1. The problem states that Jamie has a probability of $\frac{1}{3}$ to hit the target on any single attempt. 2. We are told the probability that Jamie hits the target for the first time on his $n^{th}$ attempt is $\frac{64}{2187}$. 3. The formula for the probability of hitting the target for the first time on the $n^{th}$ attempt in a geometric distribution is: $$P(X=n) = (1-p)^{n-1} p$$ where $p$ is the probability of success (hitting the target) and $1-p$ is the probability of failure (missing). 4. Here, $p = \frac{1}{3}$, so $1-p = \frac{2}{3}$. 5. Substitute the values into the formula: $$\left(\frac{2}{3}\right)^{n-1} \times \frac{1}{3} = \frac{64}{2187}$$ 6. Multiply both sides by 3 to isolate the power term: $$\left(\frac{2}{3}\right)^{n-1} = \frac{64}{2187} \times 3 = \frac{192}{2187}$$ 7. Note that $\frac{64}{2187}$ can be rewritten as $\left(\frac{2}{3}\right)^6$ because $64 = 2^6$ and $2187 = 3^7$, but let's check carefully. 8. Actually, $2187 = 3^7$, so $\frac{64}{2187} = \frac{2^6}{3^7}$. 9. From step 5, the equation is: $$\left(\frac{2}{3}\right)^{n-1} \times \frac{1}{3} = \frac{2^6}{3^7}$$ 10. Multiply both sides by 3: $$\left(\frac{2}{3}\right)^{n-1} = \frac{2^6}{3^6} = \left(\frac{2}{3}\right)^6$$ 11. Since the bases are the same, equate the exponents: $$n - 1 = 6$$ 12. Solve for $n$: $$n = 7$$ Final answer: $n = 7$