shartli ehtimol. hodisalarning bog`liqsizligi

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1576320274.doc b ) / ( b a p a b b a b a ) / ( b a p ( ) { } 6 , 1 ; 6 , 1 : , = = = j i j i w a b w (1,6)} (1,5), (1,4), (1,3), (1,2), {(1,1), a = (2,1)} (1,2), {(1,1), b = (1,2)} {(1,1), = ç b a 6 1 36 6 ) ( = = a p 12 1 36 3 ) ( = = b p 18 1 36 2 ) ( = = ç b a p ) ( ) ( 36 3 36 2 3 2 ) / ( b p b a p b a p ç = = = w n a k b r b a ç k r m r n k n m £ £ £ £ , , , n k b p = ) …

( 1 0 + = = = = n a p a p a p n i n i a b p i = ) / ( ) 1 ( 2 2 1 ) 1 ( 1 ) 1 ( 1 1 1 1 ) / ( 0 0 + = × + = + + = × + × + = å å = = n n i n n i k n n n n i n k n n i n b a p n k n k i n a shartli ehtimol. hodisalarning bog`liqsizligi agar hodisa ehtimolligini topishda kompleks shartlardan boshqa shartlar talab qilinmasa, bunday ehtimollikni shartsiz ehtimollik deyiladi ko`pgina hollarda qandaydir tasodifiy hodisa ehtimolligini musbat ehtimolga ega bo`lgan boshqa bir tasodifiy hodisasi ro`y berganlik shartida topishga to`g`ri keladi. bunday ehtimollikka shartli ehtimollik deyiladi va kabi belgilanib, ning shartidagi ehtimolligi deb o`qiladi. misol: o`yin soqqasi ikki …

di. agar hodisasi hodisasidan bog`liq bo`lmasa, hodisasi ham, hoisasidan bog`liq bo`lmaydi. haqiqatan ham, ko`paytirish teoremasiga asosan hodisasi hodisasidan bog`liqmas bo`lganligi uchun ko`paytirish teoremasiga asosan . bundan kelib chiqadi, ya`ni bog`liqmaslik o`zaro ekan. agar va hodisalari bog`liqmas bo`lsalar, va , va , va hodisalar juftliklari ham bog`lanmagan bo`ladi. masalan, va hodisalari bog`liqmaslikni ko`rsatamiz. tengligidan bo`lganligi uchun kelib chiqadi. demak, va hodisalaribog`liqmas ekan. bog`liqmas hodisalar uchun ko`paytirish teoremasi ko`rinishni oladi. endi hodisalarning bog`liqsizlik tushunchasini umumlshtiramiz. ta`rif. agar har qanday va lar uchun tenglik o`rinli bo`lsa, hodisalar birgalikda bog`liqmas deyiladi. ta`rifdan ko`rinadiki, birgalikda bog`liqmas hodisalar juft-jufti bilan bog`liqmas bo`ladi, lekin hodisalarning juft-jufti bilan bog`liqmasligidan ularning birgalikda bog`liqmasligi umuman olganda kelib chiqmaydi. bunga quyidagi misol yordamida ishonch hosil qilish mumkin. s. n. bernshteyn misoli: tetraedrning birinchi yog`i qizil rangga ( ), ikkinchi yog`i ko`k rangga ( ), uchinchi yog`i sariq rangga ( ), to`rtinchi yog`i uchala rangga ( ) bo`yalgan. tetraedr tashlanganda tushgan yoqda …

ariantlaridan tasi “baxtli” birinchi variant olishga kelgan talabaning “baxtli” variant olish ehtimoli kattami, yoki ikkinchiniki. yechish. birinchi talabaning “baxti” variant olish ehtimoli ga teng. -birinchi talabaning “baxtli” variant olish hodisasi, -birinchi talabaning “baxtli” variant olmaslik hodisasi va -ikkinchi talabaning “baxtli” variant olish hodisasi bo`lsin. u holda to`la ehtimollik formulasiga asosan . demak, ikkinchi talabaning “baxtli” variant olish ehtimoli ham ga teng ekan. endi -hodisasi ro`y bergan bo`lsa, qaysi orqali ro`y berganlik ehtimoli uchun formula keltirib chiqaramiz. oldingi teorema shartlarida ko`paytirish teoremasiga asosan . bundan to`la ehtimollik formulasiga asosan ( ) (4) bu formulaga beyes formulalari deyiladi. masala. idishda n ta shar bor . oq sharlar haqida -( ) ta gipoteza bo`lishi mumkin. -idishda ta oq shar bo`lish hodisasi bo`lsa bo`ladi. idishdan olingan shar oq bo`lib chiqdi. (b hodisasi) idishda ta oq sharlar bo`lgan bo`lish ehtimoli topilsin. , u holda (4) formulaga asosan shunday qilib gipoteza katta ehtimolli ekan. _1302462123.unknown _1302524384.unknown …

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1576320274.doc b ) / ( b a p a b b a b a ) / ( b a p ( ) { } 6 , 1 ; 6 , 1 : , = = = j i j i w a b w (1,6)} (1,5), (1,4), (1,3), (1,2), {(1,1), a = (2,1)} (1,2), {(1,1), b = (1,2)} {(1,1), = ç b a 6 1 36 6 ) ( = = a p 12 1 36 3 ) ( = = b p 18 1 36 2 ) ( = = ç b a p ) ( ) ( 36 3 36 2 3 2 ) / ( b p b a p b a p ç = = = …

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