Imibuzo Eyisibonelo Exoxa Ngomthetho Wokwengeza Imicimbi Emibili Ekhethekile A kanye no-B
Ku-probability theory, umthetho we-sum rule wemicimbi emibili ungenye yezimiso eziyisisekelo ezisetshenziswa ukubala amathuba emicimbi eminingi. Lo mqondo uvame ukusetshenziswa ezimweni ezahlukene ukuqonda imiphumela engaba khona yemicimbi ethile. Kulesi sihloko, sizoxoxa ngomthetho we-sum rule wemicimbi emibili ehlukene futhi sinikeze izibonelo zokucacisa lo mqondo.
Umthetho Wokwengeza Imicimbi Emibili Ekhethekile
Okokuqala, kubalulekile ukuqonda ukuthi kusho ukuthini ngemicimbi ehlukene. Imicimbi emibili kuthiwa ayihlangani noma ayihlangani uma ingenzeki ngesikhathi esisodwa. Ngamanye amazwi, akukho sici esisethweni somcimbi owodwa esiphinde sibe yingxenye esethweni somunye umcimbi.
Umthetho wokwengeza kumathuba uthi uma imicimbi emibili \(A\) kanye \(B\) ihlukile komunye nomunye, khona-ke amathuba emicimbi \(A\) noma \(B\) ayisamba samathuba emicimbi emibili. Ngokwezibalo, lo mthetho ungachazwa kanje:
\[ P(A \inkomishi B) = P(A) + P(B) \]
lapho \(P(A \cup B)\) kuyithuba lokuthi \(A\) noma \(B\), \(P(A)\) kuyithuba lokuthi umcimbi \(A\), kanye \(P(B)\) kuyithuba lokuthi umcimbi \(B\).
Imibuzo Yengxoxo Eyisibonelo
Ake sixoxe ngezibonelo ezithile ukuze sicacise ukusetshenziswa komthetho wokwengeza izehlakalo ezimbili ezihlukene.
Isibonelo Umbuzo 1
Umbuzo:
Idayi enezinhlangothi eziyisithupha iphonswa kanye. Thola amathuba okuthi inombolo ephumayo ingu-2 noma u-4.
Ingxoxo:
Singachaza umcimbi \(A\) njengokuvela kwenani 2, kanye nomcimbi \(B\) njengokuvela kwenani 4. Ngakho-ke:
– \(P(A)\) yithuba lokuthi inani 2 livele.
– \(P(B)\) yithuba lokuthi inani elingu-4 livele.
Njengoba idayisi inezinhlangothi eziyisithupha ezinamathuba alinganayo, amathuba okuba inani elithile ligoqwe ngu-\( \frac{1}{6} \). Ngakho-ke:
\[ P(A) = \frac{1}{6} \]
\[ P(B) = \frac{1}{6} \]
Imicimbi \(A\) kanye \(B\) ihlukile ngoba amanani 2 kanye no-4 awakwazi ukuvela ngesikhathi esisodwa ku-roll eyodwa yedayi. Ngakho-ke, ukusebenzisa umthetho wokwengeza kwimicimbi emibili ehlukile:
\[ P(A \indebe B) = P(A) + P(B) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \]
Ngakho-ke, amathuba okuthi inani elivelayo lingu-2 noma u-4 lingu-\( \frac{1}{3} \) noma cishe angu-33.33%.
Isibonelo Umbuzo 2
Umbuzo:
Esikhwameni kukhona amabhola ayi-10 aqukethe amabhola abomvu amane kanye namabhola aluhlaza okwesibhakabhaka ayi-6. Uma sikhetha ibhola elilodwa ngokungahleliwe, kungenzeka yini ukuthi ibhola elidonswe libomvu noma liluhlaza okwesibhakabhaka?
Ingxoxo:
Singachaza umcimbi \(A\) njengokuthatha ibhola elibomvu, kanye nomcimbi \(B\) njengokuthatha ibhola eliluhlaza okwesibhakabhaka. Ngakho-ke:
– \(P(A)\) yithuba lokukhetha ibhola elibomvu.
– \(P(B)\) yithuba lokukhetha ibhola eliluhlaza okwesibhakabhaka.
Amathuba omcimbi ngamunye angabalwa kanje:
\[ P(A) = \frac{\text{Inani lamabhola abomvu}}{\text{Inani eliphelele lamabhola}} = \frac{4}{10} = \frac{2}{5} \]
\[ P(B) = \frac{\text{Inani lamabhola aluhlaza okwesibhakabhaka}}{\text{Inani eliphelele lamabhola}} = \frac{6}{10} = \frac{3}{5} \]
Imicimbi \(A\) kanye \(B\) ihlukile ngoba ibhola alikwazi ukuba bomvu futhi libe luhlaza okwesibhakabhaka. Ngakho-ke, ukusebenzisa umthetho wokwengeza kwimicimbi emibili ehlukene:
\[ P(A \indebe B) = P(A) + P(B) = \frac{2}{5} + \frac{3}{5} = 1 \]
Ngakho-ke, amathuba okuthi ibhola elidonswe libomvu noma liluhlaza okwesibhakabhaka angu-1, noma u-100%. Lokhu kunengqondo ngoba wonke amabhola asesikhwameni abomvu noma aluhlaza okwesibhakabhaka.
Isibonelo Umbuzo 3
Umbuzo:
Ekilasini labafundi abangu-20, abangu-7 kubo bathanda izibalo, abangu-5 kubo bathanda isayensi, futhi akekho othanda zombili. Uma umfundi oyedwa ekhethwa ngokungahleliwe, thola amathuba okuthi umfundi athande izibalo noma isayensi.
Ingxoxo:
Singachaza umcimbi \(A\) njengokuthanda izibalo, kanye nomcimbi \(B\) njengokuthanda isayensi. Ngakho-ke:
– \(P(A)\) yithuba lokuthi umfundi athande izibalo.
– \(P(B)\) yithuba lokuthi umfundi athande isayensi.
Amathuba omcimbi ngamunye angabalwa kanje:
\[ P(A) = \frac{\text{Inani labafundi abathanda izibalo}}{\text{Inani eliphelele labafundi}} = \frac{7}{20} \]
\[ P(B) = \frac{\text{Inani labafundi abathanda isayensi}}{\text{Inani eliphelele labafundi}} = \frac{5}{20} = \frac{1}{4} \]
Imicimbi \(A\) kanye \(B\) ihlukile komunye nomunye ngoba akekho umfundi oyithanda yomibili. Ngakho-ke, ukusebenzisa umthetho wokwengeza emicimbini emibili ehlukene komunye nomunye:
\[ P(A \indebe B) = P(A) + P(B) = \frac{7}{20} + \frac{5}{20} = \frac{12}{20} = \frac{3}{5} \]
Ngakho-ke, amathuba okuthi umfundi okhethwe ngokungahleliwe athande izibalo noma isayensi ngu-\( \frac{3}{5} \) noma u-60%.
Isiphetho
Umthetho wokwengeza izenzakalo ezimbili ezingabambisene ngumqondo oyisisekelo kumbono wamathuba okusiza ekubalweni kwamathuba omcimbi ohlanganyelwe. Ezibonelweni ezingenhla, sibonile ukuthi lesi simiso singasetshenziswa ezimweni zangempela njengokugingqa idayisi, ukudweba amabhola esikhwameni, noma ukukhetha abafundi ekilasini. Ngokuqonda nokuqonda kahle lo mqondo, singabala ngempumelelo amathuba emicimbi eyahlukene engabambisene empilweni yansuku zonke.
