1. **Problem statement:** Osas buys notebooks and exercise books with at least one of each, total books up to 8. Each valid combination is equally likely. 2. **Define variables:** Let $n$ = number of notebooks, $e$ = number of exercise books. 3. **Constraints:** - $n \geq 1$ - $e \geq 1$ - $n + e \leq 8$ 4. **Total number of valid combinations:** Count all $(n,e)$ pairs with $n,e \geq 1$ and $n+e \leq 8$. Number of solutions is sum over $k=2$ to $8$ of $(k-1)$ since $n+e=k$ and $n,e \geq 1$. Total combinations = $\sum_{k=2}^8 (k-1) = 1+2+3+4+5+6+7 = 28$. --- **(a) Probability Osas buys more exercise books than notebooks:** 5. Condition: $e > n$ with $n,e \geq 1$ and $n+e \leq 8$. 6. List all valid pairs $(n,e)$ satisfying $e > n$ and $n+e \leq 8$: - For $n=1$: $e=2,3,4,5,6,7$ (since $1+7=8$ max) - For $n=2$: $e=3,4,5,6$ (max $2+6=8$) - For $n=3$: $e=4,5$ (max $3+5=8$) - For $n=4$: $e=5$ (max $4+5=9$ but limit 8, so $e=5$ is invalid, max $e=4$) Check $n=4$: $e$ must be $>4$ and $n+e \leq 8$ so $e=5$ is $4+5=9>8$ invalid. So for $n=4$, no valid $e$. Count valid pairs: - $n=1$: 6 values - $n=2$: 4 values - $n=3$: 2 values - $n=4$: 0 values Total = $6+4+2=12$. 7. Probability = $\frac{12}{28} = \frac{3}{7}$. --- **(b) Probability Osas buys exactly 5 books in total:** 8. Condition: $n+e=5$ with $n,e \geq 1$. Number of positive integer solutions to $n+e=5$ is $5-1=4$: $(1,4), (2,3), (3,2), (4,1)$. 9. Total valid combinations are 28. 10. Probability = $\frac{4}{28} = \frac{1}{7}$. --- **(c) Probability number of notebooks is prime:** 11. Prime numbers possible for $n$ with $n \geq 1$ and $n+e \leq 8$ are 2, 3, 5, 7. For each prime $n$, count valid $e$: - $n=2$: $e \geq 1$, $2+e \leq 8 \Rightarrow e \leq 6$ so $e=1..6$ (6 values) - $n=3$: $e \leq 5$ (5 values) - $n=5$: $e \leq 3$ (3 values) - $n=7$: $e \leq 1$ (1 value) Total valid pairs with prime $n$ = $6+5+3+1=15$. 12. Probability = $\frac{15}{28}$. --- **Final answers:** (a) $\boxed{\frac{3}{7}}$ (b) $\boxed{\frac{1}{7}}$ (c) $\boxed{\frac{15}{28}}$