1. The problem states that Jamie has a probability of $\frac{1}{3}$ to hit a 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 that the first success occurs on the $n^{th}$ trial in a sequence of independent trials with success probability $p$ is: $$P(X=n) = (1-p)^{n-1} p$$ 4. Here, $p = \frac{1}{3}$, so the probability is: $$P(X=n) = \left(1 - \frac{1}{3}\right)^{n-1} \times \frac{1}{3} = \left(\frac{2}{3}\right)^{n-1} \times \frac{1}{3}$$ 5. We set this equal to the given probability: $$\left(\frac{2}{3}\right)^{n-1} \times \frac{1}{3} = \frac{64}{2187}$$ 6. Multiply both sides by 3: $$\left(\frac{2}{3}\right)^{n-1} = \frac{64}{2187} \times 3 = \frac{192}{2187}$$ 7. Note that $64 = 2^6$ and $2187 = 3^7$, so: $$\frac{192}{2187} = \frac{64 \times 3}{3^7} = \frac{2^6 \times 3}{3^7} = 2^6 \times 3^{1-7} = 2^6 \times 3^{-6}$$ 8. Also, rewrite the left side: $$\left(\frac{2}{3}\right)^{n-1} = 2^{n-1} \times 3^{-(n-1)}$$ 9. Equate the powers of 2 and 3: $$2^{n-1} = 2^6 \implies n-1 = 6 \implies n = 7$$ $$3^{-(n-1)} = 3^{-6} \implies n-1 = 6 \implies n = 7$$ 10. Both bases agree on $n=7$, so the value of $n$ is 7. **Final answer:** $$n = 7$$