(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
