隨機變量和離散概率分佈問題

1(a) 2k(1) + 2k(2) + 2k(3) + 2k(4) = 1
2k+4k+6k+8k = 1
20k = 1
k = 1/20

1(b) p(4<3X+1<11)
=p(3<3X<10)
=p(1<X<10/3)
=p(X=2 or 3)
=p(X+2) + p(X+3)
=2k(2) + 2k(3)
=4k + 6k
=10k
=10(1/20)
=0.5

2. R=紅, Y=黃, B=藍
total balls=2R+3Y+5B
R=1分
Y=2分
B=3分
現從袋中隨意抽出2個球，而每次均把球放回。
可能結果:
RR(2分), RY(3分), RB(4分),
YR(3分), YY(4分), YB(5分),
BR(4分), BY(5分), BB(6分)

得分的期望值( E(X) )
=2(2/10)^2 + 3(2/10)(3/10) + 4(2/10)(5/10)
+3(3/10)(2/10) + 4(3/10)(3/10) + 5(3/10)(5/10)
+4(5/10)(2/10) + 5(5/10)(3/10) + 6(5/10)(5/10)
=(2/10)(4/10 + 9/10 + 2) + (3/10)(6/10 + 12/10 + 25/10)
+(5/10)(8/10 + 15/10 + 3)
=0.66+1.29+2.65
=4.6分

E(X^2)=(分數^2)*(概率), E(X)=得分的期望值
得分的方差
=E(X^2) - [E(X)]^2
=(2^2)(2/10)^2 + (3^2)(2/10)(3/10) + (4^2)(2/10)(5/10)
+(3^2)(3/10)(2/10) + (4^2)(3/10)(3/10) + (5^2)(3/10)(5/10)
+(4^2)(5/10)(2/10) + (5^2)(5/10)(3/10) + (6^2)(5/10)(5/10) - 4.6^2
=2.3+5.73+14.35-21.16
=1.22平方分 (留意單位)

3(a) 2k+(0.2-k)+0.1+0.1+0.2=1
k+0.6=1
k=0.4
但是p(Y=2)=(0.2-0.4)= -0.2
(正常來說,概率不會是負數,是不是打錯了?)
第3條問題的其餘部分可能都有問題,所以先不回答

4(a)
p(D=1) = (6C1)(3/20)(17/19)(16/18)(15/17)(14/16)(13/15) = 91/190
p(D=2) = (6C2)(3/20)(2/19)(17/18)(16/17)(15/16)(14/15) = 7/570
p(D=3) = (6C3)(3/20)(2/19)(1/18) = 1/1140

4(b) D的期望 = 1(91/190) + 2(7/570) + 3(1/1140) = 577/1140部
D的標準差 = D的方差的平方根
=[(1^2)(91/190) + (2^2)(7/570) + (3^2)(1/1140) - (577/1140)^2]的平方根
=0.5289平方部(4位小數)

5(a) 平均晴天的天數=30(0.7)=21天

5(b) 有不多於5個雨天的概率
=(30C5)(0.3^5)(0.7^25)+(30C4)(0.3^4)(0.7^26)+(30C3)(0.3^3)(0.7^27)
+(30C2)(0.3^2)(0.7^28)+(30C1)(0.3)(0.7^29)+(0.7^30)
=0.0803(4位小數)

5(c) 晴天and雨天的日數相等的概率
=晴天and雨天的日數都是15天的概率
=(30C15)(0.3^15)(0.7^15)=0.0106(4位小數)

5(d) 當天是雨天的概率(已知麗紗曾於四月一日出 外購物)
= (0.4)(0.3) / [(0.4)(0.3) + (0.8)(0.7)]
=3/17

6 設最少發射T-101導彈的數目是X
X [(1-0.4)^(X-1)](0.4) = 0.9
X (0.6)^(X-1) = 2.25
log(底數0.6) [X (0.6)^(X-1)] = log(底數0.6) 2.25
log(底數0.6) X + log(底數0.6) (0.6) = log(底數0.6) 2.25
(log X) / (log0.6) + 1 = (log 2.25) / (log0.6)
X = 3.75
最少發射T-101導彈的數目是4枚