Question

一个盒子里有 $3$ 枚有光泽的便士和 $4$ 枚无光泽的便士。从盒中逐枚随机抽取便士，抽出后不放回。直到第三枚有光泽的便士出现需要抽取超过四次的概率为 $\frac{a}{b}$ 。若 $\frac{a}{b}$ 是最简分数，则 $a+b=$

A box contains $3$ shiny pennies and $4$ dull pennies. One by one, pennies are drawn at random from the box and not replaced. If the probability is $\frac{a}{b}$ that it will take more than four draws until the third shiny penny appears and $\frac{a}{b}$ is in lowest terms, then $a+b=$

$11$

$20$

$35$

$58$

$66$

答案：E

知识点：对立事件概率组合 random ordering

难度评级：1900

小提示：

将三枚有光泽的便士看作占据七个抽取位置中的三个。

View the three shiny pennies as occupying three of seven draw positions

大提示：

减去三枚有光泽的便士都出现在前四个位置的情形。

Subtract the cases in which all three shiny pennies occur among the first four positions

解答：

三枚有光泽的便士所占的 $3$ 元子集，是从 $7$ 个抽取位置中等可能选出的，共有 $\binom73=35$ 种可能。第三枚有光泽的便士不迟于第 $4$ 次抽取出现，当且仅当三个有光泽便士的位置都在前四个位置中，这有 $\binom43=4$ 种情形。所求概率为 $1-\frac4{35}=\frac{31}{35}\text{。}$ 因此 $a+b=31+35=66$ 。故正确答案是 E。

The shiny pennies occupy a uniformly chosen $3$ -element subset of the $7$ draw positions. There are $\binom73=35$ possibilities. The third shiny penny appears by draw $4$ exactly when all three shiny positions lie among the first four, which occurs in $\binom43=4$ cases. The requested probability is $1-\frac4{35}=\frac{31}{35}.$ Therefore $a+b=31+35=66.$ Thus the correct answer is E.

1993 AMC 12 第 24 题

答案：E

难度评级：1900

小提示：

大提示：

解答：

其他年份的第 24 题