1. **State the problem:** A box contains 4 chocolates and 1 fruit chew. Clark and Chloe take turns drawing a treat without replacement. Clark draws first. The one who draws the fruit chew wins. We need to find the probability that Chloe wins. 2. **Understand the setup:** - Total treats: 5 (4 chocolates + 1 fruit chew) - Clark draws first, then Chloe, then Clark, then Chloe, then Clark (if needed). - Drawing is without replacement. 3. **Key idea:** Chloe wins if she draws the fruit chew on her turn before Clark does. 4. **Possible turns Chloe can draw:** 2nd or 4th draw (since Clark draws 1st, 3rd, 5th). 5. **Calculate probability Chloe wins on 2nd draw:** - Clark must NOT draw the fruit chew on 1st draw. - Probability Clark draws chocolate first: $\frac{4}{5}$ - Then Chloe draws the fruit chew on 2nd draw: $\frac{1}{4}$ (since 1 fruit chew and 3 chocolates remain) - So, probability Chloe wins on 2nd draw: $\frac{4}{5} \times \frac{1}{4} = \frac{1}{5}$ 6. **Calculate probability Chloe wins on 4th draw:** - Clark and Chloe both must NOT draw the fruit chew in their first turns. - Probability Clark draws chocolate 1st: $\frac{4}{5}$ - Probability Chloe draws chocolate 2nd: $\frac{3}{4}$ (3 chocolates left out of 4 treats) - Probability Clark draws chocolate 3rd: $\frac{2}{3}$ (2 chocolates left out of 3 treats) - Then Chloe draws fruit chew 4th: $\frac{1}{2}$ (1 fruit chew and 1 chocolate left) - Multiply all: $\frac{4}{5} \times \frac{3}{4} \times \frac{2}{3} \times \frac{1}{2} = \frac{1}{5}$ 7. **Total probability Chloe wins:** $$\frac{1}{5} + \frac{1}{5} = \frac{2}{5}$$ **Final answer:** Chloe's probability of winning is $\boxed{\frac{2}{5}}$.