1. **State the problem:** We have a table showing the distribution of students by gender and major with some missing probabilities. We need to complete the table and find certain probabilities. 2. **Complete the table:** Given probabilities: - Male Agri (M,A) = 0.10 - Male Bio (M,B) = 0.24 - Male Education (M,E) = 0.14 - Female Bio (F,B) = 0.29 - Female Education (F,E) = 0.17 Let the missing probability Female Agri (F,A) = $x$. Since total probability must sum to 1: $$0.10 + 0.24 + 0.14 + x + 0.29 + 0.17 = 1$$ $$0.94 + x = 1$$ $$x = 1 - 0.94 = 0.06$$ So, Female Agri (F,A) = 0.06. 3. **Calculate required probabilities:** (i) Probability student is an Education major: $$P(E) = P(M,E) + P(F,E) = 0.14 + 0.17 = 0.31$$ (ii) Probability student is male or an Agri major: Use formula for union: $$P(M \cup A) = P(M) + P(A) - P(M \cap A)$$ Calculate each: $$P(M) = 0.10 + 0.24 + 0.14 = 0.48$$ $$P(A) = 0.10 + 0.06 = 0.16$$ $$P(M \cap A) = 0.10$$ So: $$P(M \cup A) = 0.48 + 0.16 - 0.10 = 0.54$$ (iii) Probability student is female given Education major: Conditional probability formula: $$P(F|E) = \frac{P(F \cap E)}{P(E)} = \frac{0.17}{0.31} \approx 0.55$$ 4. **Check independence of events "Female" and "Bio Major":** Events are independent if: $$P(F \cap B) = P(F) \times P(B)$$ Calculate: $$P(F) = 0.06 + 0.29 + 0.17 = 0.52$$ $$P(B) = 0.24 + 0.29 = 0.53$$ $$P(F \cap B) = 0.29$$ Calculate product: $$P(F) \times P(B) = 0.52 \times 0.53 = 0.2756$$ Since $$0.29 \neq 0.2756$$, the events are **not independent**. **Final answers:** - Missing probability Female Agri = 0.06 - (i) $P(E) = 0.31$ - (ii) $P(M \cup A) = 0.54$ - (iii) $P(F|E) \approx 0.55$ - Events "Female" and "Bio Major" are not independent.