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2015-01-06T12:15:43Z

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	~~In [[probability theory]], the '''complement'''~~ of ~~any [[event (probability theory~~)~~\|event]] ''A''~~ is ~~the event [not ''A'']~~, i.e. ~~the event that ''~~A~~'' does not occur~~.<~~ref~~>~~Robert R. Johnson, Patricia J. Kuby: ''Elementary Statistics''. Cengage Learning 2007, ISBN 978-0-495-38386-4, p. 229 ({{Google books\|DWCAh7jWO98C\|restricted online copy\|page=229}})~~<~~/ref~~> ~~The event ''A'' and its complement [not ''A'']~~ are ~~[[mutually exclusive]] and [[Collectively exhaustive events\|exhaustive]]. Generally~~, ~~there~~ is ~~only one event ''B'' such that ''A'' and ''B''~~ are ~~both mutually exclusive~~ and ~~exhaustive; that event is~~ the ~~complement of ''A''~~. ~~The complement of an event ''A'' is usually denoted~~ as ~~''A′'', ''A<sup>c</sup>'' or ''{{overline\|A}}''~~.

	~~For example~~, ~~if a typical coin is tossed~~ and ~~one assumes that~~ it ~~cannot land on its edge, then it can either land showing "heads"~~ or "tails." Because these two [[Outcome (probability)\|outcomes]] are mutually exclusive (i.e. the coin cannot simultaneously show both heads and tails) and collectively exhaustive (i.e. there are no other possible outcomes not represented between these two), ~~they are therefore each other~~'s ~~complements~~. ~~This means that [heads]~~ is ~~logically equivalent~~ to ~~[not tails]~~, ~~and [tails] is equivalent~~ to ~~[not heads].~~

	~~==Complement rule==~~
	~~In a [[Experiment (probability theory)\|random experiment]], the probabilities of all possible events (the [[sample space]]) must total~~ to ~~1— that is, some outcome must occur on every trial. For two events~~ to ~~be complements, they must be [[Collectively exhaustive events\|collectively exhaustive]], together filling the entire sample space~~. ~~Therefore~~, ~~the probability of an event's complement must be [[Unity (mathematics)\|unity]] minus the~~
	probability of the event.<ref name="yates">{{cite book \| last = Yates \| first = Daniel S. \| last2 = Moore \| first2 = David S \| last3 = Starnes \| first3 = Daren S. \| year = 2003 \| title = The Practice of Statistics \| edition = 2nd \| publisher = [~~[W. H. Freeman and Company\|Freeman]] \| location = New York \| url =~~ http://~~bcs~~.~~whfreeman~~.~~com~~/~~yates2e~~/ ~~\| isbn = 978~~-0-~~7167~~-~~4773~~-~~4}}</ref>{{rp\|342~~-~~343}} That is, for an event ''A'~~',
	~~:<math>P(A^c) = 1 - P(A)~~.~~</math>~~
	~~Equivalently~~, the ~~probabilities of an event~~ and ~~its complement must always total~~ to ~~1. This does not~~, ~~however~~, ~~mean that ''any'' two events whose probabilities total~~ to ~~1 are each other's complements; complementary events must also fulfill the condition of [[Mutually exclusive events\|mutual exclusivity]]~~.

	~~===Example of~~ the ~~utility of this concept===~~
	~~Suppose one throws an ordinary six-sided die eight times~~. ~~What~~ is the ~~probability that one sees~~ a ~~"1"~~ at ~~least once?~~

	~~It may be tempting~~ to ~~say that~~

	~~: Pr~~(~~["1" on 1st trial] or ["1" on second trial] or~~ .~~.. or ["1" on 8th trial])~~

	~~:= Pr("1" on 1st trial) + Pr~~(~~"1" on second trial~~) ~~+ .~~.. ~~+ P("1"~~ on ~~8th trial)~~

	~~:= 1/6 + 1/6 +~~ ... + 1~~/6.~~

	~~:= 8/6 = 1.3333... (..~~.and ~~this is clearly wrong~~.)

	~~That cannot~~ be ~~right because a probability cannot be more than 1~~. ~~The technique is wrong because~~ the ~~eight events whose probabilities got added are not mutually exclusive~~.

	~~Instead one may find the probability~~ of the ~~complementary event and subtract it from 1, thus:~~

	~~: Pr(~~at ~~least one "1") = 1 − Pr(no "1"s)~~

	~~:= 1 − Pr([no "1" on 1st trial] and [no "1" on 2nd trial] and~~ ... and ~~[no "1" on 8th trial])~~

	~~:= 1 − Pr~~(~~no "1" on 1st trial~~) ~~× Pr(no "1" on 2nd trial) × ... × Pr~~(~~no "1" on 8th trial~~)

	~~:= 1 −~~(~~5/6~~) ~~× (5/6) × ..~~. ~~× (5/6)~~

	~~:= 1 − (5/6)~~<~~sup~~>8<~~/sup~~>

	~~:= 0~~.~~7674~~...

	~~==See also==~~
	*[[Negation\|Logical complement]]
	*[[Exclusive disjunction]]
	*[[Binomial probability]]

	~~==References==~~
	~~{{reflist}}~~

	~~==External links==~~
	*[http://highered.mcgraw-hill.com/sites/dl/free/0072549076/79746/ch04_p175.pdf ''Complementary events''] - (free) page from probability book of [[McGraw-Hill]]

	~~{{DEFAULTSORT:Complementary Event}}~~
	~~[[Category:Probability theory]]~~

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2013-12-20T08:10:18Z

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← Older revision		Revision as of 10:10, 20 December 2013
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	~~Biotechnologist Clarence Verderber from Deep River~~, ~~really likes fast~~, ~~new launch property singapore~~ and ~~crochet~~. ~~Gets inspiration through travel~~ and ~~just spent 3 weeks at Iguazu National Park~~.<br><br>~~Feel free~~ to ~~surf~~ to ~~my webpage ::~~ [http://~~customercare~~.~~zentelecom~~.ie/~~entries~~/~~30920185~~-~~Residential~~-~~Web~~-~~Site~~-At-~~Yishun~~-~~Avenue~~-9-~~Attracts~~-5-~~Bids new property launch singapore~~]		{{Probability fundamentals}}

			In [[probability theory]], the '''complement''' of any [[event (probability theory)\|event]] ''A'' is the event [not ''A''], i.e. the event that ''A'' does not occur.<ref>Robert R. Johnson, Patricia J. Kuby: ''Elementary Statistics''. Cengage Learning 2007, ISBN 978-0-495-38386-4, p. 229 ({{Google books\|DWCAh7jWO98C\|restricted online copy\|page=229}})</ref> The event ''A'' and its complement [not ''A''] are [[mutually exclusive]] and [[Collectively exhaustive events\|exhaustive]]. Generally, there is only one event ''B'' such that ''A'' and ''B'' are both mutually exclusive and exhaustive; that event is the complement of ''A''. The complement of an event ''A'' is usually denoted as ''A′'', ''A<sup>c</sup>'' or ''{{overline\|A}}''.

			For example, if a typical coin is tossed and one assumes that it cannot land on its edge, then it can either land showing "heads" or "tails." Because these two [[Outcome (probability)\|outcomes]] are mutually exclusive (i.e. the coin cannot simultaneously show both heads and tails) and collectively exhaustive (i.e. there are no other possible outcomes not represented between these two), they are therefore each other's complements. This means that [heads] is logically equivalent to [not tails], and [tails] is equivalent to [not heads].

			==Complement rule==
			In a [[Experiment (probability theory)\|random experiment]], the probabilities of all possible events (the [[sample space]]) must total to 1— that is, some outcome must occur on every trial. For two events to be complements, they must be [[Collectively exhaustive events\|collectively exhaustive]], together filling the entire sample space. Therefore, the probability of an event's complement must be [[Unity (mathematics)\|unity]] minus the
			probability of the event.<ref name="yates">{{cite book \| last = Yates \| first = Daniel S. \| last2 = Moore \| first2 = David S \| last3 = Starnes \| first3 = Daren S. \| year = 2003 \| title = The Practice of Statistics \| edition = 2nd \| publisher = [[W. H. Freeman and Company\|Freeman]] \| location = New York \| url = http://bcs.whfreeman.com/yates2e/ \| isbn = 978-0-7167-4773-4}}</ref>{{rp\|342-343}} That is, for an event ''A'',
			:<math>P(A^c) = 1 - P(A).</math>
			Equivalently, the probabilities of an event and its complement must always total to 1. This does not, however, mean that ''any'' two events whose probabilities total to 1 are each other's complements; complementary events must also fulfill the condition of [[Mutually exclusive events\|mutual exclusivity]].

			===Example of the utility of this concept===
			Suppose one throws an ordinary six-sided die eight times. What is the probability that one sees a "1" at least once?

			It may be tempting to say that

			: Pr(["1" on 1st trial] or ["1" on second trial] or ... or ["1" on 8th trial])

			:= Pr("1" on 1st trial) + Pr("1" on second trial) + ... + P("1" on 8th trial)

			:= 1/6 + 1/6 + ... + 1/6.

			:= 8/6 = 1.3333... (...and this is clearly wrong.)

			That cannot be right because a probability cannot be more than 1. The technique is wrong because the eight events whose probabilities got added are not mutually exclusive.

			Instead one may find the probability of the complementary event and subtract it from 1, thus:

			: Pr(at least one "1") = 1 − Pr(no "1"s)

			:= 1 − Pr([no "1" on 1st trial] and [no "1" on 2nd trial] and ... and [no "1" on 8th trial])

			:= 1 − Pr(no "1" on 1st trial) × Pr(no "1" on 2nd trial) × ... × Pr(no "1" on 8th trial)

			:= 1 −(5/6) × (5/6) × ... × (5/6)

			:= 1 − (5/6)<sup>8</sup>

			:= 0.7674...

			==See also==
			*[[Negation\|Logical complement]]
			*[[Exclusive disjunction]]
			*[[Binomial probability]]

			==References==
			{{reflist}}

			==External links==
			*[http://highered.mcgraw-hill.com/sites/dl/free/0072549076/79746/ch04_p175.pdf ''Complementary events''] - (free) page from probability book of [[McGraw-Hill]]

			{{DEFAULTSORT:Complementary Event}}
			[[Category:Probability theory]]

en>LoremIpsumDolorSitAsmet: fixed a few grammar errors and made examples uniform

2012-08-20T21:29:03Z

fixed a few grammar errors and made examples uniform

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