2021 AMC 12B Spring 第 23 题

Question

三个球被随机且相互独立地投入编号为正整数的箱子中。对每个球，它被投入箱子

i

的概率为

2^{-i}

，其中

i=1

，

2

，

3

，

\ldots

。每个箱子可以有多个球。三个球最终落在不同且等间距的箱子中的概率为

\dfrac{p}{q}

，其中

p

和

q

是互质正整数。（例如，若球被投入箱子

3

、

17

和

10

，则这些箱子是等间距的。）

p+q

是多少？

Three balls are randomly and independently tossed into bins numbered with the positive integers so that for each ball, the probability that it is tossed into bin

i

is

2^{-i}

for

i=1,

2,

3,

\ldots.

More than one ball is allowed in each bin. The probability that the balls end up evenly spaced in distinct bins is

\dfrac{p}{q},

where

p

and

q

are relatively prime positive integers. (For example, the balls are evenly spaced if they are tossed into bins

3,

17,

and

10.

) What is

p+q?

55

56

57

58

59

答案：A

知识点：基本概率等比数列等差数列

难度评级：2390

小提示：

等间距表示箱子编号为

n, n+d, n+2d

，其中

n,d\ge 1

；它们的编号和为

3(n+d)

Evenly spaced means the bins are

n, n+d, n+2d

for some

n,d\ge 1;

their labels sum to

3(n+d)

大提示：

每种这样的有序投放概率为

2^{-3(n+d)}

，三个球有

3!

种排列方式

Each such ordered assignment has probability

2^{-3(n+d)},

and there are

3!

orderings of the three balls

解答：

不同且等间距的箱子形成等差数列

n,n+d,n+2d

，其中

n,d\ge 1

。三个编号之和为

3(n+d)

，因此把球固定分配到这三个箱子的概率为

2^{-3(n+d)}

。

三个球可以按

3!=6

种方式对应到这些箱子，所以总概率为

6\sum_{n\ge 1}\sum_{d\ge 1}2^{-3(n+d)}

=6\left(\sum_{n\ge 1}\tfrac{1}{8^n}\right)^2

=6\cdot\tfrac17\cdot\tfrac17=\tfrac{6}{49}

。

因为

\gcd(6,49)=1

，得

p+q=6+49=55

。

所以正确答案是 A。

Evenly spaced distinct bins form an arithmetic progression

n,n+d,n+2d

with

n,d\ge 1.

The three labels sum to

3(n+d),

so a fixed assignment of balls to these bins has probability

2^{-3(n+d)}.

The three balls can be ordered in

3!=6

ways, so the total probability is

6\sum_{n\ge 1}\sum_{d\ge 1}2^{-3(n+d)}

=6\left(\sum_{n\ge 1}\tfrac{1}{8^n}\right)^2

=6\cdot\tfrac17\cdot\tfrac17=\tfrac{6}{49}.

Since

\gcd(6,49)=1,

we get

p+q=6+49=55.

Thus, the correct answer is A.

2021 AMC 12B Spring 第 23 题

答案：A

难度评级：2390

小提示：

大提示：

解答：

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