2024 AMC 10A 第 24 题

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24.

一只蜜蜂在三维空间中移动。掷一个公平六面骰,其面标为 A+,A,B+,B,C+A^+, A^-, B^+, B^-, C^+CC^-。若蜜蜂位于点 (a,b,c)(a, b, c),骰子显示 A+A^+ 时蜜蜂移动到 (a+1,b,c)(a + 1, b, c),显示 AA^- 时移动到 (a1,b,c)(a - 1, b, c)。其余四种结果作类似移动。

假设蜜蜂从 (0,0,0)(0, 0, 0) 出发,骰子掷四次。蜜蜂走过某个单位立方体的四条不同边的概率是多少?

A bee is moving in three-dimensional space. A fair six-sided die with faces labeled A+,A,B+,B,C+,A^+, A^-, B^+, B^-, C^+, and CC^- is rolled. Suppose the bee occupies the point (a,b,c).(a, b, c). If the die shows A+,A^+, then the bee moves to the point (a+1,b,c),(a + 1, b, c), and if the die shows A,A^-, then the bee moves to the point (a1,b,c).(a - 1, b, c). Analogous moves are made with the other four outcomes.

Suppose the bee starts at the point (0,0,0)(0, 0, 0) and the die is rolled four times. What is the probability that the bee traverses four distinct edges of some unit cube?

154\dfrac{1}{54}

754\dfrac{7}{54}

16\dfrac{1}{6}

518\dfrac{5}{18}

25\dfrac{2}{5}

答案:B
知识点:基本概率正方体分类讨论
难度评级:2380
解答:

每次掷骰都会让蜜蜂沿 ±x,±y\pm x,\pm y±z\pm z 方向移动一个单位,所以共有 64=12966^4=1296 个等可能移动序列。有效路径分成两类。第一类走过某个正方形面的四条边:坐标方向对有 33 种选择,两个方向的正负号有 44 种选择,第一个方向有 22 种选择,共 2424 条路径。第二类用到三个坐标方向,其中一个方向使用两次。重复方向有 33 种选择;它在四步方向序列中的两个位置不能相邻,这两个位置有 33 种选法,其余两个方向的顺序有 22 种。对每个这样的方向序列,恰有 23=82^3=8 种正负号选择能使四条不同的边位于同一个立方体上。因此第二类有 3328=1443\cdot3\cdot2\cdot8=144 条路径。总共有 24+144=16824+144=168 个有利序列,所以概率为 168/1296=7/54168/1296=7/54,正确答案是 B

Every roll moves the bee one unit along ±x,±y,\pm x,\pm y, or ±z,\pm z, so there are 64=12966^4=1296 equally likely sequences. There are two types of valid paths. A path around one square face has 33 choices of coordinate plane, 44 choices for the signs of its two axes, and 22 choices for which axis is used first, giving 24.24. Otherwise all three coordinate directions are used, with one repeated: choose that axis in 33 ways, choose its two nonadjacent positions in 33 ways, order the other two axes in 22 ways, and choose their three initial signs in 23=82^3=8 ways. (The second step on the repeated axis must have the opposite sign.) This gives 3328=144.3\cdot3\cdot2\cdot8=144. Hence there are 24+144=16824+144=168 favorable sequences, and the probability is 168/1296=7/54.168/1296=7/54. Therefore, the answer is B.

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