Totally probability rule

[匿名]

2011-03-31 1:04 am

Independent trials,consisting of rolling a pair of fair dice, are performed. What is theprobability that an outcome of 5 appears before an outcome of 7 when theoutcome of a roll is the sum of the dice? Ans is 0.4 but why? Thanks.

回答 (1)

用戶2053

2011-03-31 1:52 am

✔ 最佳答案

Total outcome result = 6 x 6 = 36 cases.P(Outcome 5) = P( 1+4 , 2+3 , 3+2 , 4+1) = 4/36 = 1/9P(Outcome 7) = P( 1+6 , 2+5 , 3+4 , 4+3 , 5+2 , 6+1) = 6/36 = 1/6P(Outcome is not 5 nor 7) = 1 - 1/9 - 1/6 = 13/18SoP(An outcome of 5 appears before an outcome of 7)= P(The 1st outcome = 5)
+ P(The 1st outcome is not 5 nor 7 and the 2nd outcome is 5)
+ P(The first 2 times outcome is not 5 nor 7 and the 3rd outcome is 5)
+ P(The first 3 times outcome is not 5 nor 7 and the 4th outcome is 5)
+ ...........................................................= 1/9
+ (13/18)(1/9)
+ (13/18)²(1/9)
+ (13/18)³(1/9)
+ .............................= (1/9) / (1 - 13/18)= 2/5 = 0.4
Alternatively :P(E) = (1/9) / (1/9 + 1/6) = 0.4

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