1. **State the problem:** We have 12 pupils in a class (set $\varepsilon$). Among them, 9 have a brother (set $B$), 7 have a sister (set $S$), and 2 have neither. 2. **Goal:** Find the number of pupils who have both a brother and a sister, i.e., the size of the intersection $B \cap S$. 3. **Use the principle of inclusion-exclusion:** $$|B \cup S| = |B| + |S| - |B \cap S|$$ 4. **We know:** - Total pupils $|\varepsilon| = 12$ - Pupils with neither brother nor sister = 2 So, pupils with either brother or sister or both: $$|B \cup S| = 12 - 2 = 10$$ 5. **Substitute values:** $$10 = 9 + 7 - |B \cap S|$$ 6. **Solve for $|B \cap S|$:** $$|B \cap S| = 9 + 7 - 10 = 16 - 10 = 6$$ 7. **Interpretation:** 6 pupils have both a brother and a sister. 8. **Fill in the Venn diagram:** - Intersection $B \cap S = 6$ - Only brother $= |B| - |B \cap S| = 9 - 6 = 3$ - Only sister $= |S| - |B \cap S| = 7 - 6 = 1$ - Neither $= 2$ **Final answer:** - $|B \cap S| = 6$ - Only brother = 3 - Only sister = 1 - Neither = 2