Imfundiso yokuziphendukela okungenzeka. Amathuba umcimbi, umcimbi ngezikhathi ezithile (okungenzeka theory). ngentuthuko Independent futhi akuhambisani imfundiso yokuziphendukela okungenzeka

Cishe, akunakwenzeka ukuba abantu abaningi bacabanga ukuthi kungenzeka ukubala izenzakalo, okuyinto ngezinga elingakanani ngephutha. Ukuze ukubeke ngamazwi alula, ingabe kunengqondo ukuthi iyiphi uhlangothi cube e idayisi ziyokuwa ngokuzayo. Kwaba lo mbuzo ukubuza ososayensi ezimbili ezinkulu, wabeka isisekelo salokhu isayensi, imfundiso ecatshangelwayo, ematfuba umcimbi lapho wafunda kabanzi ngokwanele.

isizukulwane

Uma uzama ukuchaza umqondo ezifana imfundiso yokuziphendukela okungenzeka, sithola lokhu okulandelayo: lokhu ingenye amagatsha wezibalo ukuthi etadisha ukuqinisela imicimbi okungahleliwe. Ngokusobala, lo mqondo ngempela akazivezi essence, ngakho kudingeka ucabangele ke kabanzi.

Ngithanda ukuqala abasunguli kwemfundiso yokuziphendukela. Njengoba kubonisiwe ngenhla, kwakukhona amabili, ukuthi Per Ferma futhi Blez Paskal. Yibona abaqala Uzame ngokusebenzisa amafomula izibalo ukuze abale umphumela umcimbi. Ngokuvamile, iziqalo yalesi isayensi ngisho Ephakathi. Nakuba ongqondongqondo ehlukahlukene ososayensi baye bazama sihlaziye imidlalo yasekhasino ezifana roulette, Craps, nokunye, ngokwenza njalo ukuze abasebenzi nomkhuba, futhi ukulahlekelwa iphesenti inombolo. Isisekelo ukuthi wamnika ekhulwini lesikhombisa kwaba izazi esesikhulume ngazo.

Ekuqaleni, umsebenzi wabo ayikwazanga chasiselwe impumelelo enkulu kulo mkhakha, phela lokho abakwenza, bathi zazimane amaqiniso enokwehla futhi ucwaningo ayeyizidalwa ngaphandle kokusebenzisa amafomula. Ngokuhamba kwesikhathi, baphendukela kuzuzwe imiphumela elikhulu, eyabonakala ngenxa okwashiwo ukhonkolo we amathambo. It is lokhu ithuluzi iye yasiza ukuletha ifomula lokuqala ahlukile.

abasekeli

Ingasaphathwa indoda efana Christiaan Huygens, enqubeni ukutadisha isihloko ukuthi iqanjwe "okungenzeka theory" (ematfuba umcimbi uqokomisa kule isayensi). Lo muntu ithakazelisa kakhulu. Yena, kanye ososayensi ethulwa ngenhla, balingwa ngendlela zezibalo baphethe iphetheni izenzakalo ezingahleliwe. Kuyaphawuleka ukuthi akazange usabe Pascal futhi Fermat, okungukuthi wonke umsebenzi wakhe akusho zenqane labo izingqondo. Huygens etholakala nemiqondo eyisisekelo okungenzeka theory.

Kuvele Eqinisweni ezithakazelisayo wukuthi umsebenzi wakhe kwaphuma eside ngaphambi imiphumela imisebenzi lamaphayona, ukuba ngqo, engamashumi amabili ekuqaleni. Kukhona kuphela phakathi imiqondo ezikhonjwe kwakuyilawa:

• njengoba nomqondo okungenzeka amanani ithuba;
• silindele kunjalo ezehlukene;
• theorems kokuhlanganisa ukuphindwaphindwa kwenzeke.

Futhi, omunye singakhohlwa Yakoba Bernulli, naye waba negalelo elikhulu cwaningo yenkinga. Ngokusebenzisa bebodwa, futhi abangabamemezeli ukuhlolwa ezimele, wakwazi ukunikeza ubufakazi umthetho izinombolo ezinkulu. Ngakwelinye ihlangothi, ososayensi Poisson futhi Laplace, owayesebenza ekuqaleni kwekhulu nesishiyagalolunye, yakwazi ukuveza ubufakazi bokuthi theorem yasekuqaleni. Kusukela ngaleso sikhathi ukuhlaziya amaphutha kokuma saqala ukusebenzisa okungenzeka theory. Party kule isayensi bahluleka futhi Russian ososayensi, kunalokho Markov, Chebyshev futhi Dyapunov. Zisekelwe umsebenzi owenziwa geniuses ezinkulu, esethole ezifana nokuthatha isifundo njengamaphuzu legatsha wezibalo. Sasisebenza lezi zibalo ekupheleni yekhulu nesishiyagalolunye, futhi sibonga umnikelo wabo, ziye kungase kutholakale ezinjengokungcoliswa:

• umthetho izinombolo ezinkulu;
• Theory ka Markov lazo ngisemaketangeni;
• Emaphakathi umkhawulo theorem.

Ngakho, umlando wokuzalwa isayensi ne nobuntu ezinkulu ezibe nomthelela ke, konke ucishe ecacile. Manje sekuyisikhathi sokuba inyama aphume wonke amaqiniso.

imiqondo eyisisekelo

Ngaphambi uthinte imithetho futhi theorems bafanele bafunde imiqondo eyisisekelo okungenzeka theory. Izinkundla ke unendima evelele. Lesi sihloko kunalokho ebanzi, kodwa ngeke bakwazi ukuqonda lonke ngaphandle kwalo.

Izinkundla e okungenzeka theory - ke Noma yimuphi isethi yemiphumela ukuhlola. Concepts zalesi simo kukhona akwanele. Ngakho, Lotman usosayensi ukusebenza kule ndawo, uzwakalise ukuthi kuleli cala sikhuluma ngalokho "okwenzekile, nakuba kwakungase kungenzeki."

Izenzakalo ezingahleliwe (okungenzeka theory unaka ngokukhethekile kubo) - ngumqondo esibandakanya ngokuphelele yimuphi lomkhuba kokuba kungenzeka kwenzeke. Noma, kunalokho, lesi simo ngeke kwenzeke ukusebenza ezihlukahlukene izimo. Kubalulekile ukwazi ukuthi luthathe wawuqeda wonke lo mqulu we izenzakalo ezenzeka izenzakalo nje okungahleliwe. Okungenzeka mbono zonke izimo ziphindwe njalo. Kuwukuziphatha yabo iye yabizwa ngokuthi "umuzwa" noma "ukuhlolwa."

isenzakalo esibalulekile - lena lomkhuba wokuthi kuyinto eziyikhulu amaphesenti kulolu vivinyo kwenzeke. Ngakho, lo mcimbi akunakwenzeka - lokhu kuyinto kungenzeki.

Ukuhlanganisa ngazimbili Action (evamile kunjalo A no icala B) iyisici okuyinto eyenzeka ngesikhathi esisodwa. Yiwo okuthiwa njengoba AB.

Inani ngazimbili imicimbi A no B - C, ngamanye amazwi, uma okungenani omunye wabo (A noma B), uthola C. Ifomula mkhuba echazwe kulotshiwe njengoba C = A + B

ngentuthuko Akuhambisani imfundiso yokuziphendukela okungenzeka kusho ukuthi izimo ezimbili kukhona alizimele. Ngesikhathi esifanayo zingabantu kunoma yiliphi icala kungenzeki. izenzakalo ngokuhlanganyela okungenzeka theory - kuba antipode yabo. Lokhu kusho ukuthi uma A kwenzeke, akuvimbeli C.

Ukumelana umcimbi (okungenzeka theory ubabheka ngokuningiliziwe enkulu), kulula ukuqonda. Kungcono ukubhekana nazo uma uziqhathanisa nabo. Sekusele kancane efanayo ngentuthuko njengoba engahambisani imfundiso yokuziphendukela okungenzeka. Nokho, umehluko yabo wukuthi omunye sebuningini imihlola noma kunjalo kumele kwenzeke.

Ngokulinganayo Cishe izenzakalo - lezo zenzo, kungenzeka ukuphindaphinda ilingana. Ukuze ngikwenze kucace, ungacabanga bephonsa uhlamvu: ukulahlekelwa elilodwa ezinhlangothini zalo kuyinto ukulahlekelwa ngokulinganayo kungenzeka nezinye.

kulula cabanga ngesibonelo besekela umcimbi. Ake sithi kukhona yisenzakalo isiqephu A. lokuqala - umqulu die nge obhekwe inombolo engalingani, kanti eyesibili - ukubukeka inombolo emihlanu idayisi. Khona-ke kuvela ukuthi A V lokwamukelwa

izenzakalo Independent e okungenzeka theory kuthiwa esivezwa kuphela izikhathi ezimbili noma ngaphezulu futhi kuhilelani esizimele noma yisiphi isenzo kusuka kwenye. Ngokwesibonelo, A - at ukulahlekelwa imisila uhlamvu Yasebusuku, no-B - dostavanie Jack kusukela emphemeni. Sinenkosi izenzakalo ngekutimela okungenzeka theory. Kusukela kulo mzuzwana kwaba sobala.

izenzakalo oncikile e okungenzeka theory futhi kuvunyelwe kuphela iqoqo yabo. Asikisela ukwencika lomunye, okungukuthi, lomkhuba singenzeka kuphela esimweni lapho A selwenzekile kakade noma, kunalokho, akwenzekanga uma - isimo esiyinhloko B.

Umphumela ukuhlolwa okungahleliwe ehlanganisa ingxenye eyodwa - kungcono izenzakalo aphansi. Okungenzeka theory lithi iyisici ukuthi kwenziwa kanye kuphela.

ifomula eyisisekelo

Ngakho, la mazwi angenhla zazibhekwa welithi "umcimbi", "okungenzeka theory", izincazelo magama ayinhloko kwale sayensi Naye wanikelwa. Manje sekuyisikhathi sokuba yazi ngokwayo ne amafomula ezibalulekile. La magama izibalo waqinisekisa bonke imiqondo esemqoka a subject ezinjalo kuba nzima njengoba imfundiso yokuziphendukela okungenzeka. Amathuba umcimbi futhi kunendima enkulu.

Kungcono ukuqala nge amafomula eziyisisekelo combinatorics. Futhi ngaphambi kokuba uqale kubo, kuhle ukucabangela ukuthi kuyini.

Combinatorics - Ngokuyinhloko legatsha wezibalo, ubelokhu ukutadisha lenqwaba integers, futhi permutations ezihlukahlukene kokubili izinombolo kanye izakhi zabo, idatha ahlukahlukene, njll, okuholela eziningi inhlanganisela ... Ngaphezu imfundiso yokuziphendukela okungenzeka, lokhu umkhakha Kubalulekile izibalo, ikhompyutha isayensi indlela yokubhala efihla incazelo.

Ngakho manje ungakwazi udlulele kokwethula ngokwabo and definition yabo amafomula.

Eyokuqala yalezi uwukubonakaliswa ngesibalo permutations, kunjengoba kulandela:

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

Isibalo sisebenza kuphela endabeni uma izakhi kuhluke kuphela oda ilungiselelo.

Manje ukubekwa ifomula, kubukeka sengathi lokhu kuzoxoxwa ngakho:

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

Le nkulumo kusebenza hhayi kuphela kwingxenye kuphela oda ukubekwa, kodwa futhi ukuba ngezithako zawo.

I kwesibalo wesithathu combinatorics, futhi lesi samuva ngokuthi ifomula ngesibalo inhlanganisela:

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

Inhlanganisela ngokuthi izibonelo zalokho, ezingalotshwanga ngaphambilini, ngokulandelana, futhi bazisebenzise lo mthetho.

Nge amafomula of combinatorics bawaqonde kalula, ungakwazi manje ukuya definition zasendulo okungenzeka. Kubonakala sengathi le nkulumo ngale ndlela:

P (A) = m: n.

Kulesi ifomula, m - inombolo izimo kunikela umcimbi A, kanye n - nenani ngokulinganayo futhi ngokuphelele yonke imicimbi aphansi.

Kukhona amagama amaningi esihlokweni ayoshaywa indiva lutho kodwa abathintekayo kuyoba yiwo obaluleke kakhulu ezifana, isibonelo, ematfuba izenzakalo kufana:

P (A + B) = P (A) + P (B) - lokhu theorem yokwengeza kuphela izenzakalo alizimele;

P (A + B) = P (A) + P (B) - P (AB) - kodwa lokhu kuphela yokwengeza ehambisanayo.

Amathuba wemisebenzi kumcimbi:

P (A ⋅ B) = P (A) ⋅ P (B) - lokhu theorem yemicimbi ezimele;

(P (A ⋅ B) = P (A) ⋅ P (B | A); P (A ⋅ B) = P (A) ⋅ P (A | B)) - futhi lokhu ngenxa nasekuthengeni.

uhlu Kuphelile imicimbi ifomula. Imfundiso yokuziphendukela okungenzeka kusitshela theorem Bayes, okuyinto libukeka kanje:

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

Kulesi ifomula, H ₁ H _2, ..., H _n - kuyinto isethi ephelele imicabango.

Ngalesi stop, amasampula amafomula isicelo manje kucatshangelwe imisebenzi ethize kusukela umkhuba.

izibonelo

Uma ngokucophelela ukufunda noma igatsha le-IiMbalo, akusiyo ngaphandle ukuzilolonga kanye isampula izixazululo. Futhi imfundiso yokuziphendukela okungenzeka: izenzakalo, izibonelo nazi ebalulekile eqinisekisa izibalo ngokwesayensi.

Le formula for isibalo permutations

Ngokwesibonelo, e emphemeni ikhadi ukuba namakhadi amathathu, uthome eyodwa bokuzisholo. Umbuzo olandelayo. Zingaki izindlela phinda emphemeni ukuze amakhadi nge value ubuso bombusi oyedwa kwathi ababili ungekho esilandelayo?

Umsebenzi isethiwe, manje ake udlulela ukubhekana naso. Okokuqala udinga ukucacisa inombolo permutations izakhi amathathu, le njongo sithatha ifomula ngenhla, kuvela P_30 = 30!.

Kususelwa lo mthetho, siyazi bangaki ongakhetha kukhona ukubeka phansi emphemeni ngezindlela eziningi, kodwa thina kumele emholweni kwazo yilezo lapho ikhadi lokuqala nelesibili ozolandela. Ukuze wenze lokhu, qala nge okuhlukile, lapho abokuqala isendaweni yesibili. It kuvela ukuthi ibalazwe kokuqala kungase kuthathe izindawo amabili nesishiyagalolunye - kusukela ngosuku lokuqala kuya amabili nesishiyagalolunye, kanye ikhadi lesibili kusukela yesibili amathathu, iyakunqanda izihlalo amabili nesishiyagalolunye ngazimbili of amakhadi. Futhi, abanye kungathatha izihlalo ezingamatshumi amabili lesificaminwembili, futhi noma ngayiphi indlela. Okusho ukuthi, ngokuba rearrangement amakhadi amabili nesishiyagalombili baye amabili nesishiyagalombili ongakhetha P_28 = 28!

Umphumela uba ukuthi uma sicabanga isinqumo, lapho ikhadi sokuqala olwesibili ithuba eyengeziwe ukuze uthole 29 ⋅ 28! = 29!

Ukusebenzisa indlela efanayo, nawe kudingeka ukubala inani lokukhethwa kukho eziphindaphindekayo ukuze icala lapho ikhadi lokuqala itholakala ngaphansi yesibili. Futhi etholwe 29 ⋅ 28! = 29!

Kule ke kulandela ukuthi okunye okukhethwa kukho 2 ⋅ 29!, Kuyilapho izindlela ezidingekayo wokuqoqa emphemeni 30! - 2 ⋅ 29!. It uhlala kuphela ukubala.

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

Manje sidinga ukuba azale ande ndawonye zonke izinombolo kusukela omunye amabili nesishiyagalolunye, bese ekupheleni zonke iphindwe 28. Impendulo etholakale 2,4757335 ⋅ 〖〗 10 ^ 32

Izibonelo izixazululo. Le formula for isibalo yokuhlala

Kulesi nkinga, udinga ukuthola ukuthi mangaki izindlela ukubeka amavolumu nanhlanu eshalofini, kodwa ngaphansi kombandela wokuthi amavolumu amathathu kuphela.

Kulesi msebenzi, isinqumo lubelula kuka owedlule. Ukusebenzisa ifomula eyaziwa kakade, kubalulekile ukubala inani eliphelele izindawo amathathu amavolumu nanhlanu.

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

Impendulo, ngokulandelana, uyoba elilingana 202 843 204 931 727 360 000.

Manje thatha onzima kancane xaxa. Udinga ukwazi ukuthi zingaki kunezindlela ukuhlela izincwadi amathathu nambili emashalofini, nge amalunga amavolumu nanhlanu kuphela ukuhlala phezu eshalofini efanayo.

Ngaphambi ekuqaleni isinqumo ungathanda ukucacisa ukuthi ezinye izinkinga zingaxazululwa ngezindlela eziningana, futhi kule kukhona izindlela ezimbili, kodwa kokubili indlela efanayo sisetshenziswa.

Kulesi msebenzi, ungathatha impendulo esuka sangaphambilini, ngoba kukhona siye kubalwa inani lezikhathi ungakwazi ugcwalise ishalofu amabhuku nanhlanu ngezindlela ezahlukene. Kwatholakala A_30 ^ 15 = 30 ⋅ 29 ⋅ 28 ⋅ ... ⋅ (30 - 15 + 1) = 30 ⋅ 29 ⋅ 28 ⋅ ... ⋅ 16.

I ibutho yesibili ebalwe izinguquko ifomula, ngoba ibekwe amabhuku nanhlanu, ngenkathi esele nanhlanu. Sisebenzisa ifomula P_15 = 15!.

It kuvela ukuthi isamba ngeke A_30 ^ 15 ⋅ P_15 izindlela, kodwa ngaphezu kwalokho, umkhiqizo zonke izinombolo kusukela kwabaneminyaka engamashumi amathathu nesithupha yayiyobizwa iphindwe umkhiqizo izinombolo kusuka omunye nanhlanu, ekugcineni Ihambe umkhiqizo zonke izinombolo kusukela omunye amathathu, okungukuthi impendulo 30!

Kodwa le nkinga ingaxazululeka ngendlela ehlukile - kulula. Ukuze wenze lokhu, ungase ucabange ukuthi kukhona omunye ishalofu amabhuku amathathu. Bonke ezibekwe kule indiza, kodwa ngoba isimo sidinga ukuthi kwakukhona emashalofini ezimbili, eyodwa isikhathi eside thina sawing ngesigamu, badedelane amabili nanhlanu. Kule kuvela ukuthi leli lungiselelo kungaba P_30 = 30!.

Izibonelo izixazululo. Le formula for inani inhlanganisela

Ubani obhekwa njengesikhonzi okufana inkinga wesithathu combinatorics. Udinga ukwazi ukuthi zingaki izindlela kukhona ukuhlela izincwadi nanhlanu ngombandela wokuthi kumelwe ukhethe kusukela kwabaneminyaka engamashumi amathathu nse.

Ukuze isinqumo, eqinisweni, ukusebenzisa ifomula ngesibalo inhlanganisela. Kusukela isimo ukuthi kuba sobala ukuthi i-oda lezincwadi ezifanayo nanhlanu ayibalulekile. Ngakho ekuqaleni udinga ukuthola inombolo ephelele inhlanganisela izincwadi amathathu nanhlanu.

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

Yilokho. Ukusebenzisa kule formula, esikhathini esifishane ukuxazulula inkinga enjalo, impendulo, ngokulandelana, ilingana 155.117.520.

Izibonelo izixazululo. Inchazelo zakudala ematfuba

Ukusebenzisa ifomula letinikwe lapha ngenhla, umuntu angathola impendulo ku nomsebenzi olula. Kodwa-ke uyobona ngokucacile ukuthi bese ulandela inkambo.

Umsebenzi unikezwa ukuthi i isitsha aseyishumi amabhola ezifanayo ngokuphelele. Kulawa, ezine ophuzi kanye eziyisithupha blue. Othathwe isitsha esisodwa ibhola. Kuyadingeka ukwazi okungenzeka dostavaniya okwesibhakabhaka.

Ukuze kuxazululwe le nkinga kubalulekile ukuze kubhekiselwe endaweni eyikhaya dostavanie blue ibhola umcimbi A. Lokhu okuhlangenwe nakho kungase kudingeke imiphumela eziyishumi, okuyinto yena, aphansi futhi cishe ngokulinganayo. Ngesikhathi esifanayo, eziyisithupha eziyishumi mahle A. umcimbi Xazulula le ndlela elandelayo:

P (A) = 6: 10 = 0.6

Ukusebenzisa kule formula, siye safunda ukuthi kungenzeka dostavaniya ibhola blue 0.6.

Izibonelo izixazululo. Amathuba izenzakalo lemali

Ubani ozoba okufana okuyinto ixazululwe ngokusebenzisa formula Amathuba izenzakalo lemali. Ngakho, inikezwe isimo ukuthi kukhona amacala amabili, owokuqala grey futhi amabhola amahlanu ezimhlophe, kanti eyesibili - eziyisishiyagalombili grey futhi ezine amabhola amhlophe. Ngenxa yalokho, amabhokisi lokuqala nelesibili baye bamukela omunye wabo. Kuyadingeka ukuba uthole ukuthi yini amathuba okuba ukuthi ayengenalo amabhola kukhona grey nomhlophe.

Ukuze uxazulule le nkinga, kubalulekile ukukhomba umcimbi.

• Ngakho, A - siba ibhola ezimpunga ebhokisini lokuqala: P (A) = 1/6.
• A '- Isibani mhlophe futhi othathwe ibhokisi lokuqala: P (A) = 5/6.
• I - kakade ekhishwe grey ibhola komsele yesibili: P (B) = 2/3.
• B '- wathatha ibhola ezimpunga ekhabetheni yesibili: P (B) = 1/3.

Ngokusho inkinga kubalulekile ukuthi omunye izenzakalo kwenzeka: AB 'noma' B. Ukusebenzisa ifomula, sithola: P (AB) = 1/18, P (A'B) = 10/18.

Manje ifomula wokuphindaphinda ematfuba yayisetshenziswa. Okulandelayo, ukuze uthole impendulo, kudingeka sisebenzise kwesibalo yabo unezela:

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

Yileyo indlela, kusetshenziswa indlela, ungakwazi ukuxazulula izinkinga ezinjalo.

yi

Leli phephandaba wethulwa ulwazi "okungenzeka theory", ematfuba izenzakalo zidlala indima ebalulekile. Yiqiniso, akuyona yonke into iye yabhekwa, kodwa ngesisekelo umbhalo owethulwe, ungakwazi kwakuthiwa ukubazi nale legatsha wezibalo. isayensi Kubhekwe kungaba wusizo hhayi kuphela ibhizinisi professional, kodwa futhi ekuphileni kwansuku zonke. Ungayisebenzisa ukubala yini ithuba lokuphinde umcimbi.

Lo mbhalo wabhalwa futhi ehlaselwe izinsuku abalulekile emlandweni ukuthuthukiswa okungenzeka theory njengoba isayensi, kanye namagama abantu imisebenzi kabani sezifakiwe ke. Yileyo indlela ilukuluku lomuntu kuye kwaholela yokuthi abantu baye bafunda ukubala, ngisho zenzakalo ezivele zenzeke nje. Ngaphambili they are nje nentshisekelo kulo, kodwa namuhla isivele laziwe kubo bonke. Futhi akekho ongasho kuzokwenzekani kithi esikhathini esizayo, yini ezinye okutholakele brilliant ezihlobene imfundiso yokuziphendukela akhuluma ngayo, ngabe uzibophezele. Kodwa kunento eyodwa ngempela - isifundo namanje ngeke kukusize ngalutho-ke!