Question

A large cube is painted on all six faces. It is then divided into 27 smaller, congruent cubes.

a) Use a table and a histogram to show the probability distribution for the number of painted faces on a randomly selected cube.

b) If you select 10 of the cubes at the same time, what is the probability that at least half of them will have two painted faces?

c) If you select a cube 10 times, with replacement, what is the probability that at least half of them will have two painted faces?

  d) Justify your choice of distributions in parts a), b), and c).

e) For each distribution, calculate the expectation and interpret its meaning.

Answer 1

There is 1 cube in the very center (middle of 3 in each axis direction -- pitch, roll, and yaw), so 1 cube has no paint.

On each of the 6 sides of the cube, there is a central smaller cube that is painted once.

Also on each of the 6 sides of the cube, there are 4 cubes (at the middle of the edges). These are shared with one other side of the cube (or we could just count the 12 edge lines -- 12). So, 6 * 4 / 2, or just 12, smaller cubes are painted on 2 edges.

That leaves the 8 corners of the original cube, which are painted on 3 surfaces.

1 (no paint) + 6 (painted 1) + 12 (painted 2) + 8 (painted 3) = 27 total

a)

b)
Y = number of two-painted faces

This is hypergeometric distribution

P(Y= y) = 12Cy * 15C(10-y) / 27C10

P(Y >= 5) = 0.4807

c)

Z = number of two painted faces

Z follow binomial with n = 10 and p = 0.4444

P(Z >= 5) =

d)

E(X) = 2

E(Y) = 10 * 12/27 = 4.4444

E(Z) = np = 10*0.4444 = 4.444