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当前位置:首页 > 商业/管理/HR > 信息化管理 > 概率论与数理统计(茆诗松)第二版第三章课后习题3.1参考答案
1第三章多维随机变量及其分布习题3.11.一批产品中有一等品50%,二等品30%,三等品20%.从中有放回地抽取5件,以X、Y分别表示取出的5件中一等品、二等品的件数,求(X,Y)的联合分布列.解:X,Y的全部可能取值分别为0,1,2,3,4,5,且ijijijijYiXPjiji−==×××−−⋅⋅===−−5,,0;5,4,3,2,1,0,2.03.05.0)!5(!!!5},{5L,故(X,Y)的联合分布列为0000003125.05000009375.00625.040001125.015.005.03000675.0135.009.002.02002025.0054.0054.0024.0004.0100243.00081.00108.00072.00024.000032.00543210XY2.100件商品中有50件一等品,30件二等品,20件三等品.从中不放回地抽取5件,以X、Y分别表示取出的5件中一等品、二等品的件数,求(X,Y)的联合分布列.解:X,Y的全部可能取值分别为0,1,2,3,4,5,且ijijijijYiXP−==⎟⎟⎠⎞⎜⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛−−⎟⎟⎠⎞⎜⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛===5,,0;5,4,3,2,1,0,51005203050},{L,故(X,Y)的联合分布列为000000281.0500000918.00612.040001132.01562.00495.03000661.01416.00927.00185.0200182.00539.00549.00227.00032.010019.00073.00102.00066.00019.00002.00543210XY3.盒子里装有3个黑球、2个红球、2个白球,从中任取4个,以X表示取到黑球的个数,以Y表示取到红球的个数,试求P{X=Y}.解:35935335647222347221213}2,2{}1,1{}{=+=⎟⎟⎠⎞⎜⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛+⎟⎟⎠⎞⎜⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛===+====YXPYXPYXP.4.设随机变量Xi,i=1,2的分布列如下,且满足P{X1X2=0}=1,试求P{X1=X2}.225.05.025.0101PXi−解:因P{X1X2=0}=1,有P{X1X2≠0}=0,即P{X1=−1,X2=−1}=P{X1=−1,X2=1}=P{X1=1,X2=−1}=P{X1=1,X2=1}=0,分布列为25.05.025.025.00015.0025.000110112jippXX⋅⋅−−25.05.025.025.0025.0015.025.0025.0025.0025.00110112jippXX⋅⋅−−故P{X1=X2}=P{X1=−1,X2=−1}+P{X1=0,X2=0}+P{X1=1,X2=1}=0.5.设随机变量(X,Y)的联合密度函数为⎩⎨⎧−−=.,0,42,20),6(),(其他yxyxkyxp试求(1)常数k;(2)P{X1,Y3};(3)P{X1.5};(4)P{X+Y≤4}.解:(1)由正则性:1),(=∫∫+∞∞−+∞∞−dxdyyxp,得18)6()26(26)6(20220204222042==−=−=⎟⎟⎠⎞⎜⎜⎝⎛−−⋅=−−∫∫∫∫kxxkdxxkyxyykdxdyyxkdx,故81=k;(2)∫∫∫⎟⎟⎠⎞⎜⎜⎝⎛−−⋅=−−=1032210322681)6(81}3,1{yxyydxdyyxdxYXP8322781278110210=⎟⎟⎠⎞⎜⎜⎝⎛−=⎟⎠⎞⎜⎝⎛−=∫xxdxx;(3)∫∫∫⎟⎟⎠⎞⎜⎜⎝⎛−−⋅=−−=5.104225.10422681)6(81}5.1{yxyydxdyyxdxXP3227)6(81)26(815.1025.10=−=−=∫xxdxx;(4)∫∫∫−−⎟⎟⎠⎞⎜⎜⎝⎛−−⋅=−−=+2042220422681)6(81}4{xxyxyydxdyyxdxYXP3262681246812032202=⎟⎟⎠⎞⎜⎜⎝⎛+−=⎟⎟⎠⎞⎜⎜⎝⎛+−=∫xxxdxxx.x0224yx0224y13x0224y1.5x0224y36.设随机变量(X,Y)的联合密度函数为⎩⎨⎧=+−.,0,0,0,e),()43(其他yxkyxpyx试求(1)常数k;(2)(X,Y)的联合分布函数F(x,y);(3)P{0X≤1,0Y≤2}.解:(1)由正则性:1),(=∫∫+∞∞−+∞∞−dxdyyxp,得112e12e4e41e030300)43(00)43(==−==⎥⎦⎤⎢⎣⎡−⋅=+∞−∞+−∞++∞+−∞+∞++−∫∫∫∫kkdxkkdxdykdxxxyxyx,故k=12;(2)当x≤0或y≤0时,F(x,y)=P(∅)=0,当x0且y0时,∫∫∫∫−−+−+−−=−⋅==xyuxyvuxyvudududvduyxF04300)43(00)43()e1(e3]e3[e12),()e1)(e1()e1(e43043yxxyu−−−−−−=−−=故(X,Y)的联合分布函数为⎩⎨⎧−−=−−.,0,0,0),e1)(e1(),(43其他yxyxFyx(3)P{0X≤1,0Y≤2}=P{X≤1,Y≤2}=F(1,2)=(1−e−3)(1−e−8).7.设二维随机变量(X,Y)的联合密度函数为⎩⎨⎧=.,0,10,10,4),(其他yxxyyxp试求(1)P{0X0.5,0.25Y1};(2)P{X=Y};(3)P{XY};(4)(X,Y)的联合分布函数.解:(1)∫∫∫⋅==5.00125.025.00125.024}125.0,5.00{xydxxydydxYXP641516158155.0025.00===∫xxdx;(2)P{X=Y}=0;(3)∫∫∫∫−=⋅==1031012101)22(24}{dxxxxydxxydydxYXPxx21211042=⎟⎠⎞⎜⎝⎛−=xx;(4)当x0或y0时,F(x,y)=P(∅)=0,当0≤x1且0≤y1时,x0yx0y12x0y11x0y110.50.25x0y114220220200200224},{),(yxyuduuyuvduuvdvduyYxXPyxFxxxyxy===⋅==≤≤=∫∫∫∫;当0≤x1且y≥1时,20200102010224},{),(xuuduuvduuvdvduyYxXPyxFxxxx===⋅==≤≤=∫∫∫∫;当x≥1且0≤y1时,210221021002100224},{),(yyuduuyuvduuvdvduyYxXPyxFyy===⋅==≤≤=∫∫∫∫;当x≥1且y≥1时,F(x,y)=P(Ω)=1,故(X,Y)的联合分布函数为⎪⎪⎪⎩⎪⎪⎪⎨⎧≥≥≤≥≥≤≤≤=.1,1,1,10,1,,1,10,,10,10,,00,0),(2222yxyxyyxxyxyxyxyxF或8.设二维随机变量(X,Y)的联合密度函数为⎩⎨⎧=.,0,10,),(2其他xyxkyxp(1)试求常数k;(2)求P{X0.5}和P{Y0.5}.解:(1)由正则性:1),(=∫∫+∞∞−+∞∞−dxdyyxp,得1632)(1032102101022==⎟⎟⎠⎞⎜⎜⎝⎛−=−=⋅=∫∫∫∫kxxkdxxxkykdxkdydxxxxx,故k=6;(2)∫∫∫∫−=⋅==15.0215.015.0)66(66}5.0{22dxxxydxdydxXPxxxx5.0)23(15.032=−=xx;∫∫∫∫−=⋅==5.005.005.00)66(66}5.0{dyyyxdydxdyYPyyyy432)34(5.00223−=−=yy.9.设二维随机变量(X,Y)的联合密度函数为⎩⎨⎧−=.,0,10),1(6),(其他yxyyxp(1)求P{X0.5,Y0.5};(2)求P{X0.5}和P{Y0.5};(3)求P{X+Y1}.解:(1)81)1()1(3])1(3[)1(6}5.0,5.0{15.0315.0215.01215.01=−−=−=−−⋅=−=∫∫∫∫xdxxydxdyydxYXPxx;x0y11x0y110.5x0y110.5x0y110.50.55(2)∫∫∫−−⋅=−=5.00125.001])1(3[)1(6}5.0{xxydxdyydxXP87)1()1(35.0035.002=−−=−=∫xdxx;∫∫∫−−⋅=−=5.005.025.005.0])1(3[)1(6}5.0{xxydxdyydxYP21)1(43)1(3435.0035.002=⎥⎦⎤⎢⎣⎡−−−=⎥⎦⎤⎢⎣⎡−+−=∫xxdxx;(3)∫∫∫−−−−⋅=−=+5.00125.001])1(3[)1(6}1{xxxxydxdyydxYXP43])1([])1(33[5.00335.0022=−−−=−+−=∫xxdxxx.10.设随机变量Y服从参数为λ=1的指数分布,定义随机变量Xk如下:2,1.,1,,0=⎩⎨⎧≤=kkYkYXk.求X1和X2的联合分布列.解:因Y的密度函数为⎩⎨⎧≥=−.0,0,0,e)(yyypyY且X1和X2的全部可能取值为0,1,则1101021e1ee}1{}2,1{}0,0{−−−−=−==≤=≤≤===∫yydyYPYYPXXP,P{X1=0,X2=1}=P{Y≤1,Y2}=P(∅)=0,21212121eeee}21{}2,1{}0,1{−−−−−=−==≤=≤===∫yydyYPYYPXXP,22221eee}2{}2,1{}1,1{−+∞−+∞−=−======∫yydyYPYYPXXP,故X1和X2的联合分布列为221112eee10e1010−−−−−−XX11.设二维随机变量(X,Y)的联合密度函数为⎪⎩⎪⎨⎧+=.,0,20,10,3),(2其他yxxyxyxp求P{X+Y≥1}.解:∫∫∫−−⎟⎟⎠⎞⎜⎜⎝⎛+⋅=⎟⎠⎞⎜⎝⎛+=≥+1021221021263}1{xxxyyxdxdyxyxdxYXP72652459441653421104321032=⎟⎠⎞⎜⎝⎛++=⎟⎠⎞⎜⎝⎛++=∫xxxdxxxx.x0y110.50.5x0y110.50.5x0y110.50y12x1612.设二维随机变量(X,Y)的联合密度函数为⎩⎨⎧=−.,0,0,e),(其他yxyxpy试求P{X+Y≤1}.解:∫∫∫∫−−−−−−+−=−⋅==≤+5.0015.0015.001)ee()e(e}1{dxdxdydxYXPxxxxyxxy5.015.001e2e1)ee(−−−−−+=−−=xx.13.设二维随机变量(X,Y)的联合密度函数为⎩⎨⎧=.,0,20,10,2/1),(其他yxyxp求X与Y中至少有一个小于0.5的概率.解:85831431211}5.0,5.0{1}5.0},{min{15.015.025.0=−=−=−=≥≥−=∫∫∫dxdydxYXPYXP.14.从(0,1)中随机地取两个数,求其积不小于3/16,且其和不大于1的概率.解:设X、Y分别表示“从(0,1)中随机地取到的两个数”,则(X,Y)的联合密度函数为⎩⎨⎧=.,0,10,10,1),(其他yxyxp故所求概率为∫∫∫⎟⎠⎞⎜⎝⎛−−==≤+≥−4341434111631631}1,163{dxxxdydxYXXYPxx3ln16341ln1632143412−=⎟⎠⎞⎜⎝⎛−−=xxx.x0y110.50y12x0.50.5x0y113/41/4
本文标题:概率论与数理统计(茆诗松)第二版第三章课后习题3.1参考答案
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