+0  
 
0
44
1
avatar+1533 

The six faces of a cube are painted black.  The cube is then cut into $3^3$ smaller cubes, all the same size.

 

(a) How many of the smaller cubes have exactly one black face?

(b)  How many of the smaller cubes do not have any black faces?

(c) One of the small cubes is chosen at random, and rolled.  What is the probability that when it lands, the face on the top is black?

 Jan 11, 2024
 #1
avatar+37159 
+1

If you start with a  3 x 3 x 3   cube   , when you cut it up there will be 

ONE cube on each face with only one painted side = six cubes with one painted side

Only the very center cube will have no painted sides 

EIGHT cubes will have THREE painted sides

TWELVE cubes will have TWO painted sides

there are 27 cubes, randomly drawing a cube:   

 

of the cubes you have a   6/27   chance of picking a 1 painted side                   prob of rolling paint side up = 1/6

                                          1 / 27                                zero painted side                                                          = 0/6

                                             12/27                             two painted sides                                                          = 2/6

                                               8/27                             three painted sides                                                       = 3/6 

 

 

The the total probability of a painted side up is 

             6/27 * 1/6    + 1/27 * 0/6     +   12/27 * 2/6      +   8/27 * 3/6    =   1/3 chance of painted side up 

 

 

Maybe a bit simpler : there are a total of 54 painted sides out of  162  smaller-cube sides    black =  54/162 = 1/3 

 Jan 11, 2024

1 Online Users

avatar