Iingcamango ezinokwenzeka. Ubungakanani bokuba isiganeko, iziganeko ezingenangqungquthela (inkolelo enokwenzeka). Iziganeko ezizimeleyo nezingahambelaniyo kwiingcamango ezinokwenzeka

Akunakwenzeka ukuba abantu abaninzi bacinga ukuba kunokwenzeka ukubala iziganeko, leyo elithile ngengozi. Ukuze wayibeka amagama alula, kusengqiqweni ukuba yeyiphi icala tyhubhu kwi idayisi ziya kuwa kwixesha elilandelayo. Yaba lo mbuzo ukubuza oosonzululwazi ezimbini ezinkulu, wabeka isiseko kule inzululwazi, ithiyori lilonke, amathuba isiganeko apho ngokubanzi ngokwaneleyo kufundwa.

esizukulwaneni

Ukuba uzama ukucacisa ingqiqo efana kwithiyori ngokuqinisekileyo, sifumana oku kulandelayo: le yenye amasebe yemathematika ukuba ufunda rhoqo iziganeko random. Ngokucacileyo, le ngcamango ngokwenene ayichazi kakuhle, ngoko ke kufuneka siqonde ukuba iinkcukacha ngakumbi.

Ndingathanda ukuqala wabaseki le ngcamango. Njengoko kukhankanyiwe ngasentla, kwakukho ababini, ukuba Per Funa kunye Blez Paskal. Baba wokuqala nokuzama usebenzisa ifomula kunye izibalo ukubala kwisiphumo sesenzo. Ngokubanzi, iziqalelo le isayensi nkqu bamaXesha Aphakathi. Nangona iinkcuba ezahlukeneyo kunye izazinzulu baye bazama ekuhlalutyeni ekuyo ivideo imidlalo efana roulette, craps, njalo njalo, ngaloo ukuseka ipateni, kunye nepesenti ilahleko inani. Isiseko nalo babekwa ngenkulungwane elinesixhenxe yaba abaphengululi ngentla.

Ekuqaleni, umsebenzi wabo akukho sezandleni impumelelo enkulu kule entsimini, emva bonke, oko wakwenzayo, baba nje izibakala ngokwamava imifuniselo baba ngokucacileyo ngaphandle usebenzisa ifomula. Ekuhambeni kwexesha, yajika ukufezekisa iziphumo ezinkulu, owayebonakele ngenxa kusakhelwa maka emathanjeni. Oku le sixhobo uye wanceda ukuzisa indlela yokuqala enomahluko.

abalandeli

Singasathethi indoda enjengam nje Christiaan Huygens, sikwiphulo abenza esi sifundo elibizwa igama "theory amathuba" (okuba isiganeko sibalaselisa ngale science). Lo mntu umdla kakhulu. Yena, ngokunjalo izazinzulu uchazwe apha ngasentla zizanywayo ngohlobo iifomyula zemathematika baphethe uthotho lweziganeko random. Kuyabonakala ukuba akazange sabelane kunye Pascal kunye Fermat, oko kukuthi yonke imisebenzi yakhe ziyadibana ezo ezingqondweni. Huygens yayifumana nezimaphambili ezingundoqo kwethiyori lilonke.

An into umdla kukuba umsebenzi wakhe weza elide phambi kokuba iziphumo imisebenzi oovulindlela, ukuba ngqo, kwiminyaka engamashumi ngaphambili. Kukho kuphela phakathi iikhonsepthi ezichongiweyo:

• njengoko ingqiqo lwamaxabiso kungenzeka ithuba;
• silindele ukuba ityala ekhethekileyo;
• theorems yokudibanisa kunye phindo ebezinokwenzeka.

Kwakhona, umntu asingekhe silibale Yakoba Bernulli, naye negalelo isifundo ngxaki. Ngenxa yazo, nokuba ngubani na iimvavanyo ezizimeleyo, wakwazi ukuba ukunika ubungqina umthetho amanani amakhulu. Kwelinye icala, oosonzululwazi Poisson kunye Laplace, owayesebenza benkulungwane yeshumi elinesithoba, sakwazi ukubonisa theorem yokuqala. Ukususela kulaa mzuzu ukuya ukuhlalutya zeempazamo imigqaliselo saqala ukusebenzisa ithiyori lilonke. Party malunga isayensi kwaye Russian izazinzulu, kunokuba Markov, Chebyshev kunye Dyapunov. Bona isekelwe kumsebenzi owenziwa buchopho omkhulu, balinqabisa kwisifundo njengesebe lemathematika. Sasisebenza la manani ekupheleni kwenkulungwane yeshumi elinesithoba, kunye enkosi ngegalelo labo, ziye izenzakalo obuvunyiweyo ezifana:

• umthetho amanani amakhulu;
• Theory of amatsheyini Markov;
• Umda theorem engundoqo.

Ngoko ke, imbali ngokuzalwa inzululwazi kunye babantu ezinkulu negalelo kuyo, yonke into ngaphezulu okanye kucacile ngaphantsi. Ngoku lixesha lokuba inyama kuzo zonke izibakala.

nezimaphambili ezingundoqo

Phambi kokuba uthinte imithetho theorems kufuneka bafunde nezimaphambili ezingundoqo kwethiyori lilonke. Iindawo singene indima ephezulu. Le isihloko kunokuba ebanzi, kodwa akayi kuba nako ukuqonda bonke abanye ngaphandle kwayo.

Iindawo e theory okuba - it Nayiphi iseti zeziphumo nokuhlola. Concepts yale meko akukho ngokwaneleyo. Ngenxa yoko, Lotman Scientist ukusebenza kule ndawo, uye wabonisa ukuba kule meko sithetha ngento "Kwathi, nangona ingakwazanga kwenzeke."

iziganeko Random (theory amathuba byi ingqalelo ekhethekileyo kubo) - yingcamango obandakanya ngokupheleleyo nayiphi na isenzeko ukuba kungenzeka ukuba zenzeke. Okanye, phezu koko, le meko ngeke kwenzeke ukusebenza ezahlukeneyo iimeko. Kwakhona kubalulekile ukwazi ukuba uphethe wonke umthamo nesenzeko olwenzeke iziganeko nje random. theory kwenzeke icebisa ukuba zonke iimeko kungaba waphinda rhoqo. Yihambo kwabo kuye ngokuba "amava" okanye "uvavanyo."

umcimbi ebalulekileyo - lo mba ukuba likhulu ekhulwini kolu vavanyo kwenzeka. Ngako oko, isiganeko akunakwenzeka - le yinto alwenzeki.

Ukudibanisa ngababini Action (ngokuqhelekileyo ityala A and B) na isenzeko leyo eyenzeka ngexesha elinye. Bona ngokuthi AB.

Isixa leeperi iziganeko A and B - C, ngamanye amazwi, ukuba ubuncinane omnye wabo uya (A okanye B), ufumana C. Ifomula nesenzeko echazwe kubhaliwe njengoko C = A + B

Uphuhliso ezingahambelaniyo imfundiso lilonke kuthetha ukuba amatyala amabini azikhethe kuphela. Ngelo xesha linye xa kunokwenzeka naluphi na ayikwazi kwenzeka. iziganeko Joint theory amathuba - oku antipode zabo. Isiphumo soku kukuba xa A Kwathi, loo nto ayikuthinteli C.

Ekuchaseni isiganeko (theory amathuba iqwalasela ngokweenkcukacha ezinkulu), kulula ukuyiqonda. Kungcono ukujongana nabo uthelekiso. Sekukancinane ukuba uphuhliso njengoko engahambelaniyo efanayo kwi kwithiyori lilonke. Nangona kunjalo, umahluko yabo kukuba sinye ezininzi ngezenzeko kuyo nayiphi na imeko ukuba yenzeke.

Enye kusenokwenzeka iziganeko - ezo izenzo, ithuba phindo uyalingana. Ukwenza kucace, ungabona ukujula imali: ukulahlekelwa enye kumacala zayo ilahleko mhlawumbi ngokulinganayo abanye.

kulula ukuqwalasela umzekelo ngokukhetha siganeko. Masithi kukho isiqephu kwi episode A. I lokuqala - kumqulu bafa kunye kokufika inani elingumnqakathi, kwaye eyesibini - imbonakalo inani ezintlanu idayisi. Emva koko kwenzeka ukuba A yi V. ababefuna

iziganeko Independent theory amathuba kulindeleke kuphela kwizihlandlo ezimbini okanye ngaphezulu nokubandakanya elizimeleyo naliphi na inyathelo ukusuka kwenye. Umzekelo, A - xa ilahleko imisila ezinkozo oluvunjululweyo, kunye B - dostavanie jack ukusuka kumgangatho. Baye iziganeko ezizimeleyo theory lilonke. Ukususela ngalo mzuzu kwacaca.

iziganeko abaxhomekeke theory amathuba kwakhona kuvumelekile kuphela iseti yabo. Baya kubonisa sixhomekeke omnye phezu komnye, oko kukuthi, isenzeko angenzeka kuphela xa A sele lwenzekile okanye, phezu koko, zange yenzeke xa - imeko ezingundoqo B.

Isiphumo olu lingelo random eyenziwa licandelo elinye - ukuba iziganeko nezakhiwo. theory Ingqikelelo ithi ukuba yinkolelo yokuba kwenziwa kanye kuphela.

formula esisiseko

Ngenxa yoko, ngasentla ziye ingqalelo umba "isiganeko", "ithiyori amathuba", iinkcazelo zamagama angundoqo yale inzululwazi nayo kunikelwa. Ngoku lixesha lokuba baziqhelanise ngokwayo kunye neefomula ebalulekileyo. La magama ngokwezibalo kungqinwa zonke iingqiqo eziphambili kwisifundo enzima efana kwithiyori lilonke. Amathuba isiganeko kwaye lidlala indima enkulu.

Kulungile ukuqala kunye neefomula ezingundoqo combinatorics. Kwaye phambi kokuba uqale kwabo, kubalulekile ukuqwalasela ukuba yintoni na.

Combinatorics - isikakhulu isebe imathematika, uye sele befunda inani enkulu integers, kunye lotshintsho lobeko ezahlukeneyo zombini amanani kunye nezinto zabo, iinkcukacha ezahlukeneyo, njl, ekhokelela inani lwendibaniso ... Ukongeza kwithiyori ngokuqinisekileyo, eli shishini kubalulekile ukuba manani, inzululwazi ikhompyutha zokufihlakeleyoComment.

Ngoko ke ungahambisa kwi le ntetho bona kunye neefomula zabo definition.

Eyokuqala kwezi ibinzana ukuba inani lotshintsho lobeko, oko ngolu hlobo lulandelayo:

P_n = n ⋅ (n - 1) ⋅ (n - 2) ... 3 2 ⋅ ⋅ 1 = n!

Equation usebenza kuphela kwimeko ukuba izinto azifani kuphela ngokomyalelo lungiselelo.

Ngoku ukubekwa ifomula, ikhangeleka njengalo ziya kuqwalaselwa;

A_n ^ m = n ⋅ (n - 1) ⋅ (n-2) ⋅ ... ⋅ (n - m + 1) = n! : (N - m)!

Eli binzana ingafakwa kuphela element kuphela ukubeka umyalelo, kodwa lokuqamba yayo.

Le equation sesithathu combinatorics, kwaye le yokugqibela, kuthiwa Ifomula inani lwendibaniso:

C_n ^ m = n! : ((N - m))! : M!

Kolimo ebizwa zaphandwa ndzulu, ezo ode, ngokulandelelana, kwaye isicelo kulo mthetho.

Kunye iifomyula ze combinatorics beza ukuqonda ngokulula, ngoku ukuya nkcazo weklasiki lilonke. Kubonakala ngathi eli binzana ngolu hlobo lulandelayo:

P (A) = m: n.

Kule formula, m - linani imo evumayo siganeko A, kwaye n - inani lemisitho nezakhiwo ngokulinganayo nangokupheleleyo zonke.

Kukho amabinzana amaninzi kweli nqaku aziyi kuqwalaselwa nantoni kodwa abachaphazelekayo iya kuba izezona zibalulekileyo njengoko hlobo, umzekelo, amathuba iziganeko imali:

P (A + B) = P (A) + P (B) - le theorem yokongeza kuphela iziganeko ahlobene;

P (A + B) = P (A) + P (B) - P (AB) - kodwa oku kuphela ngokongeza ehambelanayo.

Amathuba imisebenzi isiganeko:

P (A ⋅ B) = P (A) ⋅ P (B) - le theorem ukuba iziganeko ezizimeleyo;

(P (A ⋅ B) = P (A) ⋅ P (B | A): P (A ⋅ B) = P (A) ⋅ P (A | B)) - kwaye oku ngenxa zixhomekeke.

uluhlu yaphela ubisi iziganeko. Ithiyori kungenzeka isixelela theorem Bayes, leyo sijongeka njengoku:

P (H_m | A) = (P (H_m) P (A | H_m)): (Σ_ (k = 1) ^ n P (H_k) P (A | H_k)), m = 1, ..., n

Kule formula, H _1, H _2, ..., H _n - yinto iseti epheleleyo yeengcinga.

Kule stop, isicelo iisampulu fomyula ngoku kuqwalaselwa ukuba imisebenzi ethile ekusebenzeni.

izibonelo

Ukuba ufunda ngenyameko naliphi na isebe kwimathematika, asiyonto ngaphandle exercises kunye nezisombululo isampula. Kunye nethiyori lilonke: iziganeko, imizekelo apha yinxenye ebalulekileyo kakhulu eqinisekisa izibalo zenzululwazi.

Ifomula inani lotshintsho lobeko

Ngokomzekelo, xa sinendawo ikhadi amakhadi amathathu, ukuqala kunye nalowo wesiqhelo. umbuzo olandelayo. Zingaphi iindlela ulisonge kumgangatho ukuze amakhadi ixabiso ubuso enye amabini e- elilandelayo?

Lo msebenzi imiselwe, ngoku makhe Masishukume ukuze bajongane nayo. Okokuqala kufuneka imisele inani lotshintsho lobeko izakhi mathathu, ngenxa yale njongo sithatha indlela ngasentla, kungcono ninikane P_30 = 30!.

Ngokusekelwe kulo mthetho, siyazi ukuba zingaphi iindlela kukho ukuze babeke phantsi kumgangatho ngeendlela ezininzi, kodwa kufuneka kutsalwa kuzo zezo apho ikhadi yokuqala neyesibini ziya kuba elandelayo. Ukuze wenze oku, qala uhlobo, xa kuqala ukwakweli wesibini. Kubonakala ukuba imaphu yokuqala kungathatha iindawo ezimashumi mabini anesithoba - ukususela ekuqaleni ukuya lesithoba-mabini, kunye ikhadi yesibini ukusuka yesibini ukuya kumashumi amathathu, ujika izihlalo ezimashumi mabini anesithoba izibini cards. Ngenxa yoko, abanye bakwazi ukuthatha izihlalo ezimashumi mabini anesibhozo, kwaye nawuphi na umyalelo. Oko kukuthi, ukuba ngokutsha amakhadi ezingamashumi amabini anesibhozo iinketho mabini anesibhozo P_28 = 28!

Isiphumo kukuba xa siqwalasela isigqibo, xa ikhadi lokuqala phezu ithuba extra yesibini ukufumana 29 ⋅ 28! = 29!

Ukusebenzisa indlela enye, kufuneka abale inani iinketho njengomntu ongenasithuba sangqesho ngenxa ityala xa ikhadi lokuqala ibekwe phantsi yesibini. Kwakhona wafumana 29 ⋅ 28! = 29!

Kulo kulandela ukuba iinketho ezi-2 ⋅ 29!, Ngelixa iindlela eziyimfuneko ukuqokelela kumgangatho-30! - 2 ⋅ 29!. Kuhleli kuphela ukubala.

30! = 29! ⋅ 30; 30 - 2 ⋅ 29! = 29! ⋅ (30 - 2) = 29! ⋅ 28

Ngoku kufuneka ukuba ukwanda kunye bonke amanani ukusuka kwelinye ukuya mabini anesithoba, yaye ngoko, ekupheleni kwazo zonke iphinda phindiweyo ngo 28. Impendulo wafumana 2,4757335 ⋅ 〖〗 10 ^ 32

Imizekelo izisombululo. Ifomula inani yokuhlala

Kule ngxaki, kufuneka ufumanise ukuba zingaphi kukho iindlela ukubeka imiqulu kweshumi elinesihlanu kwithala, kodwa phantsi komqathango wokuba imiqulu amathathu kuphela.

Kulo msebenzi, isigqibo lula kancinane odlulileyo. Ukusebenzisa ifomula sele kwaziwa, kuyimfuneko ukuba ukubala inani elipheleleyo kwiindawo ezimashumi imiqulu elinesihlanu.

A_30 ^ 15 = 30 ⋅ 29 ⋅ ... ⋅ 28⋅ (30 - 15 + 1) = 30 ⋅ 29 ⋅ 28 ⋅ ... ⋅ 16 = 202 843 204 931 727 360 000

Impendulo, ngokulandelelanayo, iya kulingana 202 843 204 931 727 360 000.

Ngoku thatha lo msebenzi unzima ngakumbi kancinane. Kufuneka wazi ukuba zingaphi kukho iindlela amalungiselelo iincwadi ezingamashumi amathathu anesibini ezishelfini, phantsi kwelungiselelo elithi imiqulu elinesihlanu kuphela kuhlala phezu kwithala efanayo.

Phambi ekuqaleni isigqibo ungathanda ukucacisa ukuba ezinye iingxaki ingasombululwa ngeendlela ezininzi, yaye kule kukho iindlela ezimbini, kodwa kokubini, sona fomula inye.

Kulo msebenzi, ungenza uthabathe impendulo kowangaphambili, ngenxa yokuba siye kubalwa inani yamaxesha aninako uzalise kwithala yeencwadi ezilishumi elinesihlanu ngeendlela ezahlukeneyo. Kwacaca A_30 ^ 15 = 30 ⋅ 29 ⋅ 28 ⋅ ... ⋅ (30 - 15 + 1) = 30 ⋅ 29 ⋅ 28 ⋅ ... ⋅ 16.

Webutho yesibini ibalwa le obekulindelekile ifomula, ngokuba ibekwe iincwadi elinesihlanu, ngoxa intsalela elinesihlanu. Sisebenzisa P_15 ifomula = 15!.

Kubonakala ukuba le mali uya A_30 ^ 15 ⋅ iindlela P_15, kodwa, ukongeza, wayeza udibaniso lwamaxabiso wonke amanani ukusuka mathathu ukuya elinesithandathu liphindaphindwe imveliso amanani ukusuka kwelinye ukuya kwishumi elinesihlanu, ekugqibeleni kuphumelela udibaniso lwamaxabiso wonke amanani ukusuka kwelinye ukuya mathathu, oko kukuthi impendulo 30!

Kodwa le ngxaki ingasombululwa ngendlela eyahlukileyo - lula. Ukuze wenze oku, ungabona ukuba kukho into enye kwi yeencwadi mathathu. Bonke bephela abekwe kulo moya, kodwa ngenxa yokuba imeko ifuna ukuba kwakukho iishelfu ezimbini, enye ubude thina namarhela kubini, ubhekisa ezimbini elinesihlanu. Ukusuka oku kwenzeka ukuba eli lungiselelo kuba P_30 = 30!.

Imizekelo izisombululo. Ifomula inani edibeneyo

Ngubani kugqalwa eyongeziweyo lobeko lwe ingxaki wesithathu combinatorics. Kufuneka wazi ukuba zingaphi iindlela ezikhoyo ukuba amalungiselelo iincwadi elinesihlanu kwi imeko ukuba umele ukhethe ukusuka mathathu ncam.

Kuba isigqibo ziya, Kakade ke, sebenzisa Ifomula inani lwendibaniso. Ukususela inqobo kuye kucace ukuba umyalelo yeencwadi elinesihlanu efanayo ayibalulekanga. Ngoko ekuqaleni kufuneka ufumanise inani lilonke lwendibaniso iincwadi amathathu elinesihlanu.

C_30 ^ 15 = 30! : ((30-15))! : 15! = 155117520

Kuko konke. Ukusebenzisa le formula, kweli xesha lifutshane kangangoko zokusombulula ingxaki enjalo, impendulo, ngokulandelelana, lingana ne ukuya 155.117.520.

Imizekelo izisombululo. Le nkcazelo classic lilonke

Ukusebenzisa ifomula ezinikwe apha ngentla, umntu unako ukufumana impendulo kumsebenzi elula. Kodwa uya kubona ngokucacileyo kwaye ulandela ikhondo amanyathelo.

Lo msebenzi wanikwa ukuba i ingqayi kukho iibhola ezilishumi twatse ngokupheleleyo. Kwezi, ezine tyheli ezintandathu oluhlaza. Ithathwe ukusuka ala ibhola enye. Kuyimfuneko ukwazi amathuba dostavaniya oluhlaza.

Ukusombulula le ngxaki kuyimfuneko ukuba ityumbe dostavanie blue isiganeko ibhola A. Lo amava abe iziphumo ezilishumi, mvisiswano leyo, kwelinye icala, cebetshu mhlawumbi ngokulinganayo. Ngelo xesha, ezintandathu ezilishumi mahle kwesiganeko A. Sombulula le fomyula ilandelayo:

P (A) = 6: 10 = 0.6

Ukusebenzisa le formula, siye safumanisa ukuba kungenzeka dostavaniya ibhola blue ngu 0.6.

Imizekelo izisombululo. Namathuba okuba mali iziganeko

Ngubani na oya kuba eyongeziweyo lobeko leyo isonjululwe ngokusebenzisa indlela ye amathuba mali iziganeko. Ngoko ke, enikwe imeko ukuba kukho iimeko ezimbini, eyokuqala ndiyingwevu neebhola ezintlanu ezimhlophe, ngoxa yesibini - iibhola ezimhlophe ezisibhozo engwevu ezine. Ngenxa yoko, iibhokisi yokuqala neyesibini baye bathatha omnye kubo. Kuyimfuneko ukuba ukufumanisa ukuba yintoni na amathuba wayeswele iibhola ezi zingwevu namhlophe.

Ukusombulula le ngxaki, kuyimfuneko ukuchonga siganeko.

• Ngenxa yoko, A - sibe ibhola izimvi kwibhokisi lokuqala: P (A) = 1/6.
• A '- ibhalbhu abamhlophe asuswe kwayiyo esuka kwibhokisi lokuqala: P (A') = 5/6.
• I - sele kufunxwa ibhola izimvi le conduit yesibini: P (B) = 2/3.
• B '- wathabatha ibhola izimvi kokha yesibini: P (B') = 1/3.

Ngokutsho ngxaki oko kuyimfuneko ukuba enye into eyenzekayo: AB 'okanye' B. Usebenzisa ifomula, sinokuyifumana: P (AB ') = 1/18, P (A'B) = 10/18.

Ngoku ke wasebenzisa formula phinda linokuba. Okulandelayo, ukuze ufumane impendulo, kufuneka ukuba ufake isicelo equation behlanganisa:

P = P (AB '+ A'B) = P (AB') + P (A'B) = 11/18.

Yiloo ndlela, usebenzisa ifomula, uyakwazi ukucombulula iingxaki ezinjalo.

ngenxa

Iphepha ithiwe thaca kwi-ulwazi "theory okuba", amathuba iziganeko zidlala indima ebalulekileyo. Kakade ke, yonke into othe ingqalelo, kodwa ngokusekelwe okubhaliweyo thaca, ungenza kuthiwa uqhelane nale isebe lemathematika. isayensi Izincomo kunokuba luncedo nje kuphela ishishini umsebenzi, kodwa kubomi bemihla ngemihla. Unga sebenzisa oku ukuba ukubala naziphi na isiganeko.

Isicatshulwa bayachatshazelwa imihla ebalulekileyo kwimbali kuphuhliso ingcamango amathuba njenge inzululwazi, yaye amagama labantu abaye ezifakwa kuyo imisebenzi. Yiloo ndlela ukwazi yabantu kukhokelele kwinto yokuba abantu bafunde ukubala, nkqu iziganeko random. Xa zinomdla nje oku, kodwa namhlanje sele kubo bonke. Kwaye akukho mntu uthi yintoni eza kwenzeka kuthi kwixa elizayo, yintoni enye brilliant ezifunyaniswe ezinxulumene kwithiyori phantsi ingqalelo, babeza. Kodwa inye into ngokuqinisekile - isifundo nangoku angafanele nto!