Chapter 2

• ## 例1.26

(1)

P(\text{次品})=\frac12\cdot1\%+\frac13\cdot1\%+\frac16\cdot2\%=1.167\%

(2)

%~~~~P(\text{一号车间}|\text{次品}) = \frac{P(\text{一号车间}) \cdot P(\text{次品}|\text{一号车间})}{P(\text{次品})} = \frac{0.5 \% }{1.167 \% } = \frac37

例1.30

P(\text{真阴性})=90\%\times0.99=89.1\%

P(\text{真阳性})=10\%\times0.95=9.5\%

P(\text{假阳性})=90\%\times0.01=0.9\%

P(\text{假阴性})=10\%\times0.05=0.5\%

(1)

%P(\text{真阳性}|\text{阳性})=\frac{9.5\%}{9.5\%+0.9\%}=91.3\%

(2)

%P=\frac{9.5\%^2}{9.5\%^2+0.9\%^2}=99.1\%

例1.36

P(A\cup B)=0.12, P(A\cap B)=0.1

P(A)+P(B)=P(A\cup B)-P(A\cap B)=0.02

显然不能满足 P(A)P(B)=P(AB)

故 A, B 一定不独立。

例1.37

P(A)=P(A|C)P(C)+P(A|\overline C)P(\overline C)=0.55

P(B)=P(B|C)P(C)+P(B|\overline C)P(\overline C)=0.55

P(AB)=P(AB|C)P(C)+P(AB|\overline C)P(\overline C)=0.425

可见 P(AB)\neq P(A)P(B)

故 A, B 不独立

例题1.39

(1) P_1 = p_A p_B p_C

(2) P_2 = 1-(1-p_A)(1-p_B)(1-p_C)=p_A+p_B+p_C-p_Ap_B-p_Ap_C-p_Bp_C+p_Ap_Bp_C
(3)P_3=1-(1-p_A^2)(1-p_B^2)(1-p_C^2)=p_A^2+p_B^2+p_C^2-p_A^2p_B^2-p_A^2p_C^2-p_B^2p_C^2+p_A^2p_B^2p_C^2(4)P_4=p_D^2(1-(1-p_A)(1-p_B)(1-p_C))=p_D^2(p_A+p_B+p_C-p_Ap_B-p_Ap_C-p_Bp_C+p_Ap_Bp_C)(5)P_5=p_Ap_B+p_Ap_B-p_A^2p_B^2+2\cdot p_C \cdot p_A(1-p_A)p_B(1-p_B)$

例 2.6

P(X=1)=\frac45

P(X=2)=\frac15\cdot\frac89=\frac8{45}

P(X=3)=\frac15\cdot\frac19=\frac1{45}

P(X\leq1)=\frac45

P(X\leq2)=\frac45+\frac8{45}=\frac{44}{45}

P(X\leq3)=1

F(x)=

例 2.9

(1)

P(0A)=0.7^4=0.2401

P(1A)=0.7^3\times0.3\times 4=0.4116

P(2+A)=1-P(0A)-P(1A)=0.3483

P(B)=0.6P(1A)+P(2A)=0.59526

(2)

P(1A|B)=\frac{P(B|1A)P(1A)}{P(B)}=0.41488

例 2.10

p_4=0.6^4=0.130

p_5=0.6^4\times0.4\times C^3_4=0.207 （最后一场必为甲胜）

p_6=0.6^4\times0.4^2\times C^3_5=0.207

p_7=0.6^4\times0.4^3\times C^3_6=0.166

p_4+p_5+p_6+p_7=0.71

甲在三局或以前以“三局两胜”制成为冠军的概率为： p’_3=3*0.6^2*0.4+0.6^3=0.648

可见三局两胜更有利（因为需要比赛的场数越多，乙赢得甲的概率越小）

例 2.12

产卵 k 个的概率 p_k=\frac{\lambda^ke^{-\lambda}}{k!}

\therefore P(Y=x)=\Sigma_{k=x}^\infin(p^x(1-p)^{k-x}\cdot p_k)=p^x\cdot e^{-\lambda}\cdot\Sigma_{k=0}^\infin((1-p)^k\cdot\frac{\lambda^{x+k}}{(x+k)!})

P(Z=x)=\Sigma_{k=x}^\infin(p^{k-x}(1-p)^x\cdot p_k)=(1-p)^x\cdot e^{-\lambda}\cdot\Sigma_{k=0}^\infin(p^k\cdot\frac{\lambda^{x+k}}{(x+k)!})

例 2.13

易知出故障零件个数 X 满足二项分布。

P(X=0)=b(1000,0.001,0)\approx \frac{1^0e^-1}{0!}=\frac1e