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解构魔方

Deconstructed Rubik's Cube

专题
Probability / 概率
难度
L3

题目详情

你有一个标准的魔方，你把它分成 27 个立方体。你把这 27 个立方体放进一个袋子里，然后随机拿出一个并扔到空中。立方体落地时，你只能看到 5 个未涂漆的面。你拉出一个只有一面有颜色的立方体的概率是多少？

You have a standard Rubik's Cube and you break it apart into 27 cubes. You put those 27 cubes into a bag and you randomly pull one out and toss it in the air. The cube lands such that you can only see 5 non-painted sides. What is the probability that you pulled a cube with one colored side?

解析

这是一个条件概率问题，其中条件是落地立方体显示 55 非彩色面。只有两种类型的立方体可以满足此条件：没有彩色边的中心立方体和每个面中心的六个无边立方体，只有 11 有色边。具有 11 彩色面的立方体必须面朝下落地才能显示 55 非彩色面，因此该条件的概率为： 127+627⋅16\frac1{27}+\frac6{27}\cdot\frac16 在给定的条件下，我们正在寻找落地的立方体是带有 11 彩色面的立方体的概率，因此发生这种情况的概率为： 627⋅16127+627⋅16=12\frac{\frac6{27}\cdot\frac16}{\frac1{27}+\frac6{27}\cdot\frac16} = \boxed{\frac12}

Original Explanation

This is a conditional probability problem where the condition is the landed cube shows 55 non-colored sides. There are only two types of cubes that can satisfy this condition, the center cube with no colored sides and the six non-edge cubes on the center of each face with just 11 colored side. The cubes with 11 colored side must land face down to show 55 non-colored sides, thus the probability of the condition is:

127+627⋅16\frac1{27}+\frac6{27}\cdot\frac16

We are looking for the probability the cube that landed is one with 11 colored side given the condition so the probability of this occuring is:

627⋅16127+627⋅16=12\frac{\frac6{27}\cdot\frac16}{\frac1{27}+\frac6{27}\cdot\frac16} = \boxed{\frac12}