第一章 1. (1) ?1?{(1,1),(1,2)(1,3)...(6,6)}
(2) ?2?{x|x1?x?x2} x1:当日最低价 x2:当日最高价 (3) ?3?{0,1,2,3,} (4) ?3?{1,2,3,?} 2. (1) (3) 3. ??{1,2,3,4,5,6} A?{1,3,5,} B?{1,2,3,4,} C?{2,4,} A?B?{1,2,3,4,5} A?B?{5} B?A?{2,4,} AB?{1,3} AC?? A?B?{1,2,3,4,6}
4. (5) ABC?ABC?ABC?ABC?ABC?ABC?ABC (8) ABC?ABC?ABC?ABC (10) AB?BC?AB (11) A?B?C
9. ①?P(A?B)?P(A?AB)?P(A)?P(AB)?0.25 又?P(A)?0.4 ?P(AB)?0.15
②?P(A?B)?P(A)?P(B)?P(AB) ?0.4?0.25?0.15 ?0.5
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③?P(B?A)?P(B?AB) ?P(B)?P(AB) ?0.25?0.15 ?0.1
④P(AB)?P(A?B)?1?P(A?B) ?1?0.5 ?0.5
10. ?P(A?B?C)?1?P(ABC) 而 P(A)?1?P(A)?1?0.4?0.6 又P(A)?P(AB?AB) ?P(AB)?P(AB) ?P(AB)?P(A)?P(AB)?0.4 又 AB?ABC?ABC
?P(AB)?P(ABC)?P(ABC) ?P(ABC)?0.4?0.1?0.3 ?P(A?B?C)?0.7 11. A=“其中恰有K件” CKn?①?P(A)?1CkNN?N1Cn
N② B=“其中有次品” B?“一件次品也没有” ?P(B)?1?P(B)?Cn1?N?N1Cn
N③C=“其中至少有两件次品” C?“只有一件次品,或没有” ?P(C)?1?P(C)?1Cn?N?N1C1N1Cn?1N?N1Cn?NCnN.①: A=“男生比女生先到校”
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P(A)?24!?6!24!1?24?24 30!P30C30 ②B=“李明比王先到学校” P(B)?
13. C=“至少两人生日同一天” C?“每个人生各不同” P(C)?1?P(C)?1?365?364??(365?n?1)
365n1214. ①A=“第2站停车”
A?“不停车”
?P(A)?1?P(A)?1?()25
②B=“第i和第J站至少有一站停车 B?“第i站到J站都不停” ?P(B)?1?P(B)
7?1?()25
989 ③Ai?“第i站有人下车(停车)” Aj?“第j站有人下车” P(Ai?Aj)?1?(Ai?Aj)?1?P(Ai?Aj)
?1?[P(Ai)?P(Aj)?P(AiAj)]
?1?P(Ai)?P(Aj)?P(AiAj)
87?1?()25?2?()25
99④D=“在第i站有3人下车”
3?()3?()22 (贝努里试验) P(D)?C25198915.(1)A=“前两个邮筒没有信” P(A)?2?21? 442(2)B=“第一个邮筒恰有一封信”
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1C2?33P(B)?2?
8416. A=“前i次中恰好有取到k封信”
ki?kCa?Cb?i!(a?b?i)! P(A)?
(a?b)!ki?kCaCb ?i
Ca?b17. A3?“第三把钥匙可以开门” A2?“第二把钥匙可以开门”
① P(A3)?P(A1A2A3?A1A2A3?A1A2A3?A1A2A3)
?P(A1A2A3)?P(A1A2A3)?P(A1A2A3)?P(A1A2A3)
?432654463643??????????? 109810981098109824?120?144?
720288? 7204? 10 ② A3?“第三把钥匙才可以开门” P(A3)?6541201???? 10987206 ③ C=“最多试3把就可以开门”
464654????? 1010910985 ?
6 P(C)?18. 贝努里试验
A=“其中三次是正面”
11133P(A)?C10?()3?()7?C10?()10
22219.A=“恰有一红球,一白球,一黑球”
111C5?C3?C21 P(A)??34C101C2?3?2?2?24820. P(A)? ?13!13! 4
21. 几何概型
A=“等待时间不超过3分钟” X???到达汽车站的时间
??{xt?x?t?10} A?{xt?7?x?t?10}
?P(A)?S(A)3? S(?)1022. A=“需要等零出码头的概率”
x???第1条船到达时刻 y???第2条船到达时刻
??{(x,y)0?x?24 0?y?24? A?{(x,y)0?x?y?2 0?y?x?1?
1242?(222?232)S(A)2?P(A)?? S(?)24223. A=“第一次取出的是黑球”
B=“第二次取出的是黑球”
a?(a?1)P(AB)(a?b)?(a?b?1)a?1?? (1) P(BA)?
aP(A)a?b?1a?baa?1?P(AB)a?1a?ba?b?1 (2)P(AB)? ??aa?1baP(B)a?b?1???a?ba?b?1a?ba?b?1 (3)A=“取出两个球,有一个是黑球” B=“两个都是黑球” nA?a(a?b?1)?b?a?a?(a?2b?1) nB?a?(a?1)
P(BA)?nBa(a?1)a?1 ??nA[a(a?2b?1)]a?2b?1P(AB) P(A)5
24. (1)P(BA)? ?B?A