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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?

 Apr 23, 2024
 #1
avatar+128774 
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6 cubes have 1 black face

12 have 2 black faces

8 have 3 black faces

1 has no black faces (in the center)

 

Probability of  black face  =

 

(6/27) ( 1/6) + (12/27)(2/6) + ( 8/27) (3/6)  + (1/27) (0/6)   = 1/3

 

(The first set of parentheses contains the probability of that type of cube is chosen; the second set of parentheses contains the probability of having a black side facing up; mult/add these together for the total probability

 

cool cool cool

 Apr 23, 2024
edited by CPhill  Apr 23, 2024

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