(整理)李賢平 《概率論與數(shù)理統(tǒng)計 第二章》答案

1、第 2 章 bab nnn nnn p 1n1 p0 p 0 a p)/ pk(kk1。kAB,AB,ABCA,B,C 12P(A A A ) P(A )P(A )P(A )，12n12nii、若 A 與 B ,A,B,B,A ,A ,A P(A ) p ）12nkk12件nA n A n n ，而每個卵變?yōu)槌上x的概率為k 了i221P(A ) , P(A A ) ，P(A A A ) , P(A A A A ) 5431213214321P(A A A A A )1。54321123451213214321543211303 M m兩Mm B( ) /P B C2 C2 ，C1 C1mmmm

2、M1C2/C2m.mM122mmM（）設(shè) BP(A) C2C1 C1/C2 , P(B) C1 C1mMmC1C1/C2.mMCC C2112MmmMC /C C1C122.mmMa P()babbab ab1 ab ab1 ab b1a(abb.12322112233C2Cc1C1C.aaac2c2C22C2221i1iii ,Nab, P(A ) .1ab1abkka.kkkka.NA第i piiiP(A ) P(A | A )P(A ) P(A | A )P(A )i1i1iii1ii。111iP (2pp p),n1 n2n1P (2pp p) (2pc p21nnn當(dāng) p1 1p p

3、)n (2 (2 .1p1n1）n2 2 p p；nnn 0n2k1 p cn1p , lim p 2nn1 1c1c, limp2nnn1n1nnAiiin1n1nn1nnn1nnn1n11，44nnnnn1n1nn1nnn1nnn1n11，22nnnnnnn1n1nn1nnn1nnn1n110 p q 0r q.44nnnnn1n1n1n1n1n11144nn1nnn1rpp4，nqp0001(p 4n1n1nnn11p 32111，n1n6621 1 1nq 12p (n1 .33 2n1n11263nnnnnn A n k n1得211nkknkk22nnn 1nP(B) P(A )P

4、(B| A ) ap C a C i nk ）k2nnnni01kkkk C. a2 i0A B, B2。ppk,pk1p22a (2 p)apk22P(B) ,p22k2(2 p)p.1nna12p2A 1P(A ) ap apA C ，111ap p)(2 p)ap.111P22設(shè) A= ， P(B|)0.98P(B| A) 0.05P() 0.96P(A| B)，P()P(B| )P(A)P(B| A)0.9408 0.9979.0.960.980.040.05 0.9428），BA PC)P()P(AB PC)P(AB)，BP(AB) P(A B) 1 P(A B) 1P()P(B)P

5、AB，,P(BC) P(B)P(C).P(P(BPC P(A)P(B)P(C)，112nAAiiP(A A A ) P(A A ) P(A A ) P(A A )12n2n1n1n P(A )P(A ) P(A )p(A ) P(A )P(A )P(A )。2n1n12n kP(A A A A ) P(A )P(A )P(A A ),1kk1n1kk1nk1時1n1kk2n1kk1n1kk2n1kk1n P(A )P(A P(A )P(A )P(A )1kn P(A )P(A )P(A )P(A )P(A )1kn2 P(A A A ) P(A )P(A )P(A ),12n12n A 取 A

6、 或 。AiiiP(A A ) P(A )P(A ),1n1nP(A A A ) P(A )P(A )P(A ).1n1n1n1P.11n1n得P(A A ) P(A )P(A )。11P(A A A A ) P(A )P(A )P(A )P(A ),1n1n1n1nP(A A A ) P(A )P(A )P(A ).）11得P(A A ) P(A )P(A )P(A )。1122 n1 n1n P P P) 0 0 ) P() 0 P()P P() 1 P()P(P(B) P(B) P()P(BP() P() P()P(P(AB) P(A)P(B)P(AB) P(A AB) P(A) P(A

7、B) P(A) P(A)P(B) P(P(B P()P(B)，P(AB)P(A)P(B)、解：三次射擊恰擊中目標一10.7) n A A P(A )P(A ) 。1n1nkk11nn1n1nnk1 p p ) p ) p )p p ) p )12n123n p ) p )p1n1n nnp 2n1。iijii1，A k A k ，A k01122012012012kkAP(A A A ) P(A )P(A )P(A ) ) 。n12n12nnA1 P(A A A ) 1 ) 1 (n )。n12n所有零件均為一級品 0.83 0.72 0.2509。、解：利用二項分布得至少出現(xiàn)一次正面1 n次

8、全部出現(xiàn)反面1 p) 。n至少出現(xiàn)兩次正面1 p) C1p p) 1nnn1 p) np p)。nn11P() P(B) PC) 330020p 。np P iii0 1 1iinn Ciii2nnnni0i0，nnb C1p q 1 C3p3qn3。nnnAnn 11 1b 1(pq) 2p) 。nn2Aa。n2 21 Amk A次，m次 Cp qm1 q Ckkkmm次Ak次m次Ak次，km次Ak次m次 Ak次A1次，k 2次Am次 Ak次m次Aki次，ik,k。A經(jīng)n，A經(jīng)nnn乙袋中， C第 n 次從黑球所在的袋中取出一個白球。記nnnc P(A ) 1 p , n nnnnp P(A | A )P(A )_P(A | A )P(A )nnn1n1nn1n1nn1n1nn1n1N 11N 11,n1Nn1NN1NN 21即(n。pNNn p101nnNn1 Nnn 2k p 1時 。時1n 2k 1 p 0。n12n有 p 2n 2， 則nn C (0.7) 。jjj0p 1(0.999) 5000(0.999) 0.0011e551p11 501i2p 1 C