1996 AMC 12 第 16 题

Question

将一枚公平的标准六面骰子掷三次。已知前两次点数之和等于第三次点数，求至少有一次掷出

2

的概率。

A fair standard six-sided dice is tossed three times. Given that the sum of the first two tosses equals the third, what is the probability that at least one

2

is tossed?

\frac16

\frac{91}{216}

\frac12

\frac8{15}

\frac7{12}

答案：D

知识点：条件概率 case counting

难度评级：1730

小提示：

在给定条件下，计算前两次点数之和不超过

6

的有序数对数量

Under the stated condition, count ordered pairs for the first two tosses whose sum is at most

6

大提示：

计算有利结果时，把第三次掷出

2

的情形与前两次中至少一次掷出

2

的情形分开

For the favorable count, separate a third-toss

2

from pairs having a

2

among the first two tosses

解答：

当第三次点数依次为

2,3,4,5,6

时，前两次点数的有序数对数量依次为

1,2,3,4,5

，所以共有

15

个等可能的条件结果。第三次掷出

2

对应

(1,1,2)

。第一次掷出

2

有

4

个结果，第二次掷出

2

也有

4

个结果，而

(2,2,4)

被重复计算。因此有利结果共有

1+4+4-1=8

个，正确答案是 D。

For third tosses

2,3,4,5,6,

the numbers of ordered first-two-toss pairs are

1,2,3,4,5,

for

15

equally likely conditional outcomes. A third toss of

2

contributes

(1,1,2).

A

2

in the first position gives

4

outcomes and a

2

in the second gives

4,

with

(2,2,4)

counted twice. Thus there are

1+4+4-1=8

favorable outcomes, and the correct answer is D.

1996 AMC 12 第 16 题

答案：D

难度评级：1730

小提示：

大提示：

解答：

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