Problem:
An urn contains one red ball and one blue ball. A box of extra red and blue balls lie nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?
Answer Choices:
A.
B.
C.
D.
E.
Solution:
Suppose that we have a deck, currently containing just one black card. We then insert red cards one-by-one into the deck at random positions. It is easy to see using induction, that the black card is randomly situated in the deck.
Now, suppose that we have this deck again, with only one black card. Each time we pick a red ball, we place a card above the black card, and each time we pick a blue ball, we place a card below the black card. It is easy to see that the probability that the card is inserted into the top part of the deck is simply equal to the number of red balls divided by the total number of balls, and the probability that the card is inserted into the bottom part of the deck is equal to the number of blue balls divided by the total number of balls. Therefore, this is equivalent to inserting the card randomly into the deck.
Finally, four more red cards will be inserted into the deck, and so the black card can be in five possible positions. Only one corresponds to having three balls of each type. Our probability is thus , and so the answer is
OR
Let denote the action where George selects a red ball and denote the action where he selects a blue one. Now, in order to get balls of each color, he needs more of both and .
There are cases: (we can confirm that there are only since ). However we can clump , and together since they are equivalent by symmetry.
CASE : and
Let's find the probability that he picks the balls in the order of .
The probability that the first ball he picks is red is .
Now there are reds and blue in the urn. The probability that he picks red again is now .
There are reds and blue now. The probability that he picks a blue is .
Finally, there are reds and blues. The probability that he picks a blue is .
So the probability that the case happens is . However, since the case is the exact same by symmetry, case has a probability of chance of happening.
CASE : and
Let's find the probability that he picks the balls in the order of .
The probability that the first ball he picks is red is .
Now there are reds and blue in the urn. The probability that he picks blue is .
There are reds and blues now. The probability that he picks a red is .
Finally, there are reds and blues. The probability that he picks a blue is .
So the probability that the case happens is . However, since the case is the exact same by symmetry, case has a probability of chance of happening.
CASE : and
Let's find the probability that he picks the balls in the order of .
The probability that the first ball he picks is red is .
Now there are reds and blue in the urn. The probability that he picks blue is .
There are reds and blues now. The probability that he picks a blue is .
Finally, there are reds and blues. The probability that he picks a red is .
So the probability that the case happens is . However, since the case is the exact same by symmetry, case has a probability of chance of happening.
Adding up the cases, we have .
The problems on this page are the property of the MAA's American Mathematics Competitions