Painted Cube

Probability Level 5

The outside of a wooden cube is painted blue. The cube is cut into 27 small cubies, arranged as in a Rubik's Cube. The 27 cubes are placed into a bag. You close your eyes, randomly select a cube and place it on a table.

You then open your eyes and see 5 unpainted faces. What is the probability that the last face (the one against the table) is painted blue?

This problem is not original.

Image Credit: Flickr Marco Braun .

\frac{1}{2}

\frac{1}{6}

\frac{1}{7}

\frac{6}{7}

2 solutions

Maggie Miller
Jul 17, 2015

There are seven cubies that have at least five unpainted sides. Six cubies (the ones at the centers of the faces of the original cube) have one painted side, and one of the cubies (the centermost) has no painted sides.

With respect to face placed against the table, there is one way each of the six cubies with one painted side might have been placed down (as the painted side must be against the table).

With respect to face placed against the table, there are six ways that the cubie with no painted side might have been placed down (as any side could be against the table).

Therefore, out of 12 equally likely situations, six feature a cube with one painted face. Thus, the probability is $\frac{6}{12}=\boxed{\frac{1}{2}}$ .

Taylor Shobe
Jul 17, 2015

27 = total cubit population
7 = candidate cubits
6 = desired cubits

a = probability of yielding a desired cubit = 6/27
b = probability of yielding the desired orientation of a desired cubit = 1/6
c = probability of yielding a non-desired cubit = 1/27
d = probability of yielding the desired orientation of of non-desired cubit = 6/6

P = (a · b) / [(a · b) + (c · d)]

=> (1/27) / (2/27)

=> 1/2

Painted Cube

This problem is not original.

Image Credit: Flickr Marco Braun .

2 solutions

0 pending reports