## quarta-feira, 5 de maio de 2021

### Matemática - 64 cubinhos

Aqui está um problema proposto por Robert Gebhardt na edição do outono de 1999 do Pi Mu Epsilon Journal. Todas as seis faces de um cubo 4 × × 4 são pintadas de vermelho. Em seguida, o cubo é picado em 64 cubos menores 1 × 1 × 1. As faces "internas" não são pintadas. Os 64 cubos pequenos são colocados em uma caixa, e um deles é sorteado ao acaso e jogado. Encontre a probabilidade de que, quando o cubinho chegar ao repouso, sua face superior seja vermelha.

(Here is a problem proposed by Robert Gebhardt in the Fall 1999 issue of Pi Mu Epsilon Journal. All six faces of a 4 × × 4 cube are painted red. Then the cube is chopped into 64 smaller 1 × 1 × 1 cubes. The "inside" faces are left unpainted. The 64 small cubes are put into a box and one is drawn at random, and tossed. Find the probability that when it comes to rest its upper face will be red.)

PCFilho

#### 5 comentários:

1. Since the original cube is painted only on the outside, it is impossible for any of the small cubes to have more than three faces painted. So, a randomly drawn small cube could have one, two, or three faces painted, or be unpainted.

The number of small cubes with three faces painted is equal to the number of vertices of the cube, which is eight. The number of small cubes with only two faces painted is equal to the number of edges times two. A cube has twelve edges, so twenty-four cubes have only two faces painted. The number of small cubes with only one face painted is equal to the number of faces times four (because 4 = 2²). Since a cube has six faces, there are also twenty-four small cubes with only one face painted. This accounts for fifty-six of the small cubes, so the remaining eight are unpainted. So, one-eighth of the small cubes have three faces painted, three-eighths have only two faces painted, three-eighths have only one face painted, and one-eighth are unpainted.

Now, if we draw a cube with three faces painted, the probability that it will land with a painted face on top is 3/6 = 1/2. If there are only two faces painted, this probability becomes 2/6 = 1/3. If there is only one face painted, this probability becomes 1/6, and if no face is painted, then it is impossible for it to land with a painted face on top. Therefore, the probability that a randomly-drawn cube lands with a painted face on top when rolled is (1/8)(1/2) + (3/8)(1/3) + (3/8)(1/6) + (1/8)(0) = 1/16 + 1/8 + 1/16 = 1/2.

1. Your only mistake was in the very last line.

1/16 + 1/8 + 1/16 = 1/4

2. My reasoning:

The total number of faces is 64*6.

The number of painted faces is 16*6.

The asked probability is (16*6)/(64*6) = 1/4.

1. Explaining it:

The total number of faces is 64*6, because there are 64 small cubes with 6 faces each.

The number of painted faces is 16*6, because the big cube has 6 faces with 16 small faces each.

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