Tasodifiy hodisalar вajardi: 2 kurs 104-21 guruhi dasturiy injiniring talabasi umirbekov Arislanbek

Shartli ehtimollik
A va B hodisalar biror tajribadagi hodisalar bo‗lsin.  B hodisaning A hodisa ro‗y bergandagi shartli ehtimolligi deb, ( ( ) 0) ( ) ( )   P A P A P A B (1.11.1) nisbatga aytiladi. Bu ehtimollikni P(B/ A) orqali belgilaymiz. Shartli ehtimollik ham Kolmogorov aksiomalarini qanoatlantiradi: 1. P(B/ A)  0 ; 2. 1 ( ) ( ) ( ) ( ) ( / )      P A P A P A P A P A ; 3. Agar BC   bo‗lsa, u holda ( / ) ( / ) , ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ( ) ) ( ( )/ ) P B A P C A P A P C A P A P B A P A P B A P C A P A P B A C A P A P B C A P B C A                    25 chunki BC   ekanligidan, (B A)(C  A)  B A AC  BC  A   A   1.10-misol. Idishda 3 ta oq va 7 ta qora shar bor. Tavakkaliga ketma-ket bittadan 2 ta shar olinadi. Birinchi shar oq rangda bo‗lsa ikkinchi sharning qora rangda bo‗lishi ehtimolligini toping. Bu misolni ikki usul bilan yechish mumkin: 1) A={birinchi shar oq rangda}, B ={ikkinchi shar qora rangda}. A hodisa ro‗y berganidan so‗ng idishda 2 ta oq va 7 ta qora shar qoladi. Shuning uchun 9 7 P(B / A)  . 2) (1.11.1) formuladan foydalanib, hisoblaymiz: 10 3 P(A)  , 30 7 9 7 10 3 P(AB)    Shartli ehtimollik formulasiga ko‗ra: 9 7 3/10 7 / 30 ( ) ( ) ( / )     P A P A B P B A . Shartli ehtimollik formulasidan hodisalar ko‗paytmasi ehtimolligi uchun ushbu formula kelib chiqadi: P(A B)  P(A) P(B/ A)  P(B) P(A/ B) (1.11.2) (1.11.2) tenglik ko‗paytirish qoidasi(teoremasi) deyiladi. Bu qoidani n ta hodisa uchun umumlashtiramiz: ( ... ) ( ) ( / ) ( / )... ( / ... ) 1 2 1 2 1 3 1 2 1 2 1        P A A An P A P A A P A A A P An A A An . (1.11.3)  Agar P(A/ B)  P(A) tenglik o‗rinli bo‗lsa, u holda A hodisa B hodisaga bog‗liq emas deyiladi va A B orqali belgilanadi. Agar A B bo‗lsa, u holda (1.11.2) formulani quyidagicha yozish mumkin: P(A B)  P(B) P(A/ B)  P(B) P(A).  A va B hodisalar o‗zaro bog‗liq emas deyiladi, agar P(A B)  P(A) P(B) munosabat o‗rinli bo‗lsa. Lemma. Agar A B bo‗lsa, u holda A  B , A  B va A  B bo‗ladi. 26 Isboti: A B bo‗lsin. U holda P(A B)  P(A) P(B) munosabat o‗rinli bo‗ladi. P(B)  P(B) 1 tenglikdan foydalanib, quyidagiga ega bo‗lamiz: ( ) ( ) ( ) ( ) (1 ( ) ) ( ) ( ) . ( ) ( ( ) ) ( ) ( ) ( ) ( ) P A P A P B P A P B P A P B P A B P A B P A A B P A A B P A P A B                         Demak, P(A B)  P(A) P(B)  A  B . Qolganlari ham xuddi shunday isbotlanadi.