Results 1 to 7 of 7

Math Help - Portraits

  1. #1
    Newbie
    Joined
    Mar 2010
    Posts
    4

    Portraits

    Set of portraits consists of 6 boys and 4 girls. Little brother drew mustache on three of these set portraits. What is the probability that a randomly selected image will be a portrait of a girl without a mustache?

    ----------------------------------------------------------------------------------
    http://translate.google.com/translat...d,1;s,39;t,528
    Last edited by speculus; March 14th 2010 at 10:42 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello speculus

    Welcome to Math Help Forum!
    Quote Originally Posted by speculus View Post
    Photo fiber consists of 6 friends to 4 friends portraits. Little brother 3 nuotraukse drew people mustache. What is the probability that a randomly selected image will be intact companion portrait?

    ----------------------------------------------------------------------------------
    Google Translate
    Sorry. Your question does not make sense in English.

    Grandad
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2010
    Posts
    4
    excuse me for wrong explanation

    I've changed question, see above
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,617
    Thanks
    1581
    Awards
    1
    Little brother will select from none to three girls. That is the partition of the space.
    Probability of selecting a girl without a mustache when he is done is <br />
\sum\limits_{k = 0}^3 {\left[ {\frac{{\binom{4}{k}\binom{6}{3-k}}}{\binom{10}{3}} \cdot \frac{{4 - k}}{{10}}} \right]}
    Last edited by Plato; March 14th 2010 at 12:36 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2010
    Posts
    4
    thank you
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,713
    Thanks
    633
    Hello, speculus!

    Here's an explanation of Plato's excellent solution.


    There are portraits of 6 boys and 4 girls.
    Little brother drew mustaches on three of these portraits.

    What is the probability that a randomly selected portrait
    will be of a girl without a mustache?

    Assume that little brother randomly chose three portraits to deface.
    . . He has: . {10\choose3} \,=\,120 possible choices.



    He has 4 choices for the genders of the subjects:

    . . \begin{array}{cccccccccc} (1) & \text{3 boys:} & {6\choose3} &=& 20\text{ ways.} & P(\text{3B}) &=& \dfrac{20}{120} &=& \dfrac{5}{30} \\ \\<br /> <br />
(2) & \text{2 boys, 1 girl:} & {6\choose2}{4\choose1} &=& 60\text{ ways.} & P(\text{ 2B, 1G}) &=& \dfrac{60}{120} &=& \dfrac{15}{30}<br />
\end{array}

    . . \begin{array}{cccccccccc}<br />
(3) & \text{1 boy, 2 girls:} & {6\choose1}{4\choose2} &=& 36\text{ ways.} & P(\text{1B, 2G}) &=& \dfrac{36}{120} &=& \dfrac{9}{30} \\ \\<br /> <br />
(4) & \text{3 girls:} & {4\choose3} &=& 4\text{ ways.} & P(\text{3G}) &=& \dfrac{4}{120} &= &\dfrac{1}{30} \end{array}



    For each choice, there is a probability of choosing a girl without a mustache.

    . . \begin{array}{cccc}<br />
\text{Mustaches on:} & P(\text{girl w/o mustache}) \\ \hline\ \\[-3mm]<br />
\text{3 boys} & \dfrac{4}{10} \\ \\[-3mm]<br />
\text{2 boys, 1 girl} & \dfrac{3}{10} \\ \\[-3mm]<br />
\text{1 boy, 2 girls} & \dfrac{2}{10} \\ \\[-3mm]<br />
\text{3 girls} & \dfrac{1}{10}<br />
\end{array}


    Therefore: . P(\text{girl w/o mustache}) \;=\;<br />
\left(\frac{5}{30}\right)\left(\frac{4}{10}\right) +<br />
\left(\frac{15}{30}\right)\left(\frac{3}{10}\right  ) + \left(\frac{9}{30}\right)\left(\frac{2}{10}\right) + \left(\frac{1}{30}\right)\left(\frac{1}{10}\right)

    . . . . . . . . . . . . . . . . . . . . . . =\;\<br />
\frac{20}{300} + \frac{45}{300} + \frac{18}{300} + \frac{1}{300} \;\;=\;\;\frac{84}{300} \;\;=\;\;\boxed{\frac{7}{25}}

    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Mar 2010
    Posts
    4

    Thumbs up

    Hello Soroban,

    thank you very much for a detailed solution.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Critical points and phase portraits
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: September 30th 2010, 09:27 AM
  2. what is the use of phase portraits?
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: March 28th 2009, 07:57 AM
  3. sketching nonlinear GLOBAL phase portraits?
    Posted in the Calculus Forum
    Replies: 0
    Last Post: February 17th 2009, 02:45 PM

Search Tags


/mathhelpforum @mathhelpforum