2018 期中

1 选择题

1-1

B

1-2

C

1-3

$P(X\ge 1)=1-P(X<1)=1-P(X=0)=1-(1-p{)}^2=\frac{5}{9}\Rightarrow 1-p=\frac{2}{3}\Rightarrow p=\frac{1}{3}$,

$P(Y=1)=1-P(Y=0)=1-{\left(\frac{2}{3}\right)}^3=\frac{19}{27}B$

1-4

$P_1=P(-2<X<2)P_2=P(-1<X<1)P_3=P(-\frac{1}{3}<X<-1)$

$P_1>P_2>P_{3}A$

1-5

C

2 填空题

2-1

$2\cdot \frac{1}{120}=\frac{1}{60}$

2-2

$0.4\times 0.3\times 0.5+0.6\times 0.3\times 0.5+0.4\times 0.7\times 0.5=0.29$

$P(X\mid B)=0.4\times 0.5+0.6\times 0.5=0.5$

$P(X\mid B)\cdot P(B)=P(B\mid X)\cdot P(X)\Rightarrow P(B\mid X)=\frac{P(X\mid B)\cdot P(B)}{P(X)}=\frac{0.5\times 0.3}{0.29}=\frac{15}{29}$

2-3

$P_1=(0.6{)}^3{P}_2=(0.4{)}^{3}P=1-P_1-P_2=0.72$

2-4

$P(\overline{A}\cup \overline{B})=1-P(A\cap B)=1-[P(A\cup B)-P(A)-P(B)]=1-(\frac{4}{3}-1)=\frac{2}{3}$

2-5

$f(x)=\{\begin{array}{cc}\lambda e^{-\lambda x}& x\ge 0\\ 0& x<0\end{array}$

$P(3<X<4)={\int }_3^4\lambda e^{-\lambda x}dx=e^{-3\lambda }-e^{-4\lambda }P(X>2)={\int }_2^{\mathrm{\infty }}\lambda e^{-\lambda x}dx=e^{-2\lambda }$

$P(3<X<4\mid X>2)=e^{-\lambda }-e^{-2\lambda }=t-t^2=-{(t-\frac{1}{2})}^2+\frac{1}{4}t=e^{-\lambda }=\frac{1}{2}\Rightarrow \lambda =\ln 2$

2-6

$P(max(X,Y)>2)=1-P(max(X,Y)\le 2)=1-P(X\le 2,Y\le 2)=1-P(X\le 2)P(Y\le 2)$

$=1-\frac{1}{2}\cdot \frac{1}{2}=\frac{3}{4}$

2-7

$X-Y\sim N(-\mu ,2{\sigma }^2)-\mu =2\Rightarrow \mu =-2$

2-8

$P(X=k)=\frac{{\lambda }^k}{k!}{e}^{-\lambda }$

$$P(Y=k)=P(X=k^{1/3})=\frac{{\lambda }^{k^{1/3}}}{(k^{1/3})!}{e}^{-\lambda }$$

3 解答题

3-1

$P=P(X\le 2/3)={\int }_0^{2/3}3x^{2}dx={\left(\frac{2}{3}\right)}^3=\frac{8}{27}$

$P(i=1)=(\genfrac{}{}{0ex}{}{3}{1})\cdot P\cdot (1-P{)}^2=3\cdot \frac{8}{27}\cdot {(1-\frac{8}{27})}^2=\frac{2888}{6561}$

3-2

(1) $\frac{1}{6}+a+\frac{1}{18}+\frac{1}{3}+\frac{2}{9}+b=\frac{7}{9}+a+b=1\Rightarrow a+b=\frac{2}{9}$

(2) $P(X=1)=\frac{1}{2}P(Y=0)=\frac{2}{9}+a$

$P(X=1,Y=0)=\frac{1}{2}(\frac{2}{9}+a)=\frac{1}{6}\Rightarrow \frac{2}{9}+a=\frac{1}{3}\Rightarrow a=\frac{1}{9}b=\frac{1}{9}$

3-3

(1) ${\int }_{-1}^1{\int }_{x^2}^{1}c{x}^{2}ydydx={\int }_{-1}^1\frac{c}{2}{x}^2(1-x^4)dx=\frac{4c}{21}=1\Rightarrow c=\frac{21}{4}$

(2) ${\int }_0^1{\int }_{x^2}^{x}c{x}^{2}ydydx={\int }_0^1\frac{c}{2}{x}^2(x^2-x^4)dx=\frac{c}{10}-\frac{c}{14}=\frac{3c}{70}$

3-4

$P(X>x)=1-P(X\le x)\le 0.05\Rightarrow P(X\le x)\ge 0.95\Rightarrow P(\frac{X-110}{12}\le \frac{x-110}{12})\ge 0.95$

$\Rightarrow \mathrm{\Phi }\left(\frac{x-110}{12}\right)\ge 0.95\Rightarrow \frac{x-110}{12}\ge 1.645\Rightarrow x\ge 129.74$ 故最小为 129.74

3-5

(1) $f(x,y)=\{\begin{array}{cc}c& (x,y)\in G\\ 0& \text{其它}\end{array}\Rightarrow {\iint }_{(x,y)\in G}f(x,y)dxdy=c\cdot \frac{1}{2}\cdot 2\cdot 1=c=1$

$$f_X(x)=\{\begin{array}{ll}{\int }_0^{x}f(x,y)dy=x& 0\le x<1\\ {\int }_0^{2-x}f(x,y)dy=2-x& 1\le x\le 2\end{array}$$

(2) $f_Y(y)={\int }_y^{2-y}f(x,y)dx=2(1-y)0\le y\le 1$

故当 $0\le y<1$ 时，$f_{X\mid Y}(x\mid y)=\frac{f(x,y)}{f_Y(y)}=\frac{1}{2(1-y)}y\le x\le 2-y$

3-6

(1) $P(Y\le 1)=P(Y^{\prime }=1)=P(X\equiv 2)={\int }_2^3\frac{1}{9}{x}^{2}dx=\frac{19}{27}$

$P(i\le x)=P(Y\le 1)+P(1<Y<x)=\frac{19}{27}+{\int }_1^x\frac{1}{9}{t}^{2}dt=\frac{19}{27}+\frac{1}{27}(x^3-1)=\frac{1}{27}{x}^3+\frac{2}{3}$

${}^{y_{22}}P(Y\le y)=\frac{26}{27}+P(X\le 1)=1$

故

$$F_1(y)=\{\begin{array}{cc}0& y<1\\ \frac{2}{3}+\frac{1}{27}{y}^3& 1⩽y<2\\ 1& 2⩽y\end{array}$$

(2) $P\{X\le Y\}=P(Y=2)+P(Y=X)=P(X<2)={\int }_0^2\frac{1}{9}{x}^{2}dx=\frac{8}{27}$