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

Total number of available cubes to choose from = 64
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

sanjoy18 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

64 = 4^3
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

Cheers

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

there are only 24 cubes out of 64 that have only two sided painted. ---> c(24,2)/c(64,2)--> A