1. **Stating the problem:** We are given fractions of students who had measles, chicken pox, and neither during one year in a school. We need to find the fraction of students who had both measles and chicken pox. 2. **Given data:** - Fraction with measles: $\frac{5}{8}$ - Fraction with chicken pox: $\frac{1}{2}$ - Fraction with neither: $\frac{7}{8}$ 3. **Understanding the problem:** Let the total number of students be represented by 1 (the whole). 4. **Formula used:** The union of students who had measles or chicken pox plus those who had neither must equal the whole: $$ P(M \cup C) + P(\text{neither}) = 1 $$ Also, the formula for union of two sets is: $$ P(M \cup C) = P(M) + P(C) - P(M \cap C) $$ 5. **Calculate $P(M \cup C)$:** $$ P(M \cup C) = 1 - P(\text{neither}) = 1 - \frac{7}{8} = \frac{1}{8} $$ 6. **Substitute values into union formula:** $$ \frac{1}{8} = \frac{5}{8} + \frac{1}{2} - P(M \cap C) $$ 7. **Simplify the right side:** $$ \frac{5}{8} + \frac{1}{2} = \frac{5}{8} + \frac{4}{8} = \frac{9}{8} $$ 8. **Solve for $P(M \cap C)$:** $$ P(M \cap C) = \frac{9}{8} - \frac{1}{8} = \frac{8}{8} = 1 $$ 9. **Interpretation:** The fraction of students who had both measles and chicken pox is 1, meaning all students had both diseases, which contradicts the given data because $P(\text{neither}) = \frac{7}{8}$ is very high. 10. **Re-examine the problem:** Since $P(\text{neither}) = \frac{7}{8}$, only $\frac{1}{8}$ had either measles or chicken pox or both. But the fractions given for measles and chicken pox individually are larger than $\frac{1}{8}$, which is impossible. **Conclusion:** The data given is inconsistent or incorrectly stated because the sum of fractions with measles and chicken pox exceeds the fraction of students who had either disease. **If we assume the problem meant $\frac{1}{8}$ had neither, then:** - $P(\text{neither}) = \frac{1}{8}$ - $P(M \cup C) = 1 - \frac{1}{8} = \frac{7}{8}$ Then: $$ \frac{7}{8} = \frac{5}{8} + \frac{1}{2} - P(M \cap C) $$ $$ \frac{7}{8} = \frac{5}{8} + \frac{4}{8} - P(M \cap C) $$ $$ \frac{7}{8} = \frac{9}{8} - P(M \cap C) $$ $$ P(M \cap C) = \frac{9}{8} - \frac{7}{8} = \frac{2}{8} = \frac{1}{4} $$ **Final answer:** The fraction of students who had both measles and chicken pox is $\boxed{\frac{1}{4}}$.