概率论与数理统计浙大第五版 第三章 部分习题

29

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∫−∞∞∫−∞∞f(x,y)dydx=1=∫−∞0∫−∞∞0dydx+∫01∫−∞∞f(x,y)dydx+∫1∞∫−∞∞0dydx=∫01∫−∞∞f(x,y)dydx=∫01∫−∞0f(x,y)dydx+∫01∫0∞0dydx=∫01∫0∞f(x,y)dydx=∫01(−be−(x+y))∣0∞dx=∫01be−xdx=b(1−e−1)=1⇒b=11−e−1\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x,y)dydx=1\\ =\int_{-\infty}^{0}\int_{-\infty}^{\infty}0dydx+\int_{0}^{1}\int_{-\infty}^{\infty}f(x,y)dydx+\int_{1}^{\infty}\int_{-\infty}^{\infty}0dydx\\ =\int_{0}^{1}\int_{-\infty}^{\infty}f(x,y)dydx\\ =\int_{0}^{1}\int_{-\infty}^{0}f(x,y)dydx+\int_{0}^{1}\int_{0}^{\infty}0dydx\\ =\int_{0}^{1}\int_{0}^{\infty}f(x,y)dydx\\ =\int_{0}^{1}(-be^{-(x+y)})|^{\infty}_{0}dx\\ =\int_{0}^{1}be^{-x}dx\\ =b(1-e^{-1})=1 \ \ \Rightarrow b=\frac{1}{1-e^{-1}} ∫−∞∞​∫−∞∞​f(x,y)dydx=1=∫−∞0​∫−∞∞​0dydx+∫01​∫−∞∞​f(x,y)dydx+∫1∞​∫−∞∞​0dydx=∫01​∫−∞∞​f(x,y)dydx=∫01​∫−∞0​f(x,y)dydx+∫01​∫0∞​0dydx=∫01​∫0∞​f(x,y)dydx=∫01​(−be−(x+y))∣0∞​dx=∫01​be−xdx=b(1−e−1)=1⇒b=1−e−11​

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fX(x)=∫−∞∞f(x,y)dy=∫−∞00dy+∫0∞(−be−(x+y))dy=e−x1−e−1fY(y)=∫−∞∞f(x,y)dx=∫−∞00dx+∫01(−be−(x+y))dx+∫1∞0dy=e−yf_X(x)=\int^{\infty}_{-\infty}f(x,y)dy\\ =\int^{0}_{-\infty}0dy+\int^{\infty}_{0}(-be^{-(x+y)})dy\\ =\frac{e^{-x}}{1-e^{-1}}\\\ \\ \ \\ f_Y(y)=\int^{\infty}_{-\infty}f(x,y)dx\\ =\int^{0}_{-\infty}0dx+\int^{1}_{0}(-be^{-(x+y)})dx+\int^{\infty}_{1}0dy\\ =e^{-y} fX​(x)=∫−∞∞​f(x,y)dy=∫−∞0​0dy+∫0∞​(−be−(x+y))dy=1−e−1e−x​fY​(y)=∫−∞∞​f(x,y)dx=∫−∞0​0dx+∫01​(−be−(x+y))dx+∫1∞​0dy=e−y

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FX(x)={1−e−x1−e−10≤x<10otherwiseFY(y)={1−e−y0≤y<∞0otherwiseFU(u)=FU(max(x,y))=FX(x)∗FY(y)FU(u)={0u<0(1−e−u)21−e−10≤u<11−e−uu≥1F_X(x)=\left\{\begin{aligned} \frac{1-e^{-x}}{1-e^{-1}}\quad 0\leq x< 1\\ 0\quad\quad\quad\quad otherwise \end{aligned}\right. \\ F_Y(y)=\left\{\begin{aligned} {1-e^{-y}}\quad 0\leq y< \infty\\ 0\quad\quad\quad\quad otherwise \end{aligned}\right. \\ F_U(u)=F_U(max(x,y))=F_X(x)*F_Y(y) \\ F_U(u)=\left\{\begin{aligned} 0\quad\quad\quad\quad\quad\quad\quad \quad u<0\\ \frac{(1-e^{-u})^2}{1-e^{-1}}\quad \quad 0\leq u<1\\ {1-e^{-u}}\quad\quad\quad\quad\quad u\geq 1 \end{aligned}\right. FX​(x)=⎩⎪⎨⎪⎧​1−e−11−e−x​0≤x<10otherwise​FY​(y)={1−e−y0≤y<∞0otherwise​FU​(u)=FU​(max(x,y))=FX​(x)∗FY​(y)FU​(u)=⎩⎪⎪⎪⎨⎪⎪⎪⎧​0u<01−e−1(1−e−u)2​0≤u<11−e−uu≥1​

30

可知一个电子元件的寿命大于180h的概率为p=(1−0.6826)2p=\frac{(1-0.6826)}{2}p=2(1−0.6826)​

则任选四只没有一只寿命小于180的概率为C44p4=0.00063C_{4}^{4}p^4=0.00063C44​p4=0.00063

36

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P{X=2∣Y=2}=P{X=2,Y=2}P{Y=2}=0.2P{Y=3∣X=0}=P{Y=3,X=0}P{X=0}=13P\{X=2|Y=2\}=\frac{P\{X=2,Y=2\}}{P\{Y=2\}}=0.2\\ P\{Y=3|X=0\}=\frac{P\{Y=3,X=0\}}{P\{X=0\}}=\frac{1}{3} P{X=2∣Y=2}=P{Y=2}P{X=2,Y=2}​=0.2P{Y=3∣X=0}=P{X=0}P{Y=3,X=0}​=31​

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