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110AMidtermsolution

# 110AMidtermsolution - STAT 110A Midterm Solution 1(a P r(X...

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STAT 110A Midterm Solution 1. (a) Pr ( X + Y < z ) = area of { ( x,y ): x [0 , 1] ,y [0 , 1] ,x + y<z } area of unit square [0 , 1] × [0 , 1] F ( z ) = Pr ( X + Y < z ) = 0 z < 0 z 2 2 0 z 1 1 - (1 - z ) 2 2 1 < z 2 1 z > 2 (b) Pr ( X > 1 2 | X + Y < 3 2 ) = Pr ( X> 1 2 X + Y < 3 2 ) Pr ( X + Y < 3 2 ) = 1 2 - 1 2 × 1 2 × 1 2 1 - 1 2 × 1 2 × 1 2 = 2 8 7 8 = 3 8 2. Pr (fire | alarm) = Pr ( fire alarm ) Pr ( alarm ) = Pr ( fire ) Pr ( alarm | fire ) Pr ( fire ) Pr ( alarm | fire )+ Pr ( no fire ) Pr ( alarm | no fire ) = αβ αβ +(1 - α ) γ 3. (a) Pr ( X 0 = 1) = 1 Pr ( X 1 = 1) = 1 / 2 , Pr ( X 1 = 2) = 1 / 4 , Pr ( X 1 = 3) = 1 / 4 Pr ( X 2 = i ) = 3 k =1 Pr ( X 1 = k X 2 = i ) = 3 k =1 Pr ( X 1 = k ) Pr ( X 2 = i | X 1 = k ) = Pr ( X 2 = 1) = Pr ( X 1 = 1) Pr ( X 2 = 1 | X 1 = 1) + Pr ( X 1 = 2) Pr ( X 2 = 1 | X 1 = 2) + Pr ( X 1 = 3) Pr ( X 2 = 1 | X 1 = 3) = 1 / 2 × 1 / 2 + 1 / 4 × 1 / 4 + 1 / 4 × 1 / 4 = 3 / 8 Pr ( X 2 = 2) = 1 / 2 × 1 / 4 + 1 / 4 × 1 / 2 + 1 / 4 × 1 / 4 = 5 / 16 Pr ( X 2 = 3) = 1 / 2 × 1 / 4 + 1 / 4 × 1 / 4 + 1 / 4 × 1 / 2 = 5 / 16 (b) Pr ( X 1 = i | X 2 = 3) = Pr ( X 1 = i X 2 =3) Pr ( X 2 =3) = Pr ( X 1 = i ) Pr ( X 2 =3 | X 1 = i ) 5 / 16 Pr ( X 1 = 1 | X 2 = 3) = 1 / 2 × 1 / 4 5 / 16 = 2 / 5 Pr ( X 1 = 2 | X 2 = 3) = 1 / 4 × 1 / 4 5 / 16 = 1 / 5 Pr ( X 1 = 3 | X 2 = 3) = 1 / 4 × 1 / 2 5 / 16 = 2 / 5 4. (a) Pr ( X = i Y = j ) = Pr ( X = i Y = j Z = 0) + Pr ( X = i Y = j Z = 1) = Pr ( X = i Y = j | Z = 0) Pr ( Z = 0) + Pr ( X = i Y = j | Z = 1) Pr ( Z = 1) = Pr ( X = i k Z = 0) Pr ( Y = j | Z = 0) Pr ( Z = 0) + Pr ( X = i k Z = 1) Pr ( Y = j | Z = 1) Pr ( Z = 1) Pr ( X = 1 Y = 1) = αβ 1 γ 1 + (1 - α ) β 0 γ 0 Pr ( X = 1 Y = 0) = αβ 1 (1 - γ 1 ) + (1 - α ) β 0 (1 - γ 0 ) Pr ( X = 0 Y = 1) = α

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110AMidtermsolution - STAT 110A Midterm Solution 1(a P r(X...

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