Cube Problem

[paritosh_b]

Not able to visualise it.Pls help me out

A large cube 10cm on a side is composed of smaller cubes 1 cm on a side. The large cube is printed red on its outside surfaces and then all the cubes are mixed in a drum. If one cube is selected randomly, which of the following is the approximate probability that the cube will have at least one red face?
OA:
[spoiler]1:2[/spoiler]

[Bharat]

Hi Paritosh,

The original cube comprises 1000 unique 1-unit-cubes of which 488 unique 1-unit-cubes are on its outer surface.
-> The drum has 1000 cubes, of which 488 are painted red on one more surfaces.
-> Probability of desired event = 488/1000 = 1/2 (approximately)

Explanation: If you peel the surface only cubes from a cube with an edge length of n units, you will be left with a cube of edge length of n-2 units

Count of surface only cubes = difference in volumes of the two cubes
= 10^3 - 8^3
= 488 unit-cubes

PS: Note that the 8 unit-cubes on vertices are painted on 3 surfaces, the remaining 104 unit cubes on the edge are painted on 2 surfaces

Let me know if you have further questions.

Regards,
Bharat.

[paritosh_b]

Got it
Thanks.

[goyalsau]

Explanation: If you peel the surface only cubes from a cube with an edge length of n units, you will be left with a cube of edge length of n-2 units AWESOME

I Never knew that we have formulas for these kind of questions as Well.

According to me,

64 *6 = 384 Cubes with one color FACE { Square on the face of a Cube with length 8 * 8 }

8 * 12 = 96 Cubes with two color on Its FACE { ON every Edge of Cube 8 Cube Excluding Corners of Cube }

8 Cubes with three color FACE { On every Cube 8 Corners are there }

384 + 96 + 8 = 488 Cubes with At least one FACE Colored.

PS: Note that the 8 unit-cubes on vertices are painted on 3 surfaces, the remaining 104 unit cubes on the edge are painted on 2 surfaces

I think you miscounted,

[Bharat]

Hi goyalsau, thanks for the response & noticing the error.

You are correct, my count has a typographical error. Below is correction with detail.
Total surface area of cube = 600 (=6*10*10) sq. units
8 vertex cubes are painted on 3 surfaces, hence each of these is counted thrice (total 3*8) in the surface area, so need to discount 2*8 for removing duplicates.
96 cubes are colored on two surfaces, hence each of these is counted twice (total 2*96) in the surface area, so need to discount 1*96 for these:

so the final answer is: 600 - 16 - 96 = 488.

Regards,
Bharat.

[junegmat221]

8 * 12 = 96 Cubes with two color on Its FACE { ON every Edge of Cube 8 Cube Excluding Corners of Cube }

@ Goyalsu,
I can only visually imagine this.
But it becomes harder when the cube is not 1000 in volume or if the smaller cubes are not 1cm in each of its side.
It is a little bit hard in exams when the units are not easily representable.
If anyone could visually represent the same data by means of formula with a diagram offcourse ,it should be really helpful.