# Math Help - Probability problem about cubes

1. ## Probability problem - cubes

Each face of a cube is painted either Red or Blue, each with probability 1/2. The color of each face is determined independently. What is the probabilty that the painted cube can be faced on a horizontal surface so that the four vertical faces are all the same color?

there can be 64 different color cubes 2x2x2x2x2x2x2 ?

case 1 when the cube is all red or blue it will work - 2/64

I couldn't think of the cases...

Could someone just give me some hints on the cases I have to set up?

Vicky.

2. Originally Posted by Vicky1997
Each face of a cube is painted either Red or Blue, each with probability 1/2. The color of each face is determined independently. What is the probability that the painted cube can be faced on a horizontal surface so that the four vertical faces are all the same color?

there can be 64 different color cubes 2x2x2x2x2x2x2 ?

case 1 when the cube is all red or blue it will work - 2/64

I couldn't think of the cases...

Could someone just give me some hints on the cases I have to set up?

Vicky.
That's a good start. You can continue it like this. If one face is one colour and the other five faces are all the other colour then it will work (2×6 cases of that). If two faces are one colour (and the other four faces are the other colour) then it will work provided that the two faces are opposite each other (2×3 cases, because there are three pairs of opposite faces). Finally, it can't be done at all if there are three faces of each colour.

3. Hello, Vicky1997!

A fascinating problem . . .

Each face of a cube is painted either Red or Blue, each with probability 1/2.
The color of each face is determined independently.
What is the probabilty that the painted cube can be placed on a horizontal surface
so that the four vertical faces are all the same color?

There can be 64 different colored cubes: $2^6 = 64$ . . . . Right!
There are seven possible colorings:

. . $\begin{array}{|c|c|} \hline
\text{Colors} & \text{Ways} \\ \hline
\text{6 red, 0 blue} & 1 \\ \text{5 red, 1 blue} & 6 \\ \text{4 red, 2 blue} & 15 \\ \text{3 red, 3 blue} & 20 \\ \text{2 red, 4 blue} & 15 \\ \text{1 red, 5 blue} & 6 \\ \text{0 red, 6 blue} & 1 \\ \hline \end{array}$
. . You may recognize these as Binomial Coefficients.

Let $V$ = "the four vertical faces have the same color".

6 red, 0 blue: . $P(\text{6 red}) \:=\:\frac{1}{64}$
To have $V$, the cube can be placed on any face.
. . $P(\text{6 red} \wedge V)\:=\:\frac{1}{64}\cdot\frac{6}{6} \:=\:\frac{6}{384}$

5 red, 1 blue: . $P(\text{5 red}) \:=\:\frac{6}{64}$
To have $V$, the blue face must be on the top or bottom
. $P(\text{5 red} \wedge V) \:=\:\frac{6}{64}\cdot\frac{2}{6} \:=\:\frac{12}{384}$

4 red, 2 blue.
To have $V$, the 2 blues must be on opposite faces.
. . This happens only $\frac{3}{64}$ of the time.
Then the blue faces must be on the top and the bottom.
. . $P(\text{4 red} \wedge V) \:=\:\frac{3}{64}\cdot\frac{2}{6} \:=\:\frac{6}{384}$

3 red, 3 blue.
With 3 of each color, it is impossible to have $V$.

2 red, 4 blue.
This has the same probability as "4 red, 2 blue".
. . $P(\text{2 red} \wedge V) \:=\:\frac{6}{384}$

1 red, 5 blue.
This has the same probability as "5 red, 1 blue".
. . $P(\text{1 red} \wedge V) \:=\:\frac{12}{384}$

0 red, 6 blue.
This has the same probability as "6 red, 0 blue".
. . $P(\text{0 red} \wedge V) \:=\:\frac{6}{384}$

Therefore: . $P(V) \;=\;\frac{6}{384} + \frac{12}{384} + \frac{6}{384} + \frac{6}{384} + \frac{12}{384} + \frac{6}{384} \;=\;\frac{48}{384} \;=\;\frac{1}{8}$