1. **Problem statement:** Jamie has a 1 in 3 chance to hit the target each time he tries. We know the chance he hits the target for the first time on his $n^{th}$ try is $\frac{64}{2187}$. We want to find $n$. 2. **Formula used:** The chance to hit for the first time on try $n$ is given by: $$P(X=n) = (1-p)^{n-1} p$$ where $p = \frac{1}{3}$ is the chance to hit. 3. **Apply the formula:** Substitute $p$ and $P(X=n)$: $$\left(1 - \frac{1}{3}\right)^{n-1} \times \frac{1}{3} = \frac{64}{2187}$$ Simplify inside the parentheses: $$\left(\frac{2}{3}\right)^{n-1} \times \frac{1}{3} = \frac{64}{2187}$$ 4. **Isolate the power:** Multiply both sides by 3: $$\left(\frac{2}{3}\right)^{n-1} = \frac{64}{2187} \times 3 = \frac{192}{2187}$$ 5. **Rewrite numbers as powers:** $64 = 2^6$ and $2187 = 3^7$, so: $$\frac{192}{2187} = \frac{64 \times 3}{3^7} = 2^6 \times 3^{-6}$$ Also, $$\left(\frac{2}{3}\right)^{n-1} = 2^{n-1} \times 3^{-(n-1)}$$ 6. **Match powers:** $$2^{n-1} \times 3^{-(n-1)} = 2^6 \times 3^{-6}$$ This means: $$n - 1 = 6$$ 7. **Solve for $n$:** $$n = 6 + 1 = 7$$ **Final answer:** Jamie hits the target for the first time on his 7th try.