2024 AMC 12B 第 25 题

Question

Pablo 要用红色或蓝色颜料、条纹或圆点图案装饰

6

个相同的白球。他会对每个球的颜色和图案各掷一次公平硬币，共

12

次决策。颜料干后，他把

6

个球放入一个盒子。Frida 从盒子中随机取一个球，并记录其颜色和图案。事件“Frida 取到的球是红色”和“Frida 取到的球是条纹”是否独立，取决于 Pablo 的掷硬币结果。这两个事件独立的概率可写为

\dfrac{m}{n}

，其中

m

和

n

是互质正整数。

m

是多少？（回忆：若两个事件

A

和

B

满足

P(A \text{ 且 } B) = P(A)\cdot P(B)

，则它们独立。）

Pablo will decorate each of

6

identical white balls with either a striped or a dotted pattern, using either red or blue paint. He will decide on the color and pattern for each ball by flipping a fair coin for each of the

12

decisions he must make. After the paint dries, he will place the

6

balls in an urn. Frida will randomly select one ball from the urn and note its color and pattern. The events “the ball Frida selects is red” and “the ball Frida selects is striped” may or may not be independent, depending on the outcome of Pablo’s coin flips. The probability that these two events are independent can be written as

\dfrac{m}{n},

where

m

and

n

are relatively prime positive integers. What is

m?

(Recall that two events

A

and

B

are independent if

P(A \text{ and } B) = P(A)\cdot P(B).

)

243

245

247

249

251

答案：A

知识点：独立事件分类讨论

难度评级：2510

小提示：

设六个球中有

k

个红色条纹球，红球总数为

R

，条纹球总数为

S

。均匀抽取时独立条件为

\dfrac{k}{6} = \dfrac{R}{6}\cdot\dfrac{S}{6}

。

Let the six balls include

k

that are red-and-striped, with

R

red and

S

striped total. Independence for Frida’s uniform pick means

\dfrac{k}{6} = \dfrac{R}{6}\cdot\dfrac{S}{6}

大提示：

条件即

6k = RS

。在

4^6

个总分配中，用多项式系数计数满足条件的

6

个球到

4

种类型的分配。

So the condition is

6k = RS.

Count assignments of

6

balls to the

4

equally likely types satisfying this, over

4^6

total, using multinomial coefficients

解答：

每个球独立地属于四种等可能类型之一：红色条纹、红色圆点、蓝色条纹、蓝色圆点。设

6

个球中有

k

个红色条纹球，红球总数为

R

，条纹球总数为

S

。对 Frida 的均匀随机选择，

P(\text{红}) = \tfrac{R}{6}

，

P(\text{条纹}) = \tfrac{S}{6}

，且

P(\text{红且条纹}) = \tfrac{k}{6}

。独立性意味着

\dfrac{k}{6} = \dfrac{R}{6}\cdot\dfrac{S}{6}

，即

6k = RS

。

先计数满足

R\in\{0,6\}

或

S\in\{0,6\}

的分配。这四个条件各给另一种属性留下

2^6=64

种选择，而它们两两交集中的四个分配被重复计算。因此并集贡献

4\cdot64-4=252

。

当

1\le R,S\le5

时，

RS

是

6

的倍数这一条件只留下

(R,S)=(2,3),(3,2)

或

(3,4),(4,3)

。每种情形都有

k=\frac{RS}{6}

，四种类型的数量恰是

(1,1,2,2)

的一个排列。因此每个数对贡献

\dfrac{6!}{1!1!2!2!}=180

种分配。有利分配数为

252+4\cdot180=972

。在

4^6=4096

个等可能分配中，概率为

\dfrac{972}{4096}=\dfrac{243}{1024}

，所以

m=243

。

因此，正确答案是 A。

Each ball is independently one of four equally likely types: red-striped, red-dotted, blue-striped, blue-dotted. Suppose among the

6

balls there are

k

red-striped, with

R

red and

S

striped in total. For Frida’s uniform pick,

P(\text{red}) = \tfrac{R}{6},

P(\text{striped}) = \tfrac{S}{6},

and

P(\text{red and striped}) = \tfrac{k}{6}.

Independence means

\dfrac{k}{6} = \dfrac{R}{6}\cdot\dfrac{S}{6},

i.e.

6k = RS.

First count assignments with

R\in\{0,6\}

or

S\in\{0,6\}.

Each of these four conditions leaves

2^6=64

choices for the other attribute, and the four assignments at their pairwise intersections have been counted twice. Their union therefore contributes

4\cdot64-4=252.

For

1\le R,S\le5,

the condition that

RS

be divisible by

6

leaves only

(R,S)=(2,3),(3,2)

or

(3,4),(4,3).

In each case

k=\frac{RS}{6},

and the four type counts are a permutation of

(1,1,2,2).

Thus each pair contributes

\dfrac{6!}{1!1!2!2!}=180

assignments. The favorable count is therefore

252+4\cdot180=972.

Out of

4^6=4096

equally likely assignments, the probability is

\dfrac{972}{4096}=\dfrac{243}{1024},

so

m=243.

Thus, the correct answer is A.

2024 AMC 12B 第 25 题

答案：A

难度评级：2510

小提示：

大提示：

解答：

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