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In probability theory, '''conditional probability''' is a measure of the probability of an event occurring, given that another event (by assumption, presumption, assertion or evidence) has already occurred. If the event of interest is {{mvar|A}} and the event {{mvar|B}} is known or assumed to have occurred, "the conditional probability of {{mvar|A}} given {{mvar|B}}", or "the probability of {{mvar|A}} under the condition {{mvar|B}}", is usually written as {{math|P(''A''{{!}}''B'')}} or occassionally <math>P_{B}</math>(''A''). This can also be understood as the fraction of probability B that intersects with A: <math>P(A \mid B) = \frac{P(A \cap B)}{P(B)}</math>.

For example, the probability that any given person has a cough on any given day may be only 5%. But if we know or assume that the person is sick, then they are much more likely to be coughing. For example, the conditional probability that someone unwell is coughing might be 75%, in which case we would have that {{math|P(Cough)}} = 5% and {{math|P(Cough{{!}}Sick)}} = 75%. Although, there doesn't have to be relationship or dependence between {{mvar|A}} and {{mvar|B}}, and they don't have to occur simultaneously.

{{math|P(''A''{{!}}''B'')}} may or may not be equal to {{math|P(''A'')}} (the unconditional probability of {{mvar|A}}). If {{math|1=P(''A''{{!}}''B'') = P(''A'')}}, then events {{mvar|A}} and {{mvar|B}} are said to be ''independent'': in such a case, knowledge about either event does not alter the likelihood of each other. {{math|P(''A''{{!}}''B'')}} (the conditional probability of {{mvar|A}} given {{mvar|B}}) typically differs from {{math|P(''B''{{!}}''A'')}}. For example, if a person has dengue fever, they might have a 90% chance of testing positive for the disease. In this case, what is being measured is that if event {{mvar|B}} (''having dengue'') has occurred, the probability of {{mvar|A}} (''testing positive'') given that {{mvar|B}} occurred is 90%: {{math|P(''A''{{!}}''B'')}} = 90%. Alternatively, if a person tests positive for dengue fever, they may have only a 15% chance of actually having this rare disease due to high false positive rates. In this case, the probability of the event {{mvar|B}} (''having dengue'') given that the event {{mvar|A}} (''testing positive'') has occurred is 15%: {{math|P(''B''{{!}}''A'')}} = 15%.It should be apparent now that falsely equating the two probabilities can lead to various errors of reasoning, which is commonly seen through base rate fallacies.

While conditional probabilities can provide extremely useful information, limited information is often supplied or at hand. Therefore, it can be useful to reverse or convert a condition probability using Bayes' theorem: <math>P(A|B) = {{P(B|A)*P(A)}\over{P(B)}}</math>. Another option is to display conditional probabilities in conditional probability table to illuminate the relationship between events.

== Definition ==
[[File:Conditional probability.svg|thumb|Illustration of conditional probabilities with an Euler diagram. The unconditional probability P(''A'') = 0.30 + 0.10 + 0.12 = 0.52. However, the conditional probability ''P''(''A''|<math>B_{1}</math>) = 1, ''P''(''A''|<math>B_{2}</math>) = 0.12 ÷ (0.12 + 0.04) = 0.75, and P(''A''|<math>B_{3}</math>) = 0.]]

[[File:Probability tree diagram.svg|thumb|On a tree diagram, branch probabilities are conditional on the event associated with the parent node. (Here, the overbars indicate that the event does not occur.)]]

[[File:Venn Pie Chart describing Bayes' law.png|thumb|Venn Pie Chart describing conditional probabilities]]

=== Conditioning on an event ===

==== Kolmogorov definition ====
Given two events {{mvar|A}} and {{mvar|B}} from the sigma-field of a probability space, with the unconditional probability of {{mvar|B}} being greater than zero (i.e., {{math|P(''B'')>0)}}, the conditional probability of {{mvar|A}} given {{mvar|B}} is defined to be the quotient of the probability of the joint of events {{mvar|A}} and {{mvar|B}}, and the probability of {{mvar|B}}:

:<math>P(A \mid B) = \frac{P(A \cap B)}{P(B)},</math>

where <math>P(A \cap B)</math> is the probability that both events {{mvar|A}} and {{mvar|B}} occur. This may be visualized as restricting the sample space to situations in which {{mvar|B}} occurs. The logic behind this equation is that if the possible outcomes for {{mvar|A}} and {{mvar|B}} are restricted to those in which {{mvar|B}} occurs, this set serves as the new sample space.

Note that the above equation is a definition—not a theoretical result. We just denote the quantity <math>\frac{P(A \cap B)}{P(B)}</math> as <math>P(A\mid B)</math>, and call it the conditional probability of {{mvar|A}} given {{mvar|B}}.

==== As an axiom of probability ====
Some authors, such as de Finetti, prefer to introduce conditional probability as an axiom of probability:

:<math>P(A \cap B) = P(A \mid B)P(B)</math>

Although mathematically equivalent, this may be preferred philosophically; under major probability interpretations, such as the subjective theory, conditional probability is considered a primitive entity. Moreover, this "multiplication rule" can be practically useful in computing the probability of <math>A \cap B</math> and introduces a symmetry with the summation axiom for mutually exclusive events
:<math>P(A \cup B) = P(A) + P(B) - P(A \cap B)</math>

==== As the probability of a conditional event ====

Conditional probability can be defined as the probability of a conditional event
<math>A_B</math>. The Goodman–Nguyen–Van Fraassen conditional event can be defined as

:<math>A_B =
\bigcup_{i \ge 1}
\left(
\bigcap_{j<i}
\overline{B}_j,
A_i B_i
\right).
</math>

It can be shown that

:<math>P(A_B)= \frac{P(A \cap B)}{P(B)}</math>

which meets the Kolmogorov definition of conditional probability.

=== Conditioning on an event of probability zero ===
If <math> P(B)=0 </math>, then according to the definition, <math> P(A|B) </math> is undefined.

The case of greatest interest is that of a random variable {{mvar|Y}}, conditioned on a continuous random variable {{mvar|X}} resulting in a particular outcome {{mvar|x}}. The event <math>B = \{ X = x \}</math> has probability zero and, as such, cannot be conditioned on.

Instead of conditioning on {{mvar|X}} being ''exactly'' {{mvar|x}}, we could condition on it being closer than distance <math>\epsilon</math> away from {{mvar|x}}. The event <math>B = \{ x-\epsilon < X < x+\epsilon \}</math> will generally have nonzero probability and hence, can be conditioned on.
We can then take the limit
:<math>\lim_{\epsilon \to 0} P(A \mid x-\epsilon < X < x+\epsilon).</math>

For example, if two continuous random variables {{mvar|X}} and {{mvar|Y}} have a joint density <math>f_{X,Y}(x,y)</math>, then by L'Hôpital's rule:
:<math>
\begin{aligned}
\lim_{\epsilon \to 0} P(Y \in U \mid x_0-\epsilon < X < x_0+\epsilon) &=
\lim_{\epsilon \to 0} \frac{\int_{x_0-\epsilon}^{x_0+\epsilon} \int_U f_{X, Y}(x, y) \mathrm{d}y \mathrm{d}x}{\int_{x_0-\epsilon}^{x_0+\epsilon} \int_\mathbb{R} f_{X, Y}(x, y) \mathrm{d}y \mathrm{d}x} \\
&= \frac{\int_U f_{X, Y}(x_0, y) \mathrm{d}y}{\int_\mathbb{R} f_{X, Y}(x_0, y) \mathrm{d}y}.
\end{aligned}
</math>
The resulting limit is the conditional probability distribution of {{mvar|Y}} given {{mvar|X}} and exists when the denominator, the probability density <math>f_X(x_0)</math>, is strictly positive.

It is tempting to ''define'' the undefined probability <math>P(A|X=x)</math> using this limit, but this cannot be done in a consistent manner. In particular, it is possible to find random variables {{mvar|X}} and {{mvar|W}} and values {{mvar|x}}, {{mvar|w}} such that the events <math>\{X = x\}</math> and <math>\{W = w\}</math> are identical but the resulting limits are not:
:<math>\lim_{\epsilon \to 0} P(A \mid x-\epsilon \le X \le x+\epsilon) \neq \lim_{\epsilon \to 0} P(A \mid w-\epsilon \le W \le w+\epsilon).</math>
The Borel–Kolmogorov paradox demonstrates this with a geometrical argument.

=== Conditioning on a discrete random variable ===

Let {{mvar|X}} be a discrete random variable and its possible outcomes denoted {{mvar|V}}. For example, if {{mvar|X}} represents the value of a rolled die then {{mvar|V}} is the set <math>\{ 1, 2, 3, 4, 5, 6 \}</math>. Let us assume for the sake of presentation that {{mvar|X}} is a discrete random variable, so that each value in {{mvar|V}} has a nonzero probability.

For a value {{mvar|x}} in {{mvar|V}} and an event {{mvar|A}}, the conditional probability
is given by <math> P(A \mid X=x) </math>.
Writing
:<math>c(x,A) = P(A \mid X=x)</math>
for short, we see that it is a function of two variables, {{mvar|x}} and {{mvar|A}}.

For a fixed {{mvar|A}}, we can form the random variable <math> Y = c(X, A) </math>. It represents an outcome of <math> P(A \mid X=x) </math> whenever a value {{mvar|x}} of {{mvar|X}} is observed.

The conditional probability of {{mvar|A}} given {{mvar|X}} can thus be treated as a random variable {{mvar|Y}} with outcomes in the interval <math>[0,1]</math>. From the law of total probability, its expected value is equal to the unconditional probability of {{mvar|A}}.

=== Partial conditional probability ===

The partial conditional probability
<math>P(A\mid B_1 \equiv b_1, \ldots, B_m \equiv b_m)</math>
is about the probability of event
<math>A</math> given that each of the condition events
<math>B_i</math> has occurred to a degree
<math>b_i</math> (degree of belief, degree of experience) that might be different from 100%. Frequentistically, partial conditional probability makes sense, if the conditions are tested in experiment repetitions of appropriate length
<math>n</math>. Such
<math>n</math>-bounded partial conditional probability can be defined as the conditionally expected average occurrence of event
<math>A</math> in testbeds of length
<math>n</math> that adhere to all of the probability specifications
<math>B_i \equiv b_i</math>, i.e.:

:<math>P^n(A\mid B_1 \equiv b_1, \ldots, B_m \equiv b_m)=
\operatorname E(\overline{A}^n\mid\overline{B}^n_1=b_1, \ldots, \overline{B}^n_m=b_m)
</math>
Based on that, partial conditional probability can be defined as
:<math>
P(A\mid B_1 \equiv b_1, \ldots, B_m \equiv b_m)
= \lim_{n\to\infty}
P^n(A\mid B_1 \equiv b_1, \ldots, B_m \equiv b_m),</math>
where <math> b_i n \in \mathbb{N}</math>

Jeffrey conditionalization is a special case of partial conditional probability, in which the condition events must form a partition:

:<math>
P(A\mid B_1 \equiv b_1, \ldots, B_m \equiv b_m)
= \sum^m_{i=1} b_i P(A\mid B_i)
</math>

== Example ==
Suppose that somebody secretly rolls two fair six-sided dice, and we wish to compute the probability that the face-up value of the first one is 2, given the information that their sum is no greater than 5.
* Let ''D''1 be the value rolled on dice 1.
* Let ''D''2 be the value rolled on dice 2.

'''''Probability that'' ''D''1 = 2'''''

Table 1 shows the sample space of 36 combinations of rolled values of the two dice, each of which occurs with probability 1/36, with the numbers displayed in the red and dark gray cells being ''D''1 + ''D''2.

''D''1 = 2 in exactly 6 of the 36 outcomes; thus ''P''(''D''1 = 2) = <math>\frac{6}{36}</math> = <math>\frac{1}{6}</math>:

:{| class="wikitable" border="1" style="background:silver; text-align:center; width:300px"
|+ Table 1
! rowspan=2 colspan=2 | +
! colspan=6 | D2
|-
! scope="col" | 1
! scope="col" | 2
! scope="col" | 3
! scope="col" | 4
! scope="col" | 5
! scope="col" | 6
|-
! rowspan=6 scope="row" | ''D''1
! scope="row" | 1
| 2 || 3 || 4 || 5 || 6 || 7
|- style="background: red;"
! scope="row" | 2
| 3 || 4 || 5 || 6 || 7 || 8
|-
! scope="row" | 3
| 4 || 5 || 6 || 7 || 8 || 9
|-
! scope="row" | 4
| 5 || 6 || 7 || 8 || 9 || 10
|-
! scope="row" | 5
| 6 || 7 || 8 || 9 || 10 || 11
|-
! scope="row" | 6
| 7 || 8 || 9 || 10 || 11 || 12
|}

'''''Probability that'' ''D''1 + ''D''2 ≤ 5'''''

Table 2 shows that ''D''1 + ''D''2 ≤ 5 for exactly 10 of the 36 outcomes, thus ''P''(''D''1 + ''D''2 ≤ 5) = {{frac|10|36}}:

:{| class="wikitable" border="1" style="background:silver; text-align:center; width:300px"
|+ Table 2
! rowspan=2 colspan=2 | +
! colspan=6 | ''D''2
|-
! scope="col" | 1
! scope="col" | 2
! scope="col" | 3
! scope="col" | 4
! scope="col" | 5
! scope="col" | 6
|-
! rowspan=6 scope="row" | ''D''1
! 1
| style="background:red;" | 2 || style="background:red;" | 3 || style="background:red;" | 4 || style="background:red;" | 5 || 6 || 7
|-
! scope="row" | 2
| style="background:red;" | 3 || style="background:red;" | 4 || style="background:red;" | 5 || 6 || 7 || 8
|-
! scope="row" | 3
| style="background:red;" | 4 || style="background:red;" | 5 || 6 || 7 || 8 || 9
|-
! scope="row" | 4
| style="background:red;" | 5 || 6 || 7 || 8 || 9 || 10
|-
! scope="row" | 5
| 6 || 7 || 8 || 9 || 10 || 11
|- style="background: red;"
|-
! scope="row" | 6
| 7 || 8 || 9 || 10 || 11 || 12
|- style="background: red;"
|}

'''''Probability that'' ''D''1 = 2 ''given that'' ''D''1 + ''D''2 ≤ 5 '''''

Table 3 shows that for 3 of these 10 outcomes, ''D''1 = 2.

Thus, the conditional probability P(''D''1 = 2 | ''D''1+''D''2 ≤ 5) = {{frac|3|10}} = 0.3:

:{| class="wikitable" border="1" style="text-align:center; width:300px"
|+ Table 3
! rowspan=2 colspan=2 | +
! colspan=6 | ''D''2
|-
! scope="col" | 1
! scope="col" | 2
! scope="col" | 3
! scope="col" | 4
! scope="col" | 5
! scope="col" | 6
|-
! rowspan=6 scope="row" | ''D''1
! 1
| style="background:silver;" | 2 || style="background:silver;" | 3 || style="background:silver;" | 4 || style="background:silver;" | 5 || 6 || 7
|-
! scope="row" | 2
| style="background:red;" | 3 || style="background:red;" | 4 || style="background:red;" | 5 || 6 || 7 || 8
|-
! scope="row" | 3
| style="background:silver;" | 4 || style="background:silver;" | 5 || 6 || 7 || 8 || 9
|-
! scope="row" | 4
| style="background:silver;" | 5 || 6 || 7 || 8 || 9 || 10
|-
! scope="row" | 5
| 6 || 7 || 8 || 9 || 10 || 11
|-
! scope="row" | 6
| 7 || 8 || 9 || 10 || 11 || 12
|}

Here, in the earlier notation for the definition of conditional probability, the conditioning event ''B'' is that ''D''1 + ''D''2 ≤ 5, and the event ''A'' is ''D''1 = 2. We have <math>P(A\mid B)=\tfrac{P(A \cap B)}{P(B)} = \tfrac{3/36}{10/36}=\tfrac{3}{10},</math> as seen in the table.

== Use in inference ==

In statistical inference, the conditional probability is an update of the probability of an event based on new information. The new information can be incorporated as follows:

* Let ''A'', the event of interest, be in the sample space, say (''X'',''P'').
* The occurrence of the event ''A'' knowing that event ''B'' has or will have occurred, means the occurrence of ''A'' as it is restricted to ''B'', i.e. <math>A \cap B</math>.
* Without the knowledge of the occurrence of ''B'', the information about the occurrence of ''A'' would simply be ''P''(''A'')
* The probability of ''A'' knowing that event ''B'' has or will have occurred, will be the probability of <math>A \cap B</math> relative to ''P''(''B''), the probability that ''B'' has occurred.
* This results in <math display="inline">P(A|B) = P(A \cap B)/P(B)</math> whenever ''P''(''B'') > 0 and 0 otherwise.

This approach results in a probability measure that is consistent with the original probability measure and satisfies all the Kolmogorov axioms. This conditional probability measure also could have resulted by assuming that the relative magnitude of the probability of ''A'' with respect to ''X'' will be preserved with respect to ''B'' (cf. a Formal Derivation below).

The wording "evidence" or "information" is generally used in the Bayesian interpretation of probability. The conditioning event is interpreted as evidence for the conditioned event. That is, ''P''(''A'') is the probability of ''A'' before accounting for evidence ''E'', and ''P''(''A''|''E'') is the probability of ''A'' after having accounted for evidence ''E'' or after having updated ''P''(''A''). This is consistent with the frequentist interpretation, which is the first definition given above.

== Statistical independence ==

Events ''A'' and ''B'' are defined to be statistically independent if

:<math>P(A \cap B) = P(A) P(B).</math>

If ''P''(''B'') is not zero, then this is equivalent to the statement that

:<math>P(A\mid B) = P(A).</math>

Similarly, if ''P''(''A'') is not zero, then

:<math>P(B\mid A) = P(B)</math>

is also equivalent. Although the derived forms may seem more intuitive, they are not the preferred definition as the conditional probabilities may be undefined, and the preferred definition is symmetrical in ''A'' and ''B''.

'''Independent events vs. mutually exclusive events'''

The concepts of mutually independent events and mutually exclusive events are separate and distinct. The following table contrasts results for the two cases (provided that the probability of the conditioning event is not zero).
{| class="wikitable"
|+
!
!'''If statistically independent'''
!'''If mutually exclusive'''
|-
|<math>P(A\mid B)=</math>
|<math>P(A)</math>
|0
|-
|<math> P(B\mid A)=</math>
|<math>P(B)</math>
|0
|-
|<math>P(A \cap B)=</math>
|<math>P(A) P(B)</math>
|0
|}
In fact, mutually exclusive events cannot be statistically independent (unless both of them are impossible), since knowing that one occurs gives information about the other (in particular, that the latter will certainly not occur).

== Common fallacies ==
:''These fallacies should not be confused with Robert K. Shope's 1978 "conditional fallacy", which deals with counterfactual examples that beg the question''.

=== Assuming conditional probability is of similar size to its inverse ===

[[File:Bayes_theorem_visualisation.svg|thumb|300px|A geometric visualisation of Bayes' theorem. In the table, the values 2, 3, 6 and 9 give the relative weights of each corresponding condition and case. The figures denote the cells of the table involved in each metric, the probability being the fraction of each figure that is shaded. This shows that P(A<nowiki>|</nowiki>B) P(B) = P(B<nowiki>|</nowiki>A) P(A) i.e. P(A<nowiki>|</nowiki>B) = {{sfrac|P(B<nowiki>|</nowiki>A) P(A)|P(B)}} . Similar reasoning can be used to show that P(Ā<nowiki>|</nowiki>B) = {{sfrac|P(B<nowiki>|</nowiki>Ā) P(Ā)|P(B)}} etc.]]

In general, it cannot be assumed that ''P''(''A''|''B'') ≈ ''P''(''B''|''A''). This can be an insidious error, even for those who are highly conversant with statistics. The relationship between ''P''(''A''|''B'') and ''P''(''B''|''A'') is given by Bayes' theorem:

:<math>\begin{align}
P(B\mid A) &= \frac{P(A\mid B) P(B)}{P(A)}\\
\Leftrightarrow \frac{P(B\mid A)}{P(A\mid B)} &= \frac{P(B)}{P(A)}
\end{align}</math>

That is, P(''A''|''B'') ≈ P(''B''|''A'') only if ''P''(''B'')/''P''(''A'') ≈ 1, or equivalently, ''P''(''A'') ≈ ''P''(''B'').

=== Assuming marginal and conditional probabilities are of similar size ===
In general, it cannot be assumed that ''P''(''A'') ≈ ''P''(''A''|''B''). These probabilities are linked through the law of total probability:
:<math>P(A) = \sum_n P(A \cap B_n) = \sum_n P(A\mid B_n)P(B_n).</math>

where the events <math>(B_n)</math> form a countable Partition of a set|partition of <math>\Omega</math>.

This fallacy may arise through selection bias. For example, in the context of a medical claim, let ''S''{{sub|''C''}} be the event that a sequela (chronic disease) ''S'' occurs as a consequence of circumstance (acute condition) ''C''. Let ''H'' be the event that an individual seeks medical help. Suppose that in most cases, ''C'' does not cause ''S'' (so that ''P''(''S''{{sub|''C''}}) is low). Suppose also that medical attention is only sought if ''S'' has occurred due to ''C''. From experience of patients, a doctor may therefore erroneously conclude that ''P''(''S''{{sub|''C''}}) is high. The actual probability observed by the doctor is ''P''(''S''{{sub|''C''}}|''H'').

=== Over- or under-weighting priors ===
Not taking prior probability into account partially or completely is called ''base rate neglect''. The reverse, insufficient adjustment from the prior probability is ''conservatism''.

== Formal derivation ==
Formally, ''P''(''A'' | ''B'') is defined as the probability of ''A'' according to a new probability function on the sample space, such that outcomes not in ''B'' have probability 0 and that it is consistent with all original probability measures.

Let Ω be a sample space with elementary events {''ω''}, and let ''P'' be the probability measure with respect to the σ-algebra of Ω. Suppose we are told that the event ''B'' ⊆ Ω has occurred. A new probability distribution (denoted by the conditional notation) is to be assigned on {''ω''} to reflect this. All events that are not in ''B'' will have null probability in the new distribution. For events in ''B'', two conditions must be met: the probability of ''B'' is one and the relative magnitudes of the probabilities must be preserved. The former is required by the axioms of probability, and the latter stems from the fact that the new probability measure has to be the analog of ''P'' in which the probability of ''B'' is one - and every event that is not in ''B'', therefore, has a null probability. Hence, for some scale factor ''α'', the new distribution must satisfy:

#<math>\omega \in B : P(\omega\mid B) = \alpha P(\omega)</math>
#<math>\omega \notin B : P(\omega\mid B) = 0</math>
#<math>\sum_{\omega \in \Omega} {P(\omega\mid B)} = 1.</math>

Substituting 1 and 2 into 3 to select ''α'':

:<math>\begin{align}
1 &= \sum_{\omega \in \Omega} {P(\omega \mid B)} \\
&= \sum_{\omega \in B} {P(\omega\mid B)} + \cancelto{0}{\sum_{\omega \notin B} P(\omega\mid B)} \\
&= \alpha \sum_{\omega \in B} {P(\omega)} \\[5pt]
&= \alpha \cdot P(B) \\[5pt]
\Rightarrow \alpha &= \frac{1}{P(B)}
\end{align}</math>

So the new probability distribution is

#<math>\omega \in B: P(\omega\mid B) = \frac{P(\omega)}{P(B)}</math>
#<math>\omega \notin B: P(\omega\mid B) = 0</math>

Now for a general event ''A'',

:<math>\begin{align}
P(A\mid B)
&= \sum_{\omega \in A \cap B} {P(\omega \mid B)} + \cancelto{0}{\sum_{\omega \in A \cap B^c} P(\omega\mid B)} \\
&= \sum_{\omega \in A \cap B} {\frac{P(\omega)}{P(B)}} \\[5pt]
&= \frac{P(A \cap B)}{P(B)}
\end{align}</math>

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# Paulos, J.A. (1988) Innumeracy: Mathematical Illiteracy and its Consequences, Hill and Wang. ISBN 0-8090-7447-8 (p. 63 et seq.)
# Thomas Bruss, F; Der Wyatt Earp Effekt; Spektrum der Wissenschaft; March 2007
# George Casella and Roger L. Berger (1990), Statistical Inference, Duxbury Press, ISBN 0-534-11958-1 (p. 18 et seq.)
# Grinstead and Snell's Introduction to Probability, p. 134

@@ Line 356: / Line 356: @@
 &= \frac{P(A \cap B)}{P(B)}
 \end{align}</math>
 == Licensing ==
 Content obtained and/or adapted from:
 * [https://en.wikipedia.org/wiki/Conditional_probability Conditional probability, Wikipedia] under a CC BY-SA license

← Older revision		Revision as of 17:46, 24 October 2021
Line 376:		Line 376:
	# George Casella and Roger L. Berger (1990), Statistical Inference, Duxbury Press, ISBN 0-534-11958-1 (p. 18 et seq.)		# George Casella and Roger L. Berger (1990), Statistical Inference, Duxbury Press, ISBN 0-534-11958-1 (p. 18 et seq.)
	# Grinstead and Snell's Introduction to Probability, p. 134		# Grinstead and Snell's Introduction to Probability, p. 134
		+
		+	== Licensing ==
		+	Content obtained and/or adapted from:
		+	* [https://en.wikipedia.org/wiki/Conditional_probability Conditional probability, Wikipedia] under a CC BY-SA license

@@ Line 302: / Line 302: @@
 === Assuming conditional probability is of similar size to its inverse ===
-[[File:Bayes_theorem_visualisation.svg|thumb|300px|A geometric visualisation of Bayes' theorem. In the table, the values 2, 3, 6 and 9 give the relative weights of each corresponding condition and case. The figures denote the cells of the table involved in each metric, the probability being the fraction of each figure that is shaded. This shows that P(A|B) P(B) = P(B|A) P(A) i.e. P(A|B) = <math>\frac{P(B|A) P(A)}{P(B)}</math> . Similar reasoning can be used to show that P(Ā|B) = <math>\frac{P(B|Ā) P(Ā)}{P(B)}</math> etc.]]
+[[File:Bayes_theorem_visualisation.svg|thumb|300px|A geometric visualisation of Bayes' theorem. In the table, the values 2, 3, 6 and 9 give the relative weights of each corresponding condition and case. The figures denote the cells of the table involved in each metric, the probability being the fraction of each figure that is shaded. This shows that P(A<nowiki>|</nowiki>B) P(B) = P(B<nowiki>|</nowiki>A) P(A) i.e. P(A<nowiki>|</nowiki>B) = <math>\frac{P(B|A) P(A)}{P(B)}</math> . Similar reasoning can be used to show that P(Ā<nowiki>|</nowiki>B) = {{sfrac|P(B<nowiki>|</nowiki>Ā) P(Ā)|P(B)}} etc.]]
 In general, it cannot be assumed that ''P''(''A''|''B'')&nbsp;≈&nbsp;''P''(''B''|''A''). This can be an insidious error, even for those who are highly conversant with statistics. The relationship between ''P''(''A''|''B'') and ''P''(''B''|''A'') is given by Bayes' theorem:
@@ Line 317: / Line 317: @@
 :<math>P(A) = \sum_n P(A \cap B_n) = \sum_n P(A\mid B_n)P(B_n).</math>
-where the events <math>(B_n)</math> form a countable Partition of a set|partition of <math>\Omega</math>.
+where the events <math>(B_n)</math> form a countable partition of <math>\Omega</math>.
 This fallacy may arise through selection bias. For example, in the context of a medical claim, let <math>S_{C}</math> be the event that a sequela (chronic disease) ''S'' occurs as a consequence of circumstance (acute condition) ''C''. Let ''H'' be the event that an individual seeks medical help. Suppose that in most cases, ''C'' does not cause ''S'' (so that ''P''(<math>S_{C}</math>) is low). Suppose also that medical attention is only sought if ''S'' has occurred due to ''C''. From experience of patients, a doctor may therefore erroneously conclude that ''P''(<math>S_{C}</math>) is high. The actual probability observed by the doctor is ''P''(<math>S_{C}</math>|''H'').

@@ Line 302: / Line 302: @@
 === Assuming conditional probability is of similar size to its inverse ===
-[[File:Bayes_theorem_visualisation.svg|thumb|300px|A geometric visualisation of Bayes' theorem. In the table, the values 2, 3, 6 and 9 give the relative weights of each corresponding condition and case. The figures denote the cells of the table involved in each metric, the probability being the fraction of each figure that is shaded. This shows that P(A<nowiki>|</nowiki>B) P(B) = P(B<nowiki>|</nowiki>A) P(A) i.e. P(A<nowiki>|</nowiki>B) = <math>\frac{P(B|A) P(A)}{P(B)}</math> . Similar reasoning can be used to show that P(Ā<nowiki>|</nowiki>B) = <math>\frac{P(B|Ā) P(Ā)}{P(B)}</math> etc.]]
+[[File:Bayes_theorem_visualisation.svg|thumb|300px|A geometric visualisation of Bayes' theorem. In the table, the values 2, 3, 6 and 9 give the relative weights of each corresponding condition and case. The figures denote the cells of the table involved in each metric, the probability being the fraction of each figure that is shaded. This shows that P(A|B) P(B) = P(B|A) P(A) i.e. P(A|B) = <math>\frac{P(B|A) P(A)}{P(B)}</math> . Similar reasoning can be used to show that P(Ā|B) = <math>\frac{P(B|Ā) P(Ā)}{P(B)}</math> etc.]]
 In general, it cannot be assumed that ''P''(''A''|''B'')&nbsp;≈&nbsp;''P''(''B''|''A''). This can be an insidious error, even for those who are highly conversant with statistics. The relationship between ''P''(''A''|''B'') and ''P''(''B''|''A'') is given by Bayes' theorem:

@@ Line 302: / Line 302: @@
 === Assuming conditional probability is of similar size to its inverse ===
-[[File:Bayes_theorem_visualisation.svg|thumb|300px|A geometric visualisation of Bayes' theorem. In the table, the values 2, 3, 6 and 9 give the relative weights of each corresponding condition and case. The figures denote the cells of the table involved in each metric, the probability being the fraction of each figure that is shaded. This shows that P(A<nowiki>|</nowiki>B) P(B) = P(B<nowiki>|</nowiki>A) P(A) i.e. P(A<nowiki>|</nowiki>B) = <math>\frac{P(B<nowiki>|</nowiki>A) P(A)}{P(B)}</math> . Similar reasoning can be used to show that P(Ā<nowiki>|</nowiki>B) = <math>\frac{P(B<nowiki>|</nowiki>Ā) P(Ā)}{P(B)}</math> etc.]]
+[[File:Bayes_theorem_visualisation.svg|thumb|300px|A geometric visualisation of Bayes' theorem. In the table, the values 2, 3, 6 and 9 give the relative weights of each corresponding condition and case. The figures denote the cells of the table involved in each metric, the probability being the fraction of each figure that is shaded. This shows that P(A<nowiki>|</nowiki>B) P(B) = P(B<nowiki>|</nowiki>A) P(A) i.e. P(A<nowiki>|</nowiki>B) = <math>\frac{P(B|A) P(A)}{P(B)}</math> . Similar reasoning can be used to show that P(Ā<nowiki>|</nowiki>B) = <math>\frac{P(B|Ā) P(Ā)}{P(B)}</math> etc.]]
 In general, it cannot be assumed that ''P''(''A''|''B'')&nbsp;≈&nbsp;''P''(''B''|''A''). This can be an insidious error, even for those who are highly conversant with statistics. The relationship between ''P''(''A''|''B'') and ''P''(''B''|''A'') is given by Bayes' theorem:

@@ Line 167: / Line 167: @@
 '''''Probability that'' ''D''<sub>1</sub>&nbsp;+&nbsp;''D''<sub>2</sub>&nbsp;≤&nbsp;5'''''
-Table 2 shows that ''D''<sub>1</sub>&nbsp;+&nbsp;''D''<sub>2</sub>&nbsp;≤&nbsp;5 for exactly 10 of the 36 outcomes, thus ''P''(''D''<sub>1</sub>&nbsp;+&nbsp;''D''<sub>2</sub>&nbsp;≤&nbsp;5)&nbsp;=&nbsp;{{frac|10|36}}:
+Table 2 shows that ''D''<sub>1</sub>&nbsp;+&nbsp;''D''<sub>2</sub>&nbsp;≤&nbsp;5 for exactly 10 of the 36 outcomes, thus ''P''(''D''<sub>1</sub>&nbsp;+&nbsp;''D''<sub>2</sub>&nbsp;≤&nbsp;5)&nbsp;=&nbsp;<math>\frac{10}{36}</math>:
 :{| class="wikitable" border="1" style="background:silver; text-align:center; width:300px"
@@ Line 207: / Line 207: @@
 Table 3 shows that for 3 of these 10 outcomes, ''D''<sub>1</sub>&nbsp;=&nbsp;2.
-Thus, the conditional probability P(''D''<sub>1</sub>&nbsp;=&nbsp;2&nbsp;|&nbsp;''D''<sub>1</sub>+''D''<sub>2</sub>&nbsp;≤&nbsp;5)&nbsp;=&nbsp;{{frac|3|10}}&nbsp;=&nbsp;0.3:
+Thus, the conditional probability P(''D''<sub>1</sub>&nbsp;=&nbsp;2&nbsp;|&nbsp;''D''<sub>1</sub>+''D''<sub>2</sub>&nbsp;≤&nbsp;5)&nbsp;=&nbsp;<math>\frac{3}{10}</math>&nbsp;=&nbsp;0.3:
 :{| class="wikitable" border="1" style="text-align:center; width:300px"
@@ Line 319: / Line 319: @@
 where the events <math>(B_n)</math> form a countable Partition of a set|partition of <math>\Omega</math>.
-This fallacy may arise through selection bias. For example, in the context of a medical claim, let ''S''{{sub|''C''}} be the event that a sequela (chronic disease) ''S'' occurs as a consequence of circumstance (acute condition) ''C''. Let ''H'' be the event that an individual seeks medical help. Suppose that in most cases, ''C'' does not cause ''S'' (so that ''P''(''S''{{sub|''C''}}) is low). Suppose also that medical attention is only sought if ''S'' has occurred due to ''C''. From experience of patients, a doctor may therefore erroneously conclude that ''P''(''S''{{sub|''C''}}) is high. The actual probability observed by the doctor is ''P''(''S''{{sub|''C''}}|''H'').
+This fallacy may arise through selection bias. For example, in the context of a medical claim, let <math>S_{C}</math> be the event that a sequela (chronic disease) ''S'' occurs as a consequence of circumstance (acute condition) ''C''. Let ''H'' be the event that an individual seeks medical help. Suppose that in most cases, ''C'' does not cause ''S'' (so that ''P''(<math>S_{C}</math>) is low). Suppose also that medical attention is only sought if ''S'' has occurred due to ''C''. From experience of patients, a doctor may therefore erroneously conclude that ''P''(<math>S_{C}</math>) is high. The actual probability observed by the doctor is ''P''(<math>S_{C}</math>|''H'').
 === Over- or under-weighting priors ===