1. **State the problem:** Luke has red and blue marbles. Initially, the probability of choosing a blue marble is $\frac{2}{5}$. After adding 5 blue marbles and removing 5 red marbles, the probability of choosing a blue marble becomes $\frac{3}{5}$. We need to find the total number of marbles in the bag initially. 2. **Define variables:** Let $r$ be the number of red marbles and $b$ be the number of blue marbles initially. 3. **Write the initial probability equation:** $$\frac{b}{r+b} = \frac{2}{5}$$ 4. **Write the new probability equation after changes:** After adding 5 blue marbles and removing 5 red marbles, the number of blue marbles is $b+5$ and red marbles is $r-5$. The total marbles now is $(r-5)+(b+5) = r+b$. The new probability is: $$\frac{b+5}{r+b} = \frac{3}{5}$$ 5. **Solve the system:** From the first equation: $$5b = 2(r+b) \implies 5b = 2r + 2b \implies 3b = 2r \implies r = \frac{3b}{2}$$ Substitute $r$ into the second equation: $$\frac{b+5}{r+b} = \frac{3}{5} \implies 5(b+5) = 3(r+b)$$ $$5b + 25 = 3r + 3b$$ Substitute $r = \frac{3b}{2}$: $$5b + 25 = 3 \times \frac{3b}{2} + 3b = \frac{9b}{2} + 3b = \frac{9b}{2} + \frac{6b}{2} = \frac{15b}{2}$$ Multiply both sides by 2: $$2(5b + 25) = 15b \implies 10b + 50 = 15b$$ $$50 = 15b - 10b = 5b \implies b = 10$$ 6. **Find $r$:** $$r = \frac{3b}{2} = \frac{3 \times 10}{2} = 15$$ 7. **Find total marbles:** $$r + b = 15 + 10 = 25$$ **Final answer:** There are 25 marbles in the bag initially.