红色魔方

这个问题更多的是对理解力的测试，而不是对数学能力或成熟度的测试。关键是要认识到我们得到了一条信息：所讨论的立方体有 5 个可见的白色面。第六个脸是面朝下躺着的，因此我们无法知道它是白色还是红色。这意味着所讨论的立方体可以是中心立方体 $or$ 它可以是六个立方体之一，这些立方体恰好有一个红色面 $and$，红色面恰好面朝下躺着，留下 5 个可见的白色面。

理解给定信息的含义对于解决问题至关重要。给定这条信息，我们需要计算所讨论的立方体是来自中心的立方体的概率。那个有 6 张白色面的人。由于我们被要求计算给定一些信息的事件的概率，我们可以使用贝叶斯规则：

A := 我们看到的立方体是中心立方体

B := 所讨论的立方体有 5 个可见的白色面 $$P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A)}{P(B)}$$ 第二个等式来自这样一个事实：如果所讨论的立方体是中心立方体，那么无论其在地面上的方向如何，都保证有 5 个可见的白色面。结果，概率降低为 $P(A)$。

$P(A)$ 等于我们看到中心立方体的概率。由于有 27 个立方体，因此该概率为 $\frac{1}{27}$。

$P(B)$ 是地面上的立方体有 5 个可见白色面的概率。总共只有 7 个立方体可以做到这一点（单个中心立方体有 6 个白色面，每个面中心的 6 个立方体只有 1 个红色面）。 $P(B)$ 是这两个不相交集合的集体和。有 6 个具有 1 个红色面的立方体，每个出现的概率为 $\frac{1}{27}$，每个红面朝下的概率为 $\frac{1}{6}$，使它们有效地显示 5 个白色面。相乘，我们得到： $$6*\frac{1}{27}*\frac{1}{6} = \frac{1}{27}$$ 中心立方体出现并处于有效方向的概率就是 $\frac{1}{27}$，因为它会在六个方向中的任何一个方向上正确显示 5 个白面。

将这两个值加在一起我们得到 $P(B) = \frac{1}{27} + \frac{1}{27} = \frac{2}{27}$

将其代入原始方程，我们得到： $$P(A | B) = \frac{P(A)}{P(B)} = \frac{\frac{1}{27}}{\frac{2}{27}} = \boxed{\frac{1}{2}}$$

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Original Explanation

This question is more of a test of comprehension than a test of mathematical ability or maturity. It's key to realize that we are given a piece of information: that the cube in question has 5 visible white faces. The sixth face is lying face down, and as a result we cannot know whether it is white or red. This implies that the cube in question can either be the center cube, $or$ it could be one of the six cubes which have exactly one red face $and$ that red face happens to be lying face down, leaving 5 visible white faces.

Comprehending the implication of the given information is essential to solving the problem. Given this piece of information, we are asked to compute the probability that the cube in question is the one from the center ie. the one with 6 white faces. Since we are asked to compute the probability of an event given some information, we can use Bayes' Rule:

A := cube we see is center cube

B := cube in question has 5 visible white faces

$$P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A)}{P(B)}$$ The second equality comes from the fact that if the cube in question is the center one, then it is guaranteed to have 5 visible white faces showing no matter its orientation on the ground. As a result, the probability reduces to $P(A)$.

The $P(A)$ is equal to the probability that we see the center cube. Since there are 27 cubes, this probability is $\frac{1}{27}$.

The $P(B)$ is the probability that the cube on the ground has 5 visible white faces. There are only 7 total cubes for which this is possible (the single center cube with 6 white faces, and the 6 cubes in the center of each face, with only 1 red face). $P(B)$ is the collective sum of these two disjoint sets. There are 6 cubes with 1 red face, each occurring with probability $\frac{1}{27}$, and each with probability $\frac{1}{6}$ of being red face down, making them validly show 5 white faces. Multiplying through, we get: $$6*\frac{1}{27}*\frac{1}{6} = \frac{1}{27}$$ The probability of the center cube occurring and being in a valid orientation is simply $\frac{1}{27}$ because it will properly show 5 white face in any of it's six orientations.

Adding these two value together we get our $P(B) = \frac{1}{27} + \frac{1}{27} = \frac{2}{27}$

Plugging this back into our original equation, we get: $$P(A | B) = \frac{P(A)}{P(B)} = \frac{\frac{1}{27}}{\frac{2}{27}} = \boxed{\frac{1}{2}}$$