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

• Dec 15th 2005, 12:43 AM
BellieButton
Probabilty --- Help it's been awhile since I've done one of this
:o 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.

• Dec 16th 2005, 09:10 AM
ThePerfectHacker
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%
• Dec 16th 2005, 10:31 AM
CaptainBlack
Quote:

Originally Posted by BellieButton
:o 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.

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

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

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

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

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

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

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

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

Finally:

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

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

So:

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

which is $\displaystyle \sim 45.5 \%$.

RonL
• Dec 20th 2005, 02:35 PM
ThePerfectHacker
CaptianBlack, what is wrong with what I said? You and me certainly understand that problem differently.
• Dec 21st 2005, 12:01 AM
CaptainBlack
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:

$\displaystyle \left( \frac{3}{8}\ .\ \frac{4}{5} \right)+\left( \frac{5}{8}\ .\ \frac{2}{5}\right) \ \neq\ 1$

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