I have 16 marbles, three of which are blue.
I am going to select three random marbles.
What is the percentage chance that at least one marble will be blue?
It sounds like you pick 3 and keep them, rather than picking one, puting it back, picking another, etc. Is this correct?
The trick for this type of problem is to calculate the reverse case, i.e. the probability that none of the picked ones are blue. At first you have 13 non-blue out of 16, so the probability of picking a non-blue one is 13/16. For the next marble, you have 12 non-blue out of 15, and for the third, 11 out of 14. So the probability of all these happening is (13/16)(12/15)(11/14)=0.51. The probability of this not happening (i.e. at least one of the picked marbles is blue) is 1-0.51=0.49, or 49%.
Just in case you’re planning on putting them back in between…
In this case, the probability of picking no blue marbles in three tries is (13/16)[sup]3[/sup], which works out to be approximately .54. 1 - .54 = .46, which is the probability that at least one marble is blue.
Thanks a bunch for both scenarios.