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?

3/8 was the correct answer,  here's a small explanation.

For the 1st coin that was flipped red side up, there are 2 possibilities, each with 1/2 probability:

possibility 1: the R/G coin
possibility 2: the R/B coin

with #1, it is a guarantee that the next coin you pick out and place green side up is the B/G coin, and the remaining coin is the R/B coin -> which has probability 1/2 of landing blue side up.  So this final scenario has probability 1/2 * 1/2 = 1/4. 

with #2, the coin that you place green side up could be (p = 1/2) the B/G coin (in which case you have no chance of flipping blue side up for the last coin) or (p = 1/2) the R/G coin.  If you chose the R/G coin, the final coin is B/G and has p = 1/2 for being flipped blue side up.  So this final scenario has probability 1/2 * 1/2 * 1/2 = 1/8.

So the total probability of flipping the last coin blue side up is 1/4 + 1/8 = 3/8.

Posted by iamkobe on 2006-05-24 10:42:37