1. **Problem Statement:** Jamie has a probability of hitting a target as $\frac{1}{3}$. We want to find the attempt number $n$ when he hits the target for the first time, given the probability of this event is $\frac{64}{2187}$. 2. **Formula Used:** The probability that the first success occurs on the $n$th attempt in a geometric distribution is given by: $$P(X=n) = (1-p)^{n-1} p$$ where $p$ is the probability of success on each attempt. 3. **Apply the values:** Here, $p = \frac{1}{3}$, so the probability of failure is $1-p = \frac{2}{3}$. 4. **Set up the equation:** $$\left(\frac{2}{3}\right)^{n-1} \times \frac{1}{3} = \frac{64}{2187}$$ 5. **Isolate the exponential term:** $$\left(\frac{2}{3}\right)^{n-1} = \frac{64}{2187} \times 3 = \frac{192}{2187}$$ 6. **Simplify the right side:** Note that $64 = 2^6$ and $2187 = 3^7$, so: $$\frac{192}{2187} = \frac{64 \times 3}{3^7} = \frac{2^6 \times 3}{3^7} = 2^6 \times 3^{1-7} = 2^6 \times 3^{-6}$$ 7. **Rewrite the left side:** $$\left(\frac{2}{3}\right)^{n-1} = 2^{n-1} \times 3^{-(n-1)}$$ 8. **Equate powers:** $$2^{n-1} \times 3^{-(n-1)} = 2^6 \times 3^{-6}$$ 9. **Match exponents for each base:** For base 2: $n-1 = 6$ For base 3: $-(n-1) = -6$ Both give $n-1 = 6$. 10. **Solve for $n$:** $$n = 6 + 1 = 7$$ **Final answer:** $n = 7$