Section 3.3 3.3 Sample Space w HHH HHT HTH THH HTT THT TTH TTT 3 1 1 1 -1 -1 -1 -3 3.5 3 (a) 1 c( x 2 4) 30c x 0 Thus, c=1/30 2 2 3 10c (b) 1 c x 0 x 3 x Thus, c=1/10 3.8 P(w=3)=(2/3)(2/3)(2/3)=8/27 P(w=1)=3(2/3)(2/3)(1/3)=4/9 P(w=-1)=3(2/3)(1/3)(1/3)=2/9 P(w=-3)=(1/3)(1/3)(1/3)=1/27 3.10 f ( x) 1 / 6, 1 x 6 3.14 (a) P ( X 12 0.2) F (0.2) 1 e 1.6 60 (b) f ( x) F ' ( x) 8e 8 x 0.2 Thus, P ( X 0.2) 8e 8 x dx e 8 x | 00.2 1 e 1.6 0 3.21 1 (a) 1 k x dx 0 2k 3 / 2 1 2k x |0 3 3 Thus, k=3/2 (b) F ( x) P( X x) x 0 3 t dt t 3 / 2 |0x x 3 / 2 2 P (0.3 X 0.6) F (0.6) F (0.3) (0.6) 3 / 2 (0.3) 3 / 2 3.22 Let X be the number of spades in the three draws. P(X=0)=(39/52)(38/51)(37/50) P(X=1)=3(13/52)(39/51)(38/50) P(X=2)=3(13/52)(12/51)(39/50) P(X=3)=(13/52)(12/51)(11/50) OR 13 39 x 3 x , 0 x3 f ( x) 52 3 Section 3.4 3.37 3 3 (a) 1 cxy 36c x 1 y 1 Thus, c=1/36 (b) 1 c | x y | 15c x y Thus, c=1/15 3.38 (a) (b) (c) (d) P( X P( X P( X P( X 2, Y 1) f (0,1) f (1,1) f (2,1) 1 / 5 2, Y 1) f (3,0) f (3,1) 7 / 30 Y ) f (1,0) f (2,0) f (2,1) f (3,0) f (3,1) f (3,2) 3 / 5 Y 4) f (2,2) f (3,1) 4 / 15 3.42 f ( x | y) f ( x, y ) h( y ) e ( x y ) 0 e ( x y ) dx e ( x y ) ex y e 1 1 0 0 P (0 X 1 | Y 2) f ( x | y )dx e x dx 1 e 1 3.46 2 x y x 1 , 10 y 0 30 2 (a) g ( x) f ( x, y ) y 0 3 3 x 0 x 0 (b) h( y ) f ( x, y ) x y 2y 3 , 30 15 x 0,1,2,3 y 0,1,2 3.49 (a) x 1 2 3 g(x) 0.1 0.35 0.55 (b) y 1 3 5 h(y) 0.2 0.5 0.3 (c) P(Y 3 | X 2) f (2,3) 0.1 g (2) 0.35 3.51 4 4 4 x y x y 3 , (a) f ( x, y ) 12 3 0 x y3 (b) P ( X Y 2) 1 P ( X Y 2) 1 [ f (0,0) f (0,1) f (1,0)] 3.54 Since f (1,1) g (1)h(1), the two random variables are dependent 3.56 (a) f ( x | y ) f ( x, y ) h( y ) 6x 1 y 0 6 xdx 6x 2x 2 3(1 y ) (1 y ) 2 Since f(x|y) involves the variable y, X and Y are not independent OR Prove f ( x, y ) g ( x)h( y ) 1 y (b) P ( X 0.3 | Y 0.5) 0.3 f ( x | y )dx 0.5 0.3 2x dx 0.64 (1 0.5) 2 42 55 3.57 (a) 1 2 1 1 0 kxy 0 0 2 zdxdydz k 3 Thus, k=3 (b) P ( X 2 1 1/ 4 1 1 21 , Y ,1 Z 2) 3xy 2 zdxdydz 1 1/ 2 0 4 2 512 3.60 4 xyz 2 2 yz 2 dx , 0 y 1, 0 z 3 (a) g ( y, z ) 0 9 9 1 (b) h( y ) 3 0 2 yz 2 dz 2 y, 0 y 1 9 2 2 1 1 / 2 4 xyz 1 1 1 7 (c) P ( X , Y ,1 Z 2) dxdydz 1 1 / 3 1 / 4 4 2 3 9 162 f ( x, y , z ) 2 x, 0 x 1 (d) f ( x | y, z ) g ( y, z ) Thus, P (0 X 1/ 2 1 1 1 | Y , Z 2) 2 xdx 0 2 4 4 Review Exercises 3.63 (a) g ( x) ye y (1 x ) dy 0 ye y (1 x ) 1 x 0 1 y (1 x ) 1 e dy , x0 1 x 0 (1 x) 2 Hint: Integration by parts 0 0 Let u y, dv e y (1 x ) dy , then g ( x) udv uv vdu h( y ) ye y (1 x ) dx e y 1 x |0 e y , y 0 0 (b) P ( X 2, Y 2) 2 2 ye y (1 x ) dxdy e y 1 x e 6 3 3.67 5 f ( x) (0.1) x (0.9) 5 x , 0 x 5 x 3.68 (a) g ( x) 2 1 3x y 4 2y dx , 1 y 2 1 9 3 9 No. Since f ( x, y ) g ( x)h( y ) h( y ) (b) 3x y x 1 dy , 1 x 3 9 3 6 3 3 x 1 2 (c) P ( x 2) ( )dx 2 3 6 3 3.69 (a) f ( x) F ' ( x) 1 x / 50 e , x0 50 (b) P( x 70) 1 P( X 70) 1 F (70) 1 (1 e 70 / 50 )