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

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New post 29 Mar 2019, 01:39
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Each face of a cube is given a single narrow stripe painted from the center of one edge to the center of the opposite edge. The choice of the edge pairing is made at random and independently for each face. What is the probability that there is a continuous stripe encircling the cube?

(A) 1/8
(B) 3/16
(C) 1/4
(D) 3/8
(E) 1/2

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New post 29 Mar 2019, 09:00
Bunuel wrote:
Each face of a cube is given a single narrow stripe painted from the center of one edge to the center of the opposite edge. The choice of the edge pairing is made at random and independently for each face. What is the probability that there is a continuous stripe encircling the cube?

(A) 1/8
(B) 3/16
(C) 1/4
(D) 3/8
(E) 1/2


total pairs of 6 sided cube ; 3 and total faces = 8
3/8 is P of having stripe
IMO D

P.S I am not sure of answer here..
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New post 29 Mar 2019, 19:38
Bunuel wrote:
Each face of a cube is given a single narrow stripe painted from the center of one edge to the center of the opposite edge. The choice of the edge pairing is made at random and independently for each face. What is the probability that there is a continuous stripe encircling the cube?

(A) 1/8
(B) 3/16
(C) 1/4
(D) 3/8
(E) 1/2



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.

So, Probability for the strip to encircle the cube for this portion=\(\frac{1}{(2c1)*(4c1)}=\frac{1}{16}\)

There will be three such protions, in total the probability is 3/16.

IMO, Option B.
Attachments

File comment: Blue part is the portion considered, white part is not considered. 3 such parts will be there, removing each pair of opposite faces.
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New post 29 Mar 2019, 23:01
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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New post 30 Mar 2019, 19:17
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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New post 31 Mar 2019, 21:08
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..
In your previous post, the answer to the question,
Ypsychotic wrote:
So, Probability for the strip to encircle the cube for this portion \(\frac{1}{(2c1)∗(4c1)}\)=\(\frac{1}{16}\)


Shouldn't the answer be 1/8?
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