I have a bag which contains three coins. One of them is red on one side, blue on the other; another of them is red on one side and green on the other; and the third is blue on one side and green on the other.
I would like to lay them out on the table so that one of each color is facing up.
I reach into the bag and randomly select a coin. I flip it onto the table and it lands red side up (you cannot see what color is on the other side).
Not wanting to risk flipping another red side up, I reach into the bag and look for a coin with a green side, and place it green side up on the table (once again you have no idea what color is on the other side).
If I were to take the last coin out of the bag and flip it, what is the probability that it will land with a blue side facing up?
(In reply to Solution (with expl)
Another way of saying it is:
Out of 8 times this is tried, in 4 cases the original was Red-Green and 4 cases Red-Blue.
In all four Red-Green cases, the second coin was Green-Blue and the remaining one Red Blue. In two of those cases the last flip will be blue.
Of the four Red-Blue cases for the original, in 2 cases The Green-Red is chosen, leaving the Green-Blue for the final toss, and in 2 cases the Green-Blue is taken, leaving the Green-Red for the final toss. In the 2 cases of the Green-blue in the final toss, in one instance it will come up blue.
So there are 3 out of the 8 that come up blue in the final toss.
Posted by Charlie
on 2006-05-24 11:55:33