the related concepts mentioned there presently that the article stands on its own. Melcombe (talk) 14:10, 4 August 2011 (UTC)
replies: text: Apologies, I wasn't very clear on this - my preferred way forward would be to 1. remove the Terminology section as it stands 2. rewrite the article to include links to Joint probability and Marginal probability in a context that more obviously establishes relevance to the topic 3. also include these links from a navbox. While it is desirable to make an article as self sufficient as possible, a balance should be struck between readers' needs and focus. I would argue that many readers will be familiar with these concepts, and for those who are not, it will be more useful to point to the prerequisites. Gnathan87 (talk) 18:36, 4 August 2011 (UTC)
thread_title:  Terminology Section 

message: The second paragraph of this article reads: "In the Bayesian interpretation of probability, the conditioning event is interpreted as evidence for the conditioned event [...] (In fact, this is also the Frequentist interpretation.)" Is there an interpretation of probability that rejects this idea? If not, the qualification on this statement is unnecessary to the point of being misleading. Even if there are major objections to the idea, this paragraph could be better structured. Nejssor (talk) 17:00, 4 December 2013 (UTC)
thread_title:  Interpretations 

message: The lead contains discussion that belongs into the article body, please refer to WP:LEAD. Paradoctor (talk) 00:42, 17 January 2014 (UTC)
thread_title:  Lead 

message: Hi all. I've tried to write a new introduction that addresses to the "too technical" problem that this article is tagged with. I've removed the formal definition (though it could go below with the other formal definitions) and given a simple concrete example of conditional probability. This edit was reverted once before because I falsely stated that B need occur before A, but that's fixed now. 
I realize this is a complex topic and this new intro is not likely to please everyone, but can we at least agree on the following goals for the introduction:
* It should be possible to get the underlying concept even if you've never studied statistics
* Lengthy formal definitions can come later in the article
* There should be at least one really simple example of a conditional probability in concrete, everyday terms.
Thanks everyone. --Jonathan Stray (talk) 00:35, 21 November 2014 (UTC)
thread_title:  New introduction 

message: I don't think this makes sense - 
"For example, if you have cancer you might have a 90% chance of testing positive for cancer, but if you test positive for cancer you might have only a 10% of actually having cancer because cancer is very rare."
In the first case, if you have cancer you'll have a 90% chance of testing positive - so the test is 90% sensitive.  
In the second case, if you test positive for cancer you have a 10% chance of having it?  Now the test is only 10% specfic?
This doesn't seem to make sense.  Here's an example I thought of:
For example, if it rains you might have 20% chance to be involved in a motor vehicle accident, but if you are involved in a motor vehicle accident, the probability it's raining is only 5%.
Cody Woodhouse (talk) May 12, 2015 - 4:29 PST
::Check out Confusion of the inverse, which used to be part of this article, but was split off at some point. It has an Example 2 about testing for rare diseases.--Nø (talk) 12:38, 13 May 2015 (UTC)
thread_title: Cancer example doesn't make sense

message: Thanks to all who have contributed to this article.  I have two related suggestions:
The example regarding the conditional probability that the first die = 2 given that the sum of both die < 5 has two problems which reduce its explanatory value:
First, the example makes it intuitively clear that the correct conditional probability is 0.3, but does not use the defining equation to arrive at that result.  I recommend that the intermediate results should be plugged back into the defining equation, culminating in 3/36 / 10/36 = 0.3 - the answer given.
However, this brings up the second problem:  The defining equation uses A to represent one event, and B the event upon which it is conditioned.  But the example uses B to represent the value of the second die, not the value of their sum, which makes it less clear to a beginner how to plug these values into the original.  I suggest not reusing "B" in this way.
Ma-Ma-Max Headroom (talk) 18:48, 29 June 2015 (UTC)
thread_title:  First example omits the denouement 

message: The section "Definition with σ-algebra" has several problems:
# The only mention of σ-algebra in this section is "However, it is possible to define a conditional probability with respect to a σ-algebra of such events (such as those arising from a continuous random variable)." Most of the section is about dealing with the case where P(B)=0. Suggest splitting this into two sections "Definition with σ-algebra" and "Conditioning on zero-probability events".
# The article claims that the case of continuous variables where B consists of a single point can be dealt with as a ratio of two integrals. I added a reference to the Borel–Kolmogorov paradox to show that this case is still problematic.
BPets (talk) 01:22, 22 September 2016 (UTC)
replies: text: On the former, I agree. On the latter, I doubt. The Borel–Kolmogorov paradox manifests itself when a set of measure zero has non-equivalent approximations by sets of positive measure, and these approximations are equally notable. For a single point, the approximation by small intervals centered at this point is more notable than others. Boris Tsirelson (talk) 04:41, 22 September 2016 (UTC)

text: Oops, no, sorry; on the latter, I agree, too. For instance, it appears that the conditional distribution of X given Y=0 is not the same as the conditional distribution of X given Y/X=0 (even if X>0 always). On the x axis we deal with a single point, but on the (x,y) plane it is a line... Boris Tsirelson (talk) 12:08, 22 September 2016 (UTC)

text: Thus, indeed, the section (as of now) contains almost nothing. If at all there is something to be said about single point versus two (or more) points, it is this: the conditional distribution of X given X=1 is uncontroversial, but (for example) the conditional distribution of X given X\in\{\tfrac12,2\} is controversial; writing it as (X-\tfrac12)(X-2)=0 we get one result, and writing it as X+\tfrac1X=\tfrac52 we get another result (even if X≠0 always). Boris Tsirelson (talk) 12:45, 22 September 2016 (UTC)
thread_title:  Definition with σ-algebra 

message: The organization of the topic is not very clear and is not the most intuitive way for people to understand the concept. Some topics (e.g. how independence/mutually exclusive events come into play in conditional probability) are scattered around the article, and no clear distinction is made on similar/frequently confused concepts.
MulingS (talk) 15:58, 17 October 2016 (UTC)
thread_title:  Organization of contents 

message: Conditional probability where A intersection B equals zero.
In set theory, of which probability is a particular aspect, there is no such a thing as zero, it´s the empty set and that empty set does NOT equal nothing. Defacto that empty set contains all and every element not enumerated in either. The infinite set in set theory does not contain everything, it contains the recursor into the, a set.
A savant mistake, is to presume the empty set in the conditional probability where A intersection B is the empty set, is empty. It is not.
On this basis, it´s also the number one fallacy that leads to wrongfull conclusions.  — Preceding unsigned comment added by 186.92.249.31 (talk) 14:37, 10 November 2016 (UTC)
replies: text: Where in the article do you find this error? -- You are right, of course, that A intersection B cannot be zero (though it may be the empty set). But P(A intersection B) can be zero, either because A and B are mutually explucsive, or because the intersection is nonempty but has probability zero. There are several situations where the probability of a set can be zero: If you roll a fair die, one (silly) probability model is that the sample space is {0,1,2,3,4,5,6} with probabilities 0, 1/6, 1/6, 1/6, 1/6, 1/6, 1/6, so P({0})=0 though {0} is not the empty set. Also, if you spin a dial and record where it stops as an angle X in degrees between 0 and 360 (including 0 but excluding 360, say), we may model it as a continuous uniform distribution, where the probability of any single outcome like X=90 is zero, where as the probability of e.g 89.5 <= X < 90.5 is 1/360. Again, P(X=90)=0, though "X=90" (which in the notation used for stochastic variables is in fact a set) is not empty, and in this case is not even impossible, in a sense.--Nø (talk) 17:49, 10 November 2016 (UTC)
thread_title:  Small interpretation error. 

message: Do we not need a further definition as given by Jeffrey Subjective Probability p12 (originally due to de Finetti) as the price of the betting ticket "Pay $1 if A&B; refund payment if not-B." I have reformulated a bit but my reformulation is better I think. That leads to the quotient rule quite naturally. That leads on to the more difficult problem of what happens if I have P(A&B) and P(B) and I subsequently learn that B is true. That is actually the only (non-frequentist) line of discussion that justifies the word "given". That leads on to the intractable problem of what happens if I subsequently update P(B). This stuff confuses people. It's taken me 30 years. Comments?Cutler (talk) 16:30, 20 July 2017 (UTC)
thread_title:  Further definition 

message: * https://archive.fo/3Iorv  — Preceding unsigned comment added by 2A02:587:4105:10BA:80A:6CBF:746A:3F48 (talk) 09:29, 29 August 2020 (UTC)
thread_title:  Conditional probability on the connectome of states (gr)