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 Post subject: GRE Probability (Select One)
PostPosted: Mon Sep 20, 2010 9:06 am 
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A wooden cube whose edge length is 10 inches is composed of smaller cubes with edge lengths of one inch. The outside surface of the large cube is painted red and then it is split up into its smaller cubes. If one cube is randomly selected from the small cubes, what is the probability that the cube will have AT LEAST one red face?
A. 0.360
B. 0.488
C. 0.500
D. 0.512
E. 0.600

(B) The large cube is composed of 10 small cubes in each dimension, so it is composed of 1000 small cubes in all because the volume of a cube is:
Volume = length × width × height = 10 × 10 × 10 = 1000.

The cubes on the surface will have at least one face painted, but the inside cubes will not. It is difficult to directly solve for the number of small cubes along the surface, so we will calculate the number of cubes on the inside (not painted), and subtract this number from the total number of cubes, to get the number of cubes that are painted.

The number of small cubes that are on the inside (not painted) is:
8 × 8 × 8 = 512.
This is because the first and last cube in each dimension will be painted, meaning that there are rows of 8 unpainted cubes in each dimension.

Then, the number of painted cubes is:
1000 – 512 = 488.
Therefore, the probability of selecting a cube with at least one face painted is:
488/1000 = 0.488

The correct answer is choice (B).

Note: If you chose choice (A), you were probably working with a 2-dimensional figure instead of a cube (3-dimensional).
---------
I can't imagine this problem in 3D. Could you help, please?


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 Post subject: Re: math: 3D geometry, probability.
PostPosted: Tue Sep 21, 2010 5:42 am 
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Posts: 498
At the beginning we have a 10 × 10 cube which sides are painted red:
Image

We want to count number of unit cubes that are NOT painted (it's easier). For that we take off outer layer of painted unit cubes:
Image

As the result we get inner 8 × 8 cube which is made of unit cubes that are NOT painted:
Image


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 Post subject: Re: math: 3D geometry, probability.
PostPosted: Mon Dec 06, 2010 1:26 pm 
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Posts: 424
Why will the number of cubes painted at least on one side ie the cubes on the outside not be 600, 10 × 10 on each side × 6 sides?


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 Post subject: Re: math: 3D geometry, probability.
PostPosted: Mon Dec 06, 2010 1:41 pm 
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Joined: Sun May 30, 2010 2:23 am
Posts: 498
questioner wrote:
Why will the number of cubes painted at least on one side ie the cubes on the outside not be 600, 10 × 10 on each side × 6 sides?

In this case the corner cubes (marked green) will be counted 3 times (because each one belongs to three sides) and the edge cubes (marked blue) will be counted 2 times (because each one belongs to two sides).

Image

Considering that there are 8 corner cubes and 8 × 12 = 96 edge cubes, the correct counting of the painted cubes is:

(100 × 6) – (8 × 2) – (96 × 1) = 488

It's really easy to make a mistake counting the number of painted cubes, while counting the number of unpainted cubes is much simpler: 8 × 8 × 8 = 512.


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