1. **Problem statement:** Given sets $E$ and $F$ with $n(\xi) = 50$, $n(E) = 15$, $n(F) = 9$, and $n((E \cup F)') = 33$, find: (i) $n(E \cap F)$ (ii) $n(E' \cup F)$ (iii) Probability a randomly chosen student has both English and French parents (iv) Probability a student with a French parent also has an English parent Also, express $x^2 - 4x - 9$ in the form $(x - a)^2 - b$ and find $a$ and $b$. --- 2. **Recall formulas and rules:** - $n(\xi)$ is total number of students. - $n((E \cup F)')$ is number of students not in $E$ or $F$. - $n(E \cup F) = n(\xi) - n((E \cup F)')$. - Inclusion-exclusion principle: $n(E \cup F) = n(E) + n(F) - n(E \cap F)$. - Complement rule: $E' = \xi \setminus E$. - Probability $P(A) = \frac{n(A)}{n(\xi)}$. --- 3. **Calculate $n(E \cup F)$:** $$ n(E \cup F) = n(\xi) - n((E \cup F)') = 50 - 33 = 17 $$ 4. **Find $n(E \cap F)$ using inclusion-exclusion:** $$ 17 = 15 + 9 - n(E \cap F) \\ n(E \cap F) = 15 + 9 - 17 = 7 $$ 5. **Find $n(E' \cup F)$:** - $E' = \xi \setminus E$, so $n(E') = n(\xi) - n(E) = 50 - 15 = 35$. - Use inclusion-exclusion: $$ n(E' \cup F) = n(E') + n(F) - n(E' \cap F) $$ - Note $E' \cap F = F \setminus E$, so $$ n(E' \cap F) = n(F) - n(E \cap F) = 9 - 7 = 2 $$ - Therefore: $$ n(E' \cup F) = 35 + 9 - 2 = 42 $$ 6. **Probability student has both English and French parents:** $$ P(E \cap F) = \frac{n(E \cap F)}{n(\xi)} = \frac{7}{50} $$ 7. **Probability student with French parent also has English parent:** $$ P(E | F) = \frac{n(E \cap F)}{n(F)} = \frac{7}{9} $$ 8. **Express $x^2 - 4x - 9$ in the form $(x - a)^2 - b$:** - Half of coefficient of $x$ is $\frac{-4}{2} = -2$. - Square it: $(-2)^2 = 4$. - Add and subtract 4 inside expression: $$ x^2 - 4x - 9 = (x^2 - 4x + 4) - 9 - 4 + 4 = (x - 2)^2 - 13 $$ - So $a = 2$, $b = 13$. --- **Final answers:** (i) $n(E \cap F) = 7$ (ii) $n(E' \cup F) = 42$ (iii) $P(E \cap F) = \frac{7}{50}$ (iv) $P(E | F) = \frac{7}{9}$ (a) $a = 2$, $b = 13$