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# Each face of a cube is given a single narrow stripe painted from the

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Each face of a cube is given a single narrow stripe painted from the [#permalink]
Ypsychotic wrote:
Let the cube be ABCDEFGH.

For each face, there will be two strips. Lets first consider the encircling part, for that remove opposite two faces, and consider the rest of the four faces. eg. consider ABCD, CDEF, EFGH, GHBA, and don't consider ABEF and CDGH. For this portion, there will be only one time, when the strip will encircle the cube. and for each face, there will be two strips.

Can you explain why there would be two strips per face?
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Re: Each face of a cube is given a single narrow stripe painted from the [#permalink]
OhsostudiousMJ wrote:
Ypsychotic wrote:
Let the cube be ABCDEFGH.

For each face, there will be two strips. Lets first consider the encircling part, for that remove opposite two faces, and consider the rest of the four faces. eg. consider ABCD, CDEF, EFGH, GHBA, and don't consider ABEF and CDGH. For this portion, there will be only one time, when the strip will encircle the cube. and for each face, there will be two strips.

Can you explain why there would be two strips per face?

Beause it is mentioned in the question that a strip is formed by joining it from the center of first edge to opposite edge. As there are four edges in a face, with two pairs opposite to each other. For instance, in the face abcd, One strip would from ab to cd, other from ad to bc. Hence, two strips.
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Each face of a cube is given a single narrow stripe painted from the [#permalink]
Ypsychotic wrote:

Beause it is mentioned in the question that a strip is formed by joining it from the center of first edge to opposite edge. As there are four edges in a face, with two pairs opposite to each other. For instance, in the face abcd, One strip would from ab to cd, other from ad to bc. Hence, two strips.

Okay, thanks, but I still got a question unanswered..
Ypsychotic wrote:
So, Probability for the strip to encircle the cube for this portion $$\frac{1}{(2c1)∗(4c1)}$$=$$\frac{1}{16}$$

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Re: Each face of a cube is given a single narrow stripe painted from the [#permalink]
OhsostudiousMJ wrote:
Ypsychotic wrote:

Beause it is mentioned in the question that a strip is formed by joining it from the center of first edge to opposite edge. As there are four edges in a face, with two pairs opposite to each other. For instance, in the face abcd, One strip would from ab to cd, other from ad to bc. Hence, two strips.

Okay, thanks, but I still got a question unanswered..
Ypsychotic wrote:
So, Probability for the strip to encircle the cube for this portion $$\frac{1}{(2c1)∗(4c1)}$$=$$\frac{1}{16}$$

You're right. I wrote it wrong. It should be $$\frac{1}{2c1*2c1*2c1*2c1}=\frac{1}{16}$$ because for every strip on one face there are two strips on another face. So, in total, there are 2*2*2*2=16 combinations with only one combination when it encircles it. So, only once when each strip will be the one in alignment, it will encircle the cube. So, Probability=$$\frac{1}{16}$$. Hope it helps.
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Re: Each face of a cube is given a single narrow stripe painted from the [#permalink]
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Re: Each face of a cube is given a single narrow stripe painted from the [#permalink]
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