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統計機率問題，請高手幫忙！拜託！拜託！
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. 1/4
b. 1/3
12 個解答
- 麻辣Lv 77 年前最佳解答
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)=29/648
b. 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
