chapter 1

Chapter 3: Discrete Random Variables and Probability Distributions25.a. Possible X values are 1, 2, 3, …1P(1) = P(X = 1 ) = P(return home after just one visit) = 32 1⋅P(2) = P(X = 2) = P(second visit and then return home) =3 32 2 1⋅P(3) = P(X = 3) = P(three visits and then return home) = ( )3 32 x−11In general p(x) = ( ) ( ) for x = 1, 2, 3, …33b. The number of straight line segments is Y = 1 + X (since the last segment traversed2 y−21returns Alvie to O), so as in a, p(y) = ( ) ( )33for y = 2, 3, …c. Possible Z values are 0, 1, 2, 3 , …1 1 1p(0) = P(male first and then home) = ⋅ = ,2 3 6p(1) = P(exactly one visit to a female) = P(female 1 st , then home) + P(F, M, home) +P(M, F, home) + P(M, F, M, home)= ( 1 1 1 2 1 1 2 1 1 2 2 1)( ) + ( )( )( ) + ( )( )( ) + ( )( )( )( )2 3 2 3 3 2 3 3 2 3 3 3=( 1 2 1 1 2 2 1 1 5 1 1 2 5 1)( 1 + )( ) + ( )( )( + 1 )( ) = ( )( )( ) + ( )( )( )( )23323332where the first term corresponds to initially visiting a female and the second termcorresponds to initially visiting a male. Similarly,2 2 5 1 1 2 2 5 1p(2) = ( )( ) ( )( ) ( )( ) ( )( )1+ . In general,233323333323331 2 2z−25 1 1 2 2z−25 1 24 2 2z−2p(z) = ( )( ) ( )( ) ( )( ) ( )( ) ( )( )2333+2333=543for z = 1, 2, 3, …26.a. The sample space consists of all possible permutations of the four numbers 1, 2, 3, 4:outcome y value outcome y value outcome y value1234 4 2314 1 3412 01243 2 2341 0 3421 01324 2 2413 0 4132 11342 1 2431 1 4123 01423 1 3124 1 4213 11432 2 3142 0 4231 22134 2 3214 2 4312 02143 0 3241 1 4321 0b. Thus p(0) = P(Y = 0) = 9 24, p(1) = P(Y = 1) = 8 24, p(2) = P(Y = 2) = 6 24,p(3) = P(Y = 3) = 0, p(3) = P(Y = 3) = 1 24.102