We have 3 Blue balls and 4 Red balls. We will set it up just as before. However this time when one is used it is not replaced back with the other balls. So if one blue ball is drawn then there are only two left in the group and four red remain.
If a red is drawn first there are still three blue and now there are 3 red still in the group.
We can now figure out the probability.
Lets start by multiplying three sevenths and two sixths,this results in a one in three chance
now multiply three sevenths and four sixths,a two in three chance
Next multiply four sevenths and three sixths, a one in two chance
And lastly four sevenths and three sixths, a one in two chance
See that wasn't too bad :)
No comments:
Post a Comment