Results 1 to 5 of 5

Math Help - Probabilty --- Help it's been awhile since I've done one of this

  1. #1
    BellieButton
    Guest

    Question Probabilty --- Help it's been awhile since I've done one of this

    It's been awhile I can't remember how to do this

    ONE BLUE BALL AND FOUR RED BALLS ARE PLACED IN A BOX. TWO RED BALL AND THREE BLUE BALLS ARE PLACED IN A SECOND BOX. A BOX IS CHOSEN AT RANDOM, AND A BALL IS SELECTED FROM IT. THE PROBABILITY OF CHOOSING THE FIRST BOX IS 3/8. FIND THE PROBABILITY THAT THE BALL IS FROM THE SECOND BOX, GIVEN THAT IT IS RED.


    Please help!!! I need to jog my memory
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    The probability of chosing the first box is 3/8 thus the probability of chosing the second box must be 5/8. This is called the "Probabilty of not chosing is 1-p where p is chosing".

    Next, the second box has 2 red and 3 blue balls. Thus the probabilty of taking a red is 2/5 "Probabilty is equal to the ratio of favorable to possible". Favorable here is 2 possibilities and altogether there are 5 possibilities.

    Next, the "Probabilty of event A and event B is the product of the probabilties" thus (3/8)(2/5)=6/40=3/20=15%
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by BellieButton
    It's been awhile I can't remember how to do this

    ONE BLUE BALL AND FOUR RED BALLS ARE PLACED IN A BOX. TWO RED BALL AND THREE BLUE BALLS ARE PLACED IN A SECOND BOX. A BOX IS CHOSEN AT RANDOM, AND A BALL IS SELECTED FROM IT. THE PROBABILITY OF CHOOSING THE FIRST BOX IS 3/8. FIND THE PROBABILITY THAT THE BALL IS FROM THE SECOND BOX, GIVEN THAT IT IS RED.

    Please help!!! I need to jog my memory

    <<check the working - no guarantee that all the arithmetic is correct>>

    This is an application of Bayes theorem, which in this case
    says:

    \mbox{P(2-nd box}|\mbox{red)}=\frac{\mbox{P(red}|\mbox{2-nd box).P(2-nd box)}}{\mbox{P(red)}}.

    Where \mbox{P(2-nd box}|\mbox{red)} denotes the probability that the ball
    is drawn from the second box given that the ball is red, \mbox{P(red}|\mbox{2-nd box)}
    is the probability that the ball is red given that it is drawn from the 2-nd box,
    \mbox{P(red)} is the probability of drawing a red ball whichever box is chosen,
    and \mbox{P(2-box)} is the probability that the second box is chosen.

    Now from what we are told of the content of the 2-nd box:

    \mbox{P(red}|\mbox{2-nd box)}=\frac{2}{5}.

    Also we are told the probability of choosing the first box so the probability
    of choosing the second box is one minus this, hence:

    \mbox{P(2-nd box)}=\frac{5}{8}.

    Finally:

    \mbox{P(red)}=\mbox{P(red}|\mbox{1-st box)}.\mbox{P(1-st box)}+\mbox{P(red}|\mbox{2-nd box)}. \mbox{P(2-nd box)}

    =\frac{4}{5}.\frac{3}{8}+\frac{2}{5}.\frac{5}{8}
    =\frac{22}{40}
    .

    So:

    \mbox{P(2-nd box}|\mbox{red)}=\frac{10}{22},

    which is \sim 45.5 \%.

    RonL
    Last edited by CaptainBlack; December 16th 2005 at 10:33 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    CaptianBlack, what is wrong with what I said? You and me certainly understand that problem differently.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by ThePerfectHacker
    CaptianBlack, what is wrong with what I said? You and me certainly understand that problem differently.
    1. The product rule for probabilities applies only when the events
    are independent. That is not the case here.

    2. That you have the wrong answer can be shown by considering
    the following argument:

    Given that the ball selected is red then it came from either
    urn 1 or urn 2 and so the probability that the ball comes
    from urn 1 given it is red plus the probability that it comes
    from urn2 given that it is red should be 1. But using your
    method this sum is:

    \left( \frac{3}{8}\ .\ \frac{4}{5} \right)+\left( \frac{5}{8}\ .\ \frac{2}{5}\right) \ \neq\ 1
    3. If I understand your argument you should have had (5/8)(2/5)=2/8=1/4=25%.

    4. When in doubt about a probability calculation I usually simulate the
    problem. Below is an Euler session (comments in blue, answer in red)
    modelling this problem:

    (Euler is a freeware matrix numerical analysis language see:
    http://mathsrv.ku-eichstaett.de/MGF/...othmann/euler/)

    >
    >ncase=100000;.. set number of replications
    >
    >..select urn for each replication
    >xurn=random(1,ncase);
    >nurn=(xurn<3/8)*1+(xurn>=3/8)*2;
    >
    >
    >..select colour for each replication-using
    >..correct prob for each urn; 1-red ball, 0-blue bal
    l
    >xcol=(nurn==1)*(random(1,ncase)<(4/5));
    >xcol=xcol+(nurn==2)*(random(1,ncase)<(2/5));
    >
    >..find replications with red ball selected
    >idx=nonzeros(xcol==1);
    >urn=nurn(idx);
    >
    >
    >..compute proportion of the replications
    >..where a red ball selected where the urn
    >..was urn 2.

    >pp=sum(urn==2)/length(urn)
    0.455936
    >

    RonL
    Last edited by CaptainBlack; December 21st 2005 at 05:55 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Need some help with Probabilty
    Posted in the Statistics Forum
    Replies: 1
    Last Post: December 18th 2009, 02:45 PM
  2. Probabilty help
    Posted in the Statistics Forum
    Replies: 1
    Last Post: December 10th 2009, 11:21 PM
  3. probabilty
    Posted in the Statistics Forum
    Replies: 6
    Last Post: September 15th 2008, 02:39 PM
  4. Replies: 2
    Last Post: April 4th 2008, 05:11 PM
  5. Probabilty....?
    Posted in the Math Topics Forum
    Replies: 6
    Last Post: July 9th 2005, 10:40 PM

Search Tags


/mathhelpforum @mathhelpforum