概率论与数理统计浙大第五版 第三章 部分习题
29
(1)
∫?∞∞∫?∞∞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?
(2)
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
(3)
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
可知一個電子元件的壽命大于180h的概率為p=(1?0.6826)2p=\frac{(1-0.6826)}{2}p=2(1?0.6826)?
則任選四只沒有一只壽命小于180的概率為C44p4=0.00063C_{4}^{4}p^4=0.00063C44?p4=0.00063
36
| 0 | 0.00 | 0.01 | 0.03 | 0.05 | 0.07 | 0.09 | 0.25 |
| 1 | 0.01 | 0.02 | 0.04 | 0.05 | 0.06 | 0.08 | 0.26 |
| 2 | 0.01 | 0.03 | 0.05 | 0.05 | 0.05 | 0.06 | 0.25 |
| 3 | 0.01 | 0.02 | 0.04 | 0.06 | 0.06 | 0.05 | 0.24 |
| P{X=i}P\{X=i\}P{X=i} | 0.03 | 0.08 | 0.16 | 0.21 | 0.24 | 0.28 | 1 |
(1)
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?
(2)
| PkP_kPk? | 0 | 0.04 | 0.16 | 0.28 | 0.24 | 0.28 |
(3)
| PkP_kPk? | 0.28 | 0.30 | 0.25 | 0.17 |
(4)
| PkP_kPk? | 0 | 0.02 | 0.06 | 0.13 | 0.19 | 0.24 | 0.19 | 0.12 | 0.05 |
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