a. a sum of 3 before we obtain a sum of 7?
b. a sum of 4 before we obtain a sum of 7?
更新1:
a. 1/4 b. 1/3
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2014-04-14 6:05 am
Suppose that two balanced dice are tossed repeatedly and the sum of the two uppermost faces is determined on each toss. What is the probability that we obtain
a. a sum of 3 before we obtain a sum of 7?
b. a sum of 4 before we obtain a sum of 7?
a. a sum of 3 before we obtain a sum of 7?
b. a sum of 4 before we obtain a sum of 7?
回答 (12)
2014-04-16 12:10 am
✔ 最佳答案
Suppose that two balanced dice are tossed repeatedly and the sum of the two uppermost faces is determined on each toss. What is the probability that we obtaina. A sum of 3 before we obtain a sum of 7?Ans:Sum(3)={1+2,2+1}=2 => P(3)=1/18Sum(7)={1+6,2+5,3+4,4+3,5+2,6+1}=6 => P(7)=7/36=> P(3 before 7)=(1-7/36)*(1/18)=29/648b. A sum of 4 before we obtain a sum of 7?Ans:Sum(4)={1+3,2+2,3+1}=3 => P(4)=1/12=> P(4 before 7)=(1-7/36)*(1/12)=29/432
2014-04-15 16:13:29 補充：
Revision for Problem (a):
Sum(7)={1+6,2+5,3+4,4+3,5+2,6+1}=6 => P(7)=1/6
=> P(3 before 7)=(1-1/6)*(1/18)=5/108
2014-04-15 16:14:23 補充：
Revision for Problem (b):
P(4 before 7)=(1-1/6)*(1/12)=5/72
2014-05-30 4:41 pm
2014-05-29 5:22 pm
2014-05-28 5:06 pm
2014-05-27 1:34 pm
2014-05-04 3:56 pm
2014-05-02 1:54 pm
2014-04-30 9:09 pm
2014-04-29 2:43 pm
2014-04-21 12:51 pm
2014-04-16 8:42 am
a. P(3) : P(7) = 2:6 = 1:3, so P(3 before 7) = 1/4
b. P(4) : P(7) = 3:6 = 1:2, so P(4 before 7) = 1/3
b. P(4) : P(7) = 3:6 = 1:2, so P(4 before 7) = 1/3
2014-04-14 6:16 pm
收錄日期: 2021-04-30 18:38:05
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https://hk.answers.yahoo.com/question/index?qid=20140413000016KK06509