>>258
æ¡ä»¶ä»ã確ç, ç¬ç«äºè±¡, ãã€ãºã®å®ç åé¡ 1 解ç
1 ããæ©æ¢°ã¯çšŒåã㊠1 æ¥æŸçœ®ãããš 10 % ã®ç¢ºçã§æ
éãã. ãã®æ©æ¢°ã皌åã㊠3 æ¥é£
ç¶ã§æŸçœ®ãããšãã.
(1) æ©æ¢°ãæ
éããŠããªã確çãæ±ãã.
[解]: æ
éããäºè±¡ã A ãšãã.
P(Ac) = 9/10Ã9/10Ã9/10=729/1000.
(2) æ©æ¢°ãæ
éããŠãããšã, 2 æ¥ç®ã«æ
éãã確çãæ±ãã.
[解]: 2 æ¥ç®ã«æ
éããäºè±¡ã B ãšãã. ãã®ãšã,
P(A) = 271/1000 = 0.271, P(B) = 9/10 Ã 1/10 = 9/100 = 0.09,
P(A|B) = 1ã§ãã. ãã€ãºã®å®çãã, P(B|A)P(A) = P(A|B)P(B) ã§ãããã,
P(B|A) = 9/100/271/1000=90/271= 0.332.