1a) A bag contains 4 red counters and 6 green counters. Four counters are drawn at random from the bag, without replacement. Calculate the probability that
i) All the counters are green;
ii) at least one counter of each colour is drawn;
iii) at least two green counters are drawn; given that at least one of each colour is drawn.
b) Are the events "at least two green counters are drawn" and "at least one counter of each colour is drawn" independent? Explain briefly.
2) A coin and a six-faced die are thrown simultaneously. The random variable X is defined as follows:
"If the coin shows a head, then X is the score on the die. If the coin shows tail, then X is twice the score on the die."
a) Find the expected value of X.
b) Find the variance of X
If the experiment is repeated and the sum of the two values obtained for X is denoted b Y.
c) Find P (Y=4) and E(Y)
唔該哂:-(
有冇人讀過stats, 有野想問..:(
2010-04-15 2:04 am
回答 (1)
2010-04-16 5:49 pm
✔ 最佳答案
X
Pr(X)
X * Pr(X)
X2 * Pr(X)
1
0.0833
0.0833
0.0833
2
0.1667
0.3333
0.6667
3
0.0833
0.2500
0.7500
4
0.1667
0.6667
2.6667
5
0.0833
0.4167
2.0833
6
0.1667
1.0000
6.0000
8
0.0833
0.6667
5.3333
10
0.0833
0.8333
8.3333
12
0.0833
1.0000
12.0000
S
5.2500
37.9167
Y
Pr(Y)
Y * Pr(Y)
2
0.0069
0.0139
3
0.0278
0.0833
4
0.0417
0.1667
5
0.0556
0.2778
6
0.0764
0.4583
7
0.0833
0.5833
8
0.0972
0.7778
9
0.0694
0.6250
10
0.0903
0.9028
11
0.0556
0.6111
12
0.0833
1.0000
13
0.0417
0.5417
14
0.0833
1.1667
15
0.0278
0.4167
16
0.0625
1.0000
17
0.0139
0.2361
18
0.0417
0.7500
20
0.0208
0.4167
22
0.0139
0.3056
24
0.0069
0.1667
S
10.5000
2010-04-16 09:50:07 補充：
1ai
Pr = 6/10 x 5/9 x 4/8 x 3/7 = 1/14 = 0.0714
2010-04-16 09:50:18 補充：
1aii
Pr = 1 – Pr(all green) – Pr(all red) = 1 – 1/14 – (4/10 x 3/9 x 2/8 x 1/7) = 1 – 1/14 – 1/210 = 97/105 = 0.9238
2010-04-16 09:50:21 補充：
Pr(at least 2 green | at least 1 green) = Pr(at least 2 green) / Pr(at least 1 green) = [1 – Pr(all red) – Pr(1 red)] / [1 – Pr(all red)] = [1 – 1/210 – (4/10 x 6/9 x 5/8 x 4/7) x 4] / (1 – 1/210) = (1 – 1/210 – 2/21) / (1 – 1/210) = (9/10) / (209/210) = 189/209 = 0.9043
2010-04-16 09:50:34 補充：
2a: E(X) = 5.25
2b: Var(X) = E(X2) – E2(X) = 37.9167 – 5.252 = 10.3542
2010-04-16 09:50:48 補充：
X1 X2 Y Pr(Y)
1 3 4 0.0069
2 2 4 0.0278
3 1 4 0.0069
0.0417
Pr(Y=4) = 0.0417
2010-04-16 09:51:02 補充：
E(Y) = 10.5
收錄日期: 2021-04-23 18:39:03
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100414000051KK00925