Database Reference
In-Depth Information
So, if the red ball is returned to the urn after it is drawn, the calculation
above becomes the following:
If the red ball is not replaced into the urn then there is now one less ball in
the urn, and the probability becomes this:
This can be extended to deal with more complicated discrete events, such as
the probability that X of the balls drawn will be red if Y total balls are drawn.
For example, if X is 2 and Y is 5 then the brute force approach would be to
enumerate the different ways that two red balls could be drawn. If R is used
to denote red balls and N represents non-red balls (it doesn't matter what
color they are), there are 10 possible ways 2 red balls can be drawn out of 5
draws, as shown in this list:
• RRNNN
• RNRNN
• RNNRN
• RNNNR
• NRRNN
• NRNRN
• NRNNR
• NNRRN
• NNRNR
• NNNRR
If the balls are replaced after each draw, the probability of drawing a red
ball is #red/#balls, and the probability of drawing another ball is 1-(#red/
