Chapter 15, Problem 46CR

a.

To determine

To find: the probability of exactly ten of the twelve faces will be unpainted.

Answer to Problem 46CR

  $\frac{13}{56}$ .

Explanation of Solution

Given:

Sixty-four cubes $1\times 1\times 1$ are stacked together to form $4\times 4\times 4$ cube.

The larger cube is painted and then broke up into original 64 cubes.

Calculation:

According to the question,

8 cubes have 3 face painted.

24 cubes have 2 face painted.

24 cubes have 1 face painted.

8 cubes have 0 face painted.

Now, to find the probability of 2 small cubes are randomly selected,

The probability is $\frac{24}{64}\times \frac{8}{63}+\frac{8}{64}\times \frac{24}{63}+\frac{24}{64}\times \frac{24}{63}=\frac{13}{56}$ probability that exacts 10 faces will be unpainted.

b.

To determine

To find: the probability of at least ten of the twelve faces will be unpainted.

Answer to Problem 46CR

  $\frac{43}{126}$ .

Explanation of Solution

Given:

Sixty-four cubes $1\times 1\times 1$ are stacked together to form $4\times 4\times 4$ cube.

The larger cube is painted and then broke up into original 64 cubes.

Calculation:

According to the question,

8 cubes have 3 face painted.

24 cubes have 2 face painted.

24 cubes have 1 face painted.

8 cubes have 0 face painted.

Now, to find the probability of 0,1, or 2 small cubes are randomly selected,

The probability is $\frac{8}{64}\times \frac{7}{63}+\frac{24}{64}\times \frac{8}{63}+\frac{8}{64}\times \frac{24}{63}+\frac{13}{56}=\frac{43}{126}$ probability that exacts 10 faces will be unpainted.