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：

其他年份的第 16 题