A cube is painted with red colour and then cut into 64 small identical cubes. If two cubes are picked
randomly from the heap of 64 cubes, what is the probability that both of them have exactly two faces
painted red?
(a)23/168
(b)47/84
(c)1/4
(d)31/63
(e)41/63
probability
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Total number of available cubes to choose from = 64A cube is painted with red colour and then cut into 64 small identical cubes. If two cubes are picked
randomly from the heap of 64 cubes, what is the probability that both of them have exactly two faces
painted red?
(a)23/168
(b)47/84
(c)1/4
(d)31/63
(e)41/63
A cube has 12 sides and on each side we will find 2 identical cubes whose only 2 sides have red paint.
Hence total number of cubes with only 2 sides painted red = 12*2= 24
Therefore probability for picking the 2 cubes with only 2 sides painted red = (24C2)/(64C2)= 23/168
Hence choose A
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64 = 4^3sanjoy18 wrote:A cube is painted with red colour and then cut into 64 small identical cubes. If two cubes are picked
randomly from the heap of 64 cubes, what is the probability that both of them have exactly two faces
painted red?
(a)23/168
(b)47/84
(c)1/4
(d)31/63
(e)41/63
So each face is divided into 4^2 squares.
The cubes at the edges have exactly 2 faces painted red.
Number of cubes per edge with 2 faces painted red = 4 - 2 = 2 (removing the cubes at the corner)
Number of edges = 12
Total number of cubes with 2 faces painted red = 2 * 12 = 24
Favorable cases:
Select 2 out of 24 cubes : 24C2
Sample space:
Select 2 out of 64 cubes : 64C2
Probability = 24C2/64C2 = (24*23)/(64*63) = 23/168
Choose A
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alt approach
let the side of big cube is A
cube divided in 64 inditical cubes
A^3/64 = I (Int)
the smallest value of A is 4 and the value of a(side of small cube is 1)
note:- the second large cube is of side 8 and small cube's side will be 2, the required probablity will be same is either cases
now consider cube with all faces painted, only cubes at edges will have two faces painted
these cube are 4*4 + 4*2 = 24
required probablity 24c2/64c2 = 23/168
let the side of big cube is A
cube divided in 64 inditical cubes
A^3/64 = I (Int)
the smallest value of A is 4 and the value of a(side of small cube is 1)
note:- the second large cube is of side 8 and small cube's side will be 2, the required probablity will be same is either cases
now consider cube with all faces painted, only cubes at edges will have two faces painted
these cube are 4*4 + 4*2 = 24
required probablity 24c2/64c2 = 23/168
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Hi All,
In these types of situations, it's often a good idea to draw a picture for reference. To be painted on "two faces", a piece would have to be on the "edge" of the big cube but NOT be a corner piece (because corner pieces would have paint on 3 faces).
You'll find that 24 pieces fit the "2 face" description.
Then the probability math is straight-forward:
(24/64)(23/63 = 23/168
GMAT assassins aren't born, they're made,
Rich
In these types of situations, it's often a good idea to draw a picture for reference. To be painted on "two faces", a piece would have to be on the "edge" of the big cube but NOT be a corner piece (because corner pieces would have paint on 3 faces).
You'll find that 24 pieces fit the "2 face" description.
Then the probability math is straight-forward:
(24/64)(23/63 = 23/168
GMAT assassins aren't born, they're made,
Rich