Results 1 to 3 of 3

Math Help - Probability problem about cubes

  1. #1
    Member
    Joined
    Apr 2009
    From
    california
    Posts
    121

    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.
    Last edited by Vicky1997; November 15th 2009 at 06:21 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by Vicky1997 View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,660
    Thanks
    600
    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<br />
\text{Colors} & \text{Ways} \\ \hline<br />
\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}

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Cubes perspective problem
    Posted in the Geometry Forum
    Replies: 2
    Last Post: May 12th 2010, 03:36 AM
  2. sum of cubes
    Posted in the Algebra Forum
    Replies: 14
    Last Post: May 18th 2009, 09:45 PM
  3. Cubes
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 30th 2008, 08:10 PM
  4. Cubes
    Posted in the Geometry Forum
    Replies: 1
    Last Post: May 2nd 2007, 06:50 AM
  5. two cubes
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 19th 2007, 07:03 AM

Search Tags


/mathhelpforum @mathhelpforum