一袋中有

1. (8%)一袋中有⁴白球及²紅球 。 由袋中任拿³球 。 令^X 為拿出的球中^, 紅球之數目 。 求P (X = 0), P (X = 1), P (X = 2)及 E(X)。⁽各²分^,共⁸分⁾

Sol:

P (X = 0) = C₃⁴ C₃⁶ =1

5; P (X = 1) = C₂⁴C₁²

C₃⁶ =3 5; P (X = 2) = C₁⁴C₂²

C₃⁶ =1 5;

E(X) =

∑2 k=0

k· P (X = k) = 0 ·1 5 + 1·3

5 + 2· 1 5 = 1.

2. (12%) 設隨機變數^{X , Y} 其值域分別為^IX={0, 1} , IY ={0, 1, 2}其聯合機率如下表^:

P (X = 0, Y = 0) = 1

9, P (X = 0, Y = 1) = 1

18, P (X = 0, Y = 2) = 1 6, P (X = 1, Y = 0) = 1

18, P (X = 1, Y = 1) = 5

18, P (X = 1, Y = 2) = 1 3. (a) 求P (Y = 0) , P (Y = 1) , P (Y = 2) 及^{E(Y )}。⁽各²分^,共⁸分⁾

(b) 求^{P (X = 0)}。 隨機變數^{X , Y} 是否獨立^{? (}各²分^, 共⁴分⁾ Sol:

(a)

P (Y = 0) =

∑1 k=0

P (X = k, Y = 0) =1 9 + 1

18 =1 6;

P (Y = 1) =

∑1 k=0

P (X = k, Y = 1) = 1 18+ 5

18= 1 3;

P (Y = 2) =

∑1 k=0

P (X = k, Y = 2) = 1 6 +1

3 = 1 2;

E(Y ) =

∑2 k=0

k· P (Y = k) = 0 ·1 6+ 1·1

3 + 2· 1 2 = 4

3. (b)

P (X = 0) =

∑2 k=0

P (X = 0, Y = k) = 1 9+ 1

18+1 6 = 1

3. X and Y are not independent since

P (X = 0, Y = 0) = 1 9 6= 1

18 = P (X = 0)P (Y = 0).

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