Here is the solution to your question. Assumption: The probability that Forson will pass the examination is $\frac{3}{4}$, as the image cuts off the last fraction. Part 9(a): Probability Let $P(A)$, $P(I)$, and $P(F)$ be the probabilities that Ayuba, Idris, and Forson pass the examination, respectively. Given: $P(A) = \frac{2}{3}$ $P(I) = \frac{5}{8}$ $P(F) = \frac{3}{4}$ The probabilities that they fail are: $P(A') = 1 - P(A) = 1 - \frac{2}{3} = \frac{1}{3}$ $P(I') = 1 - P(I) = 1 - \frac{5}{8} = \frac{3}{8}$ $P(F') = 1 - P(F) = 1 - \frac{3}{4} = \frac{1}{4}$ i. The three of them will pass the examination Step 1: Multiply the individual probabilities of passing. $$P(\text{A and I and F pass}) = P(A) \times P(I) \times P(F)$$ $$P(\text{A and I and F pass}) = \frac{2}{3} \times \frac{5}{8} \times \frac{3}{4}$$ Step 2: Calculate the product. $$P(\text{A and I and F pass}) = \frac{2 \times 5 \times 3}{3 \times 8 \times 4} = \frac{30}{96}$$ Step 3: Simplify the fraction. $$P(\text{A and I and F pass}) = \frac{5}{16}$$ The probability that the three of them will pass is $\boxed{\frac{5}{16}}$. ii. None of them will pass the examination Step 1: Multiply the individual probabilities of failing. $$P(\text{None pass}) = P(A') \times P(I') \times P(F')$$ $$P(\text{None pass}) = \frac{1}{3} \times \frac{3}{8} \times \frac{1}{4}$$ Step 2: Calculate the product. $$P(\text{None pass}) = \frac{1 \times 3 \times 1}{3 \times 8 \times 4} = \frac{3}{96}$$ Step 3: Simplify the fraction. $$P(\text{None pass}) = \frac{1}{32}$$ The probability that none of them will pass is $\boxed{\frac{1}{32}}$. iii. Ayuba and Idris will pass the examination Step 1: Multiply the probabilities of Ayuba and Idris passing. $$P(\text{A and I pass}) = P(A) \times P(I)$$ $$P(\text{A and I pass}) = \frac{2}{3} \times \frac{5}{8}$$ Step 2: Calculate the product. $$P(\text{A and I pass}) = \frac{10}{24}$$ Step 3: Simplify the fraction. $$P(\text{A and I pass}) = \frac{5}{12}$$ The probability that Ayuba and Idris will pass is $\boxed{\frac{5}{12}}$. Part 9(b): Ratio Problem Step 1: Find the initial number of boys and girls. Total students = 432 Ratio of boys to girls = 5:4 Total ratio parts = $5 + 4 = 9$ Initial number of boys: $$B_{initial} = \frac{5}{9} \times 432 = 5 \times 48 = 240$$ Initial number of girls: $$G_{initial} = \frac{4}{9} \times 432 = 4 \times 48 = 192$$ Step 2: Calculate the new number of boys. The number of boys increases by 12. $$B_{new} = B_{initial} + 12 = 240 + 12 = 252$$ Step 3: Use the new ratio to find the new number of girls. The new ratio of boys to girls is 7:6. $$\frac{B_{new}}{G_{new}} = \frac{7}{6}$$ $$\frac{252}{G_{new}} = \frac{7}{6}$$ $$G_{new} = \frac{252 \times 6}{7}$$ $$G_{new} = 36 \times 6 = 216$$ Step 4: Find the increase in the number of girls. Increase in girls = $G_{new} - G_{initial}$ Increase in girls = $216 - 192 = 24$ The increase in the number of girls is $\boxed{24}$. Send me the next one 📸