1. **Problem statement:** We are given that Jamie's probability of hitting the target on any attempt is $\frac{1}{3}$. The probability that he hits the target for the first time on his $n^{th}$ attempt is $\frac{64}{2187}$. We need to find the value of $n$. 2. **Formula used:** The probability that the first success occurs on the $n^{th}$ trial in a sequence of independent Bernoulli trials with success probability $p$ is given by the geometric distribution formula: $$P(X=n) = (1-p)^{n-1} p$$ where $p = \frac{1}{3}$ is the probability of hitting the target. 3. **Apply the formula:** Substitute $p = \frac{1}{3}$ and $P(X=n) = \frac{64}{2187}$: $$\left(1 - \frac{1}{3}\right)^{n-1} \times \frac{1}{3} = \frac{64}{2187}$$ Simplify $1 - \frac{1}{3} = \frac{2}{3}$: $$\left(\frac{2}{3}\right)^{n-1} \times \frac{1}{3} = \frac{64}{2187}$$ 4. **Isolate the exponential term:** Multiply both sides by 3: $$\left(\frac{2}{3}\right)^{n-1} = \frac{64}{2187} \times 3 = \frac{192}{2187}$$ 5. **Express numerator and denominator as powers:** Note that $64 = 2^6$ and $2187 = 3^7$, so: $$\frac{192}{2187} = \frac{64 \times 3}{3^7} = \frac{2^6 \times 3}{3^7} = 2^6 \times 3^{1-7} = 2^6 \times 3^{-6}$$ Rewrite the left side: $$\left(\frac{2}{3}\right)^{n-1} = 2^{n-1} \times 3^{-(n-1)}$$ 6. **Set the exponents equal:** Since bases are the same, equate powers: $$2^{n-1} \times 3^{-(n-1)} = 2^6 \times 3^{-6}$$ This implies: $$n - 1 = 6$$ 7. **Solve for $n$:** $$n = 6 + 1 = 7$$ **Final answer:** $n = 7$