✔ 最佳答案
1.假設X表投擲硬幣2次出現正面的次數,Y表投擲一粒骰子出現的點數,試求:
(1) |X-Y| 的機率分配
X \ Y | 1 2 3 4 5 6
0 | 1/24 1/24 1/24 1/24 1/24 1/24
1 | 1/12 1/12 1/12 1/12 1/12 1/12
2 | 1/24 1/24 1/24 1/24 1/24 1/24
Z = |X-Y| 機率
0 | (1,1),(2,2) | 1/12+1/24=3/24
1 | (0,1),(1,2),(2,1),(2,3) | 1/24+1/12+1/24+1/24 = 5/24
2 | (0,2),(1,3),(2,4) | 1/24+1/12+1/24 = 4/24
3 | (0,3),(1,4),(2,5) | 1/24+1/12+1/24 = 4/24
4 | (0,4),(1,5),(2,6) | 1/24+1/12+1/24 = 4/24
5 | (0,5),(1,6) | 1/24+1/12 = 3/24
6 | (0,6) | 1/24
Z = |X-Y| 0 1 2 3 4 5 6
P(Z) 3/24 5/24 4/24 4/24 4/24 3/24 1/24
(2)承上題,求期望值與變異數
E(z) = Σ zp(z)
= 0*3/24 + 1*5/24 + 2*4/24 + 3*4/24 + 4*4/24 + 5*3/24 + 6*1/24
= 31/12 ..... 期望值
E(z²) = Σz²p(z)
= 0²*3/24 + 1²*5/24 + 2²*4/24 + 3²*4/24 + 4²*4/24 + 5²*3/24 + 6²*1/24
= 29/3
Var(z) = E(z²) - [E(z)]² = 29/3 - (31/12)² = 2.993 ..... 變異數
2.A學生想通過某數學考試,其每次通過數學考試的機率為0.6,另X表A學生考數學考試的次數,試問:
(1) X之機率分配為何?寫出機率函數
X ~ Geo(p)
f(x) = p(1-p)ˣ⁻¹ x = 1,2,3,.....
= 0.6*0.4ˣ⁻¹
(2) A學生預期要可幾次數學考試,才能通過?
E(X) = 1/p = 1/0.6 = 5/3
(3) A學生於第三次通過數學考試的機率為何?
f(3) = 0.6*0.4³⁻¹ = 0.096
(4) 某補習班保證學員至多考2次即可通過考試,假設A學生為這家補習班的學員,請問A學生不砸這家補習班招牌的機率為何?
f(1) + f(2) = 0.6*0.4¹⁻¹ + 0.6*0.4²⁻¹ = 0.6+ 0.24 = 0.84
對於Poisson分配近似二項分配採用時機 :
(1) 二項分配 Bin(n,p) , 當 n→∞ , p ≤ 0.01 時採用 Poisson分配
(2) 期望值 E(X) = λ = np
變異數 V(X) = λ