1. **Problem Statement:** You are given a seating layout on an aeroplane with some seats occupied (shaded) and some unoccupied. You are to find the probability of being allocated an unoccupied seat that is: i) a front row seat ii) a seat in the emergency row iii) a window seat iv) an aisle seat 2. **Formula for Probability:** The probability of an event is given by: $$\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$$ 3. **Important Rules:** - Only unoccupied seats count as possible outcomes. - Favorable outcomes depend on the seat type asked. - We assume the seating layout is known and the count of unoccupied seats for each category can be determined. 4. **Step-by-step solution:** - Let $N$ be the total number of unoccupied seats. - Let $F$ be the number of unoccupied front row seats. - Let $E$ be the number of unoccupied emergency row seats. - Let $W$ be the number of unoccupied window seats. - Let $A$ be the number of unoccupied aisle seats. Then: $$P(\text{front row seat}) = \frac{F}{N}$$ $$P(\text{emergency row seat}) = \frac{E}{N}$$ $$P(\text{window seat}) = \frac{W}{N}$$ $$P(\text{aisle seat}) = \frac{A}{N}$$ 5. **Explanation:** - Count all unoccupied seats to find $N$. - Identify which unoccupied seats belong to each category (front row, emergency row, window, aisle). - Count those seats to find $F$, $E$, $W$, and $A$ respectively. - Divide each count by $N$ to get the probability. Since the exact seating layout and shading are not provided here, the numerical values cannot be computed. However, this method will allow you to calculate the probabilities once you have the counts. **Final answer:** $$P(\text{front row seat}) = \frac{F}{N},\quad P(\text{emergency row seat}) = \frac{E}{N},\quad P(\text{window seat}) = \frac{W}{N},\quad P(\text{aisle seat}) = \frac{A}{N}$$