Imibuzo Eyisibonelo Exoxa Ngokungenzeka Kwemicimbi Ehlanganisiwe
Isingeniso Sokuthi Imicimbi Ehlanganisiwe Ingenzeka Kanjani
Amathuba ayigatsha lezibalo elifunda amathuba okuba kwenzeke isenzakalo. Amathuba okuba khona kwesenzakalo esihlanganisiwe amathuba okuba kwenzeke isenzakalo esingaphezu kwesisodwa. Isibonelo, amathuba okugoqa inombolo elinganayo kudayisi kanye ne-ace evela ekhadini lamakhadi okudlala yizibonelo zemicimbi ehlanganisiwe. Lesi sihloko sizoxoxa ngezinkinga eziningana zezibonelo futhi sixoxe ngamathuba okuba khona kwemicimbi ehlanganisiwe.
Umqondo Oyisisekelo Wokuthi Imicimbi Ehlanganisiwe Ingenzeka Kanjani
Kunezinhlobo ezimbili zemicimbi ehlanganisiwe:
1. Imicimbi Ekhethekile Ehambisanayo: Imicimbi emibili engenzeki ngasikhathi sinye. Isibonelo, lapho kugoqwa idayisi, imicimbi yokugoqwa kuka-2 no-5 iyimicimbi ekhethekile ngoba akunakwenzeka ukugoqwa kwezinombolo zombili ngasikhathi sinye.
2. Imicimbi Engakhethekile Engavumelani: Imicimbi emibili engenzeka ngesikhathi esisodwa. Isibonelo, ekudwebeni amakhadi okudlala, imicimbi yokuthola ikhadi lenhliziyo (♥) kanye nekhadi elinenombolo 10 yimicimbi engeyona eyingqayizivele ngoba kukhona ikhadi lenhliziyo elinenombolo 10.
Nazi ezinye izindlela eziyisisekelo ezisetshenziswa ekubaleni amathuba emicimbi ehlanganisiwe:
– P(A noma B) (kwemicimbi engeyona eyodwa): \(P(A \indebe B) = P(A) + P(B) – P(A \indebe B)\)
– P(A noma B) (ngemicimbi ehlukene): \(P(A \indebe B) = P(A) + P(B)\)
– P(A kanye no-B) (kwemicimbi ezimele): \(P(A \cap B) = P(A) \times P(B)\)
Imibuzo Eyisibonelo Nengxoxo
Isibonelo Umbuzo 1: Idayisi
Umbuzo:
Iyini amathuba okuthola inombolo elinganayo noma inombolo enkulu kuno-4 kudayi?
Ingxoxo:
Okokuqala, ake sichaze izenzakalo:
– Umcimbi A: Ukuthola inombolo elinganayo (2, 4, 6)
– Umcimbi B: Ukuthola inombolo enkulu kuno-4 (5, 6)
Okulandelayo, sinquma amathuba omcimbi ngamunye:
– \(P(A) = \frac{3}{6} = \frac{1}{2}\)
– \(P(B) = \frac{2}{6} = \frac{1}{3}\)
Njengoba kunenombolo 6 efakiwe kuzo zombili izehlakalo u-A no-B, sidinga ukubala \(P(A \cap B)\):
– \(P(A \cap B) = \frac{1}{6}\) (ngoba inombolo eyodwa kuphela, okungu-6, ifakiwe kokubili ku-A naku-B)
Ngokusebenzisa ifomula yemicimbi engeyona eyodwana:
\[P(A \indebe B) = P(A) + P(B) – P(A \indebe B) = \indebe{1}{2} + \indebe{1}{3} – \indebe{1}{6}\]
Ake senze ama-denominator ala ma-fraction afanayo:
\[P(A \indebe B) = \frac{3}{6} + \frac{2}{6} – \frac{1}{6} = \frac{4}{6} = \frac{2}{3}\]
Ngakho-ke, amathuba okuthola inombolo elinganayo noma inombolo enkulu kuno-4 yi-\(\frac{2}{3}\).
Isibonelo Umbuzo 2: Amakhadi Okudlala
Umbuzo:
Angakanani amathuba okuthola i-Ace noma i-spade ekhadini lamakhadi okudlala?
Ingxoxo:
Okokuqala, ake sichaze izenzakalo:
– Umcimbi A: Ukuthola ikhadi le-Ace (amane esewonke, elilodwa ngesudi ngayinye)
– Umcimbi B: Ukuthola ikhadi le-spade (inani eliphelele elingu-13)
Okulandelayo, sinquma amathuba omcimbi ngamunye:
– \(P(A) = \frac{4}{52} = \frac{1}{13}\)
– \(P(B) = \frac{13}{52} = \frac{1}{4}\)
Njengoba i-Ace of Spades ifakiwe kuzo zombili izehlakalo u-A no-B, sidinga ukubala \(P(A \cap B)\):
– \(P(A \cap B) = \frac{1}{52}\)
Ngokusebenzisa ifomula yemicimbi engeyona eyodwana:
\[P(A \indebe B) = P(A) + P(B) – P(A \indebe B) = \indebe{1}{13} + \indebe{1}{4} – \indebe{1}{52}\]
Ake senze ama-denominator ala ma-fraction afanayo:
\[
P(A \indebe B) = \frac{4}{52} + \frac{13}{52} – \frac{1}{52} = \frac{16}{52} = \frac{4}{13}
\]
Ngakho-ke, amathuba okuthola i-Ace noma i-spade yi-\(\frac{4}{13}\).
Isibonelo Inkinga 3: Ibhola Ebhokisini
Umbuzo:
Ebhokisini kukhona amabhola abomvu amathathu, amabhola aluhlaza okwesibhakabhaka amane, kanye namabhola aluhlaza ayisihlanu. Uma ibhola elilodwa lidonswa ngokungahleliwe, angakanani amathuba okuthola ibhola elibomvu noma eliluhlaza?
Ingxoxo:
Okokuqala, ake sichaze izenzakalo:
– Umcimbi A: Ukuthola ibhola elibomvu (inombolo 3)
– Umcimbi B: Ukuthola ibhola eliluhlaza (inombolo 5)
Okulandelayo, sinquma amathuba omcimbi ngamunye:
– Inani eliphelele lamabhola = 3 + 4 + 5 = 12
– \(P(A) = \frac{3}{12} = \frac{1}{4}\)
– \(P(B) = \frac{5}{12}\)
Njengoba kungekho bhola elingaba bomvu neluhlaza ngesikhathi esisodwa, lezi zenzakalo zihlukile:
\[P(A \indebe B) = P(A) + P(B) = \frac{1}{4} + \frac{5}{12}\]
Ake senze ama-denominator ala ma-fraction afanayo:
\[
P(A \indebe B) = \frac{3}{12} + \frac{5}{12} = \frac{8}{12} = \frac{2}{3}
\]
Ngakho-ke, amathuba okuthola ibhola elibomvu noma ibhola eliluhlaza ngu-\(\frac{2}{3}\).
Isibonelo Umbuzo 4: Izinhlamvu zemali ezimbili
Umbuzo:
Uma izinhlamvu zemali ezimbili zijikijelwa ngesikhathi esisodwa, kungenzeka yini ukuthi okungenani ikhanda elilodwa livele?
Ingxoxo:
Sichaza Umcimbi A: ukubona okungenani isithombe esisodwa.
Kunemiphumela emine engaba khona yokuphonsa izinhlamvu zemali ezimbili:
1. HH
2. I-HT
3. TH
4. TT
Imicimbi equkethe okungenani isithombe esisodwa yile:
– I-HT
– TH
– TT
Ake sibale amathuba ento ngayinye:
– Inani lemicimbi engenzeka (isiyonke): 4
– Inani lemicimbi equkethe okungenani isithombe esisodwa: 3
\[
P(A) = \frac{Inani lemicimbi okungenani enekhanda elilodwa}{Inani eliphelele lemicimbi} = \frac{3}{4}
\]
Ngakho-ke, amathuba okuthi okungenani isithombe esisodwa sivele angama-\(\frac{3}{4}\).
Isiphetho
Ingxoxo ngezinkinga ezingenhla ibonisa ukuthi singabala kanjani amathuba omcimbi ohlanganisiwe, kungakhathaliseki ukuthi uhlukile komunye nomunye noma awuhlukile komunye nomunye. Ngokuqonda imiqondo eyisisekelo nokusebenzisa amafomula afanele, singanquma amathuba okuhlanganiswa okuthile kwemicimbi eyenzeka ezimweni ezahlukahlukene zansuku zonke. Qhubeka uzijwayeza amakhono akho ngezinkinga ezahlukahlukene ukuze ube nekhono elingcono ekunqumeni amathuba emicimbi ehlanganisiwe.
