Jolene and Tia are playing a two-player game at a carnival. In one bin, there are five red balls numbered 5, 10, 15, 20 and 25. In another bin, there are 25 green balls numbered 1 through 25. In the first stage of the game, Jolene chooses one of the red balls at random. Next, the carnival worker removes the green ball with the same number as the ball Jolene chose. Tia then chooses one of the 24 remaining green balls at random.

Jolene and Tia win if the number on the ball chosen by Tia is a multiple of 3 . What is the probability that they will win?
The multiples of 3 from 1 through 25 are $3,6,9,12,15,18,21$, and 24 .

The only multiple of 5 in this list is 15 .

Jolene chooses the red ball numbered 15 with probability $\frac{1}{5}$ since there are exactly five red balls.

If Jolene chooses the red ball numbered with 15, then there will be 7 green balls remaining with a multiple of 3 on them. The probability that Tia chooses a ball numbered with a multiple of 3 in this situation is $\frac{7}{24}$.

The probability that Jolene and Tia win the game if Jolene chooses the ball numbered with 15 is $\frac{1}{5} \times \frac{7}{24}=\frac{7}{120}$.

Jolene chooses a red ball other than the one numbered with 15 with probability $\frac{4}{5}$.

In this situation, there are 8 green balls numbered with a multiple of 3 , so the probability that Tia chooses a green ball numbered with a multiple of 3 is $\frac{8}{24}$.

The probability that Jolene and Tia win the game if Jolene does not choose the red ball numbered by 15 is $\frac{4}{5} \times \frac{8}{24}=\frac{32}{120}$.

Since exactly one of the events "Jolene chooses the ball numbered with 15 " and "Jolene does not choose the ball numbered with 15 " must occur, the probability that Jolene and Tia win the game is equal to the sum of the probabilities that they win in each of the two situations. Therefore, the probability that they win the game is

$$
\frac{7}{120}+\frac{32}{120}=\frac{39}{120}=\frac{13}{40}
$$

ANSWER: $\frac{13}{40}$