1993 AMC 12 第 24 题

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

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

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

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.

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