Isibonelo sombuzo wengxoxo mayelana namathuba emicimbi ehlanganisiwe ezimele ngokwemibandela
I-Pendahuluan
I-Probability iyigatsha lezibalo elifunda amathuba okuba kwenzeke isenzakalo. Omunye wemibono eyisisekelo ku-probability theory yimicimbi ehlanganisiwe, engahlukaniswa njengezimele noma enemibandela. Lesi sihloko sizohlola le mibono ngokujulile ngezinkinga zezibonelo kanye nezingxoxo.
Imicimbi Ehlanganisiwe
Umcimbi ohlanganisiwe uyinhlanganisela yemicimbi emibili noma ngaphezulu eyenzeka esikhaleni sesampula. Kunezinhlobo ezimbili zemicimbi ehlanganisiwe: imicimbi ehlukene kanye nemicimbi enemibandela.
1. Imicimbi Ezimele: Imicimbi emibili kuthiwa izimele uma impumelelo noma ukwehluleka kwesenzakalo esisodwa kungathinti esinye. Isibonelo, umphumela wokuphonswa kohlamvu lwemali kanye ne-die roll.
2. Imicimbi Enemibandela: Imicimbi enemibandela yenzeka lapho impumelelo noma ukwehluleka kwesenzakalo esisodwa kuthinta amathuba okuba kwenzeke esinye isenzakalo. Isibonelo, ithuba lokuthi othile abe nesifo uma enesimo sofuzo saleso sifo.
Amathuba Emicimbi Ezimele
Ifomula yokubala amathuba emicimbi emibili ezimele u-A no-B yile:
\[ P(A \cap B) = P(A) \times P(B) \]
Kuphi:
– \(P(A \cap B)\) amathuba ezenzakalo u-A no-B ezenzeka ngesikhathi esisodwa.
– \(P(A)\) kungenzeka ukuthi umcimbi u-A uvele.
– \(P(B)\) kungenzeka ukuthi umcimbi B wenzeke.
Imibuzo Nengxoxo Yezibonelo: Imicimbi Ezimele Yobabili
Umbuzo 1: Uhlamvu lwemali luyaphonswa bese kugoqwa idayisi elinezinhlangothi eziyisithupha. Thola amathuba okuthola amakhanda ohlamvu lwemali kanye no-4 kudayisi.
Ingxoxo:
– Amathuba okubonakala kwesithombe kuhlamvu lwemali: \( P(G) = \frac{1}{2} \)
– Amathuba okuthola inombolo 4 kudayisi: \( P(4) = \frac{1}{6} \)
Njengoba uhlamvu lwemali kanye nedayi kuyimicimbi emibili ezimele, amathuba okuthi kokubili kwenzeke ngesikhathi esisodwa yilawa:
\[ P(G \cap 4) = P(G) \times P(4) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12} \]
Ngakho-ke, amathuba okuthola isithombe kuhlamvu lwemali kanye no-4 kudayi yi-\( \frac{1}{12} \).
Amathuba Emicimbi Anemibandela
Amathuba anemibandela yezehlakalo ezimbili u-A no-B yithuba lokuthi u-A avele uma kubhekwa ukuthi u-B uvele. Ifomula yokubala amathuba anemibandela iboniswa kanje:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]
Kuphi:
– \( P(A|B) \) kungenzeka ukuthi umcimbi u-A uvele uma umcimbi u-B usuvele uvele.
– \( P(A \cap B) \) amathuba ezenzakalo u-A no-B ezenzeka ngesikhathi esisodwa.
– \( P(B) \) amathuba okuba kwenzeke isenzakalo B.
Imibuzo Nengxoxo Yezibonelo: Imicimbi Enemibandela
Umbuzo 2: Ebhokisini eliqukethe amabhola abomvu amathathu namabhola aluhlaza okwesibhakabhaka amabili, amabhola amabili adonswa ngokungahleliwe ngalinye ngaphandle kokushintshwa. Thola amathuba okuthi ibhola lesibili elidonswe libomvu uma ubheka ukuthi ibhola lokuqala nalo libomvu.
Ingxoxo:
Ngokwesibonelo:
– A yisenzakalo lapho ibhola lokuqala libomvu.
– B yisenzakalo lapho ibhola lesibili libomvu.
Sifuna i- \( P(B|A) \). Okokuqala, sibala i- \(P(A)\) kanye ne- \(P(A \cap B)\):
Inani eliphelele lamabhola = 5 (3 abomvu kanye nama-2 aluhlaza okwesibhakabhaka).
Ithuba lokuqala lebhola elibomvu:
\[
P(A) = \frac{3}{5}
\]
Ngemva kokuba ibhola lokuqala elibomvu selidonswe, inani lamabhola abomvu asele lingama-2 kanti inani eliphelele lamabhola liba ngu-4.
Amathuba okuthi ibhola lesibili libe bomvu ngemva kokuba ibhola lokuqala elibomvu selidonswe:
\[
P(B|A) = \frac{2}{4} = \frac{1}{2}
\]
Ngakho-ke, amathuba okuthi ibhola lesibili libe bomvu uma ibhola lokuqala nalo libomvu yi-\( \frac{1}{2} \).
Imibuzo Yesibonelo Esihlanganisiwe
Ukuze sijulise ukuqonda kwethu, singahlanganisa imicimbi ehlanganisiwe ezimele kanye nenemibandela embuzweni owodwa.
Umbuzo 3: Isikhwama siqukethe amabhola abomvu ama-5 namabhola aluhlaza okwesibhakabhaka ama-3. Amabhola amabili adonswa ngokungahleliwe ngaphandle kokushintshwa. Thola amathuba okuthi ibhola lokuqala libomvu kanti elesibili liluhlaza okwesibhakabhaka.
Ingxoxo:
Sisebenzisa umbhalo ofanayo njengangaphambili:
– A yisenzakalo lapho ibhola lokuqala libomvu.
– B yisenzakalo lapho ibhola lesibili liluhlaza okwesibhakabhaka.
Okokuqala, sibala amathuba omcimbi ngamunye ngokulandelana.
Amathuba okuthi ibhola lokuqala libe bomvu:
\[
P(A) = \frac{5}{8}
\]
Uma ibhola lokuqala libomvu, inani lamabhola abomvu asele = 4, kanye nenani eliphelele lamabhola asele = 7.
Amathuba ebhola lesibili libe luhlaza okwesibhakabhaka ngemva kwebhola lokuqala libe bomvu:
\[
P(B|A) = \frac{3}{7}
\]
Ngakho-ke, amathuba okuthi ibhola lokuqala libomvu kanti elesibili liluhlaza okwesibhakabhaka abangelwa yilawa mathuba amabili anemibandela:
\[
P(A \cap B) = P(A) \times P(B|A) = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}
\]
Ngakho-ke, amathuba okuthi ibhola lokuqala libomvu kanti elesibili liluhlaza okwesibhakabhaka yi-\( \frac{15}{56} \).
Isiphetho
Ku-probability theory, ukuqonda umehluko phakathi kwezehlakalo ezizimele nezinemibandela kubalulekile ekuxazululeni izinkinga ezihilela izehlakalo ezihlanganisiwe. Ngezinkinga zesibonelo, sifunda ukuthi singabala kanjani amathuba ezimo ezahlukahlukene ezihilela le mibono emibili. Ukuqonda kahle le mibono kungasiza ekwenzeni izinqumo ezimweni zangempela ezifana nokuphathwa kwezingozi, imidlalo, kanye nocwaningo lwesayensi.
Ukusetshenziswa kwezibalo empilweni yansuku zonke kubonisa ukuthi lokhu kuqonda kubaluleke kangakanani ezicini ezahlukene zokuphila komuntu, kusukela kokulula kuya kokuyinkimbinkimbi kakhulu. Ngokuzijwayeza njalo nokuqonda kahle le mibono eyisisekelo, amakhono ethu ekuhlaziyeni amathuba azokhula.
