1. **State the problem:** There are 16 students in the club, all 7th, 8th, or 9th graders. The probability of choosing a 7th, 8th, or 9th grader is 1, meaning all students are in these grades. 2. The probability of choosing a 7th grader is $\frac{1}{2}$, so the number of 7th graders is half of 16: $$\text{Number of 7th graders} = \frac{1}{2} \times 16 = 8$$ 3. Let the number of 8th graders be $x$ and the number of 9th graders be $y$. We know: $$8 + x + y = 16$$ which simplifies to: $$x + y = 8$$ 4. The problem states choosing an 8th grader is more likely than choosing a 9th grader, so: $$\frac{x}{16} > \frac{y}{16} \implies x > y$$ 5. To find the greatest possible number of 9th graders, maximize $y$ while keeping $x > y$ and $x + y = 8$. 6. Since $x > y$ and $x + y = 8$, the greatest integer $y$ can be is 3 (because if $y=4$, then $x=4$ which is not greater than $y$). 7. To find the least possible number of 9th graders with $y > 0$, minimize $y$ while keeping $x > y$ and $x + y = 8$. 8. The smallest positive integer $y$ can be is 1, with $x=7$ which satisfies $x > y$. **Final answers:** - Greatest possible number of 9th graders: **3** - Least possible number of 9th graders (greater than 0): **1**