Arjun has a bag that contains 5 balls and Becca has a bag that contains 3 balls. Arjun's bag contains 2 red balls, 1 green, 1 yellow, and 1 violet ball. Becca's bag contains 2 black balls and 1 orange ball. Arjun randomly chooses 1 ball from his bag and puts it into Becca's bag. Becca then randomly chooses 1 ball from her bag and puts it into Arjun's bag. Again, Arjun randomly chooses 1 ball from his bag and puts it into Becca's bag. After these exchanges, there are 4 balls in each bag. What is the probability that each bag contains exactly 3 different colours of balls?
(A) $\frac{3}{10}$
(B) $\frac{6}{25}$
(C) $\frac{9}{50}$
(D) $\frac{3}{25}$
(E) $\frac{9}{25}$
After all exchanges there are 4 balls in each bag, and so if a bag contains exactly 3 different colours of balls, then it must contain exactly 2 balls of the same colour and 2 balls each having a colour that is different than all other balls in the bag.

Among the 8 balls in the two bags, there are 2 red balls and 2 black balls and each of the remaining balls has a colour that is different than all the other balls.

Thus after all exchanges, one bag must contain the 2 red balls and the other bag must contain the 2 black balls.

Since Becca's bag initially contains both black balls, and Becca moves only 1 ball from her bag to Arjun's bag, it is not possible for Arjun's bag to contain the 2 black balls after all exchanges. This tells us that if each bag contains exactly 3 different colours of balls after all exchanges, then Arjun's bag contains the 2 red balls and Becca's bag contains the 2 black balls.

We let the first letter of each colour represent a ball of that colour.

Then initially, Arjun's bag contains $R R G Y V$ and Becca's bag contains $B B O$.

For the first ball chosen to be moved, there are exactly two cases to consider:

Case 1: The first ball moved from Arjun's bag to Becca's bag is $R$, or

Case 2: The first ball moved from Arjun's bag to Becca's bag is not $R$ (thus it is $G, Y$ or $V$ ). We begin with Case 1 and determine the probability that after all exchanges, each bag contains exactly 3 different colours of balls.

Case 1: The first ball moved from Arjun's bag to Becca's bag is $R$

Since Arjun's bag initially contains 5 balls, 2 of which are $R$, the probability that $R$ is chosen as the first ball to move is $\frac{2}{5}$.

After $R$ is moved from Arjun's bag to Becca's, Arjun's bag contains $R G Y V$ and Becca's bag
contains BBOR.

Since Arjun's bag must contain both $R$ 's after all exchanges, and Becca moves only 1 ball from her bag to Arjun's, then the next ball chosen to move must be $R$.

Becca's bag contains 4 balls, 1 of which is $R$, and so in this case the probability that $R$ is chosen as the second ball to move is $\frac{1}{4}$.

After the first two balls are moved, Arjun's bag contains $R R G Y V$ and Becca's bag contains $B B O$.

Finally, a ball is chosen from Arjun's bag and moved to Becca's.

Since Arjun's bag must contain both $R$ 's after all exchanges, then the ball that is chosen must be $G, Y$ or $V$ which occurs with probability $\frac{3}{5}$.

If the first ball moved from Arjun's bag to Becca's bag is $R$, then the probability that each bag contains exactly 3 different colours of balls after all exchanges is given by the product of the

individual probabilities of choosing each of the 3 balls or $\frac{2}{5} \times \frac{1}{4} \times \frac{3}{5}=\frac{6}{100}=\frac{3}{50}$.

Case 2: The first ball moved from Arjun's bag to Becca's bag is $G, Y$ or $V$

Since Arjun's bag initially contains 5 balls, the probability that $G, Y$ or $V$ is chosen as the first ball to move is $\frac{3}{5}$.

After one of $G, Y$ or $V$ is moved from Arjun's bag to Becca's, Arjun's bag contains $R R$ and two of $G, Y$ and $V$, and Becca's bag contains $B B O$ and one of $G, Y$ or $V$.

For the second ball chosen to be moved, there are exactly two cases to consider and so we split Case 2 into these two separate cases.

Case 2a: The first ball moved from Arjun's bag to Becca's bag is $G, Y$ or $V$, and the ball moved from Becca's bag to Arjun's bag is $B$, or

Case 2b: The first ball moved from Arjun's bag to Becca's bag is $G, Y$ or $V$, and the ball moved from Becca's bag to Arjun's bag is not $B$ (that is, it is $O$ or the first ball moved).

We begin with Case 2a and determine the probability that after all exchanges, each bag contains exactly 3 different colours of balls.

Case 2a: The first ball moved from Arjun's bag to Becca's bag is $G, Y$ or $V$, and the ball moved from Becca's bag to Arjun's bag is $B$

We previously determined that the probability that $G, Y$ or $V$ is chosen as the first ball to move is $\frac{3}{5}$.

At this point, Becca's bag contains 4 balls, 2 of which are $B$ 's, and so the probability that $B$ is chosen is $\frac{2}{4}=\frac{1}{2}$.

After $B$ is moved from Becca's bag to Arjun's, Arjun's bag contains $R R B$ and two of $G, Y$ and $V$ and Becca's bag contains $B O$ and one of $G, Y$ or $V$.

Since Becca's bag must contain both $B$ 's after all exchanges, then the final ball chosen to move must be $B$.

Arjun's bag contains 5 balls, 1 of which is $B$, and so in this case the probability that $B$ is chosen as the final ball to move is $\frac{1}{5}$.

If the first ball moved from Arjun's bag to Becca's bag is $G, Y$ or $V$, and the ball moved from Becca's bag to Arjun's bag is $B$, then the probability that each bag contains exactly 3 different colours of balls after all exchanges is given by the product of the individual probabilities of choosing each of the 3 balls or $\frac{3}{5} \times \frac{1}{2} \times \frac{1}{5}=\frac{3}{50}$.

Case 2b: The first ball moved from Arjun's bag to Becca's bag is $G, Y$ or $V$, and the ball moved from Becca's bag to Arjun's bag is $O$ or the first ball moved $(G, Y$ or $V$ )

The probability that $G, Y$ or $V$ is chosen as the first ball to move is $\frac{3}{5}$.

At this point, Becca's bag contains 4 balls, 2 of which are $B$, and so the probability that $B$ is
not chosen is $\frac{2}{4}=\frac{1}{2}$.

After this ball $(O, G, Y$ or $V)$ is moved from Becca's bag to Arjun's, Arjun's bag contains $R R$ and three of $O, G, Y$ and $V$ and Becca's bag contains two $B$ 's and one of $O, G, Y$ or $V$.

Since Arjun's bag must contain both $R$ 's after all exchanges, then the final ball chosen to move must be one of the 3 balls in Arjun's bag that is not $R$.

Arjun's bag contains 5 balls, 2 of which are $R$, and so in this case the probability that $R$ is not chosen as the final ball to move is $\frac{3}{5}$.

If the first ball moved from Arjun's bag to Becca's bag is $G, Y$ or $V$, and the ball moved from Becca's bag to Arjun's bag is not $B$, then the probability that each bag contains exactly 3 different colours of balls after all exchanges is given by the product of the individual probabilities of choosing each of the 3 balls or $\frac{3}{5} \times \frac{1}{2} \times \frac{3}{5}=\frac{9}{50}$.

Finally, the probability that each bag contains exactly 3 different colours of balls after all exchanges is $\frac{3}{50}+\frac{3}{50}+\frac{9}{50}=\frac{15}{50}=\frac{3}{10}$.