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:

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

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