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概率统计——习题十解答 F(-∞,.0)=A(B-)C=0 B=C= 2.(1)由F(0,-∞)=AB(C-)=0,得 A F(-∞,∞)=A(B+C+)=1 (2)f(x,y)=2(xy π(4+x2)(4+y2) o <x, y<oo. (3)F1(x)=F(x,∞)=1(x+amg3),-m<x<x F()=F(o,2)=I6+ arct f(x)=Fx(x) 2 <x<0, π(4+x2) f(y)=F(y)= ∞<y<0 π(4+y2) 3.P=p{X+P<}-P(¥x48 4.(1)fx(x)=24x2(2-x),(0≤x≤1) f(y)=24y(3-4y+y2),(0≤y≤1) (2)fx(x)=e,(x>0) f(y)=ye-y,(y>0) 5.(1)k=1/8: (2)P{X<1,Y<}=「d(6x-m3 3)Px<13=46-x-y=27 (4)Px+Y≤4}=∫八(xy)=Jdj(6-x-y)k=概率统计——习题十解答 1. 4 1 。 2.(1)由          =  +  −  = + =  −  = − =  − = − ) 1, 2 )( 2 ( , ) ( ) 0, 2 (0, ) ( ) 0, 2 ( , 0) ( F A B C F AB C F A B C 得       =  = = ; 1 , 2 2 A B C (2) −      + + =    = x y x y x y F x y f x y , , (4 )(4 ) ( , ) 4 ( , ) 2 2 2 2 ; (3) ) 2 2 ( 1 ( ) ( , ) x FX x F x + arctg   =  = ,−  x  , ) 2 2 ( 1 ( ) ( , ) y FY y F y + arctg   =  = ,−   y  , −      + = = x x f x F x X X , (4 ) 2 ( ) ( ) 2 ' , , . (4 ) 2 ( ) ( ) 2 ' −      + = = y y f y F y Y Y 3. } 2 1 , 2 1 } { 2 1 } { 2 1 p = p{X  + P Y  − P X  Y  = 8 5 4. (1) ( ) 2.4 (2 ),(0 1) 2 f X x = x − x  x  ( ) ,( 0) (2) ( ) ,( 0) ( ) 2.4 (3 4 ),(0 1) 2 =  =  = − +   − − f y ye y f x e x f y y y y y y Y x X Y 5.(1) k=1/8; (2) 8 3 (6 ) 8 1 { 1, 3} 1 0 3 2   = − − =   P X Y dx x y dy ; (3)    = − − = 4 2 1.5 0 32 27 (6 ) 8 1 P{X 1.5} dx x y dy ; (4)    +  − +  = = − − = 4 4 2 4 2 . 3 2 (6 ) 8 1 { 4} ( , ) x y y P X Y f x y dxdy dy x y dx
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