Question: 1. (a) A fair die and a fair coin are thrown together once. (i) Write down the set of all possible outcomes. (ii) Find the probability of obtaining a prime number and a tail. (b) The map of a field is drawn to a scale of 1 : 100. If the width and area of the field on the map are 8 cm and 88 cm2 respectively, find in m2, the area of the actual field. (c) Copy and complete the 3 x 3 magic square such that the sum of the numbers in each row, column and diagonal is equal to 21. 10 3 7 2.

1. **Problem Statement:** (a)(i) List all possible outcomes when a fair die and a fair coin are thrown together once. (a)(ii) Find the probability of obtaining a prime number on the die and a tail on the coin. (b) Given a map scale of 1:100, width 8 cm and area 88 cm² on the map, find the actual area in m². (c) Complete the 3x3 magic square so each row, column, and diagonal sums to 21, given some numbers. --- 2. **Step 1: (a)(i) Set of all possible outcomes** - The die has outcomes: $\{1,2,3,4,5,6\}$. - The coin has outcomes: $\{H, T\}$. - The combined outcomes are pairs $(\text{die}, \text{coin})$. - So the sample space is: $$\{(1,H),(1,T),(2,H),(2,T),(3,H),(3,T),(4,H),(4,T),(5,H),(5,T),(6,H),(6,T)\}$$ --- 3. **Step 2: (a)(ii) Probability of prime number and tail** - Prime numbers on a die: $2,3,5$. - Outcomes with prime and tail: $$\{(2,T),(3,T),(5,T)\}$$ - Number of favorable outcomes = 3. - Total outcomes = 12. - Probability = $\frac{3}{12} = \frac{1}{4}$. --- 4. **Step 3: (b) Actual area of the field** - Scale: $1:100$ means $1$ cm on map = $100$ cm actual. - Width on map = $8$ cm, so actual width = $8 \times 100 = 800$ cm = $8$ m. - Area on map = $88$ cm². - Area scale factor = $(100)^2 = 10,000$ (since area scales by square of length scale). - Actual area = $88 \times 10,000 = 880,000$ cm². - Convert to m²: $880,000 \text{ cm}^2 = \frac{880,000}{10,000} = 88$ m². --- 5. **Step 4: (c) Complete the magic square** - Given partial magic square: $$\begin{matrix}10 & 3 & x \\ y & 7 & z \\ 2 & a & b \end{matrix}$$ - Each row, column, diagonal sums to 21. - Row 1: $10 + 3 + x = 21 \Rightarrow x = 8$. - Row 2: $y + 7 + z = 21 \Rightarrow y + z = 14$. - Row 3: $2 + a + b = 21 \Rightarrow a + b = 19$. - Column 1: $10 + y + 2 = 21 \Rightarrow y = 9$. - Column 2: $3 + 7 + a = 21 \Rightarrow a = 11$. - Column 3: $x + z + b = 21 \Rightarrow 8 + z + b = 21 \Rightarrow z + b = 13$. - From $y + z = 14$ and $y=9$, $z=5$. - From $z + b = 13$ and $z=5$, $b=8$. **Completed magic square:** $$\begin{matrix}10 & 3 & 8 \\ 9 & 7 & 5 \\ 2 & 11 & 8 \end{matrix}$$ --- **Final answers:** - (a)(i) Sample space: $\{(1,H),(1,T),(2,H),(2,T),(3,H),(3,T),(4,H),(4,T),(5,H),(5,T),(6,H),(6,T)\}$. - (a)(ii) Probability = $\frac{1}{4}$. - (b) Actual area = $88$ m². - (c) Completed magic square: $$\begin{matrix}10 & 3 & 8 \\ 9 & 7 & 5 \\ 2 & 11 & 8 \end{matrix}$$