To find the value of \( n \) where the probability that Jamie hits the target for the first time on his \( n \)th attempt is \( \frac{64}{2187} \), we need to understand the conditions of hitting a target for the first time.
The probability that Jamie hits the target is \( p = \frac{1}{3} \). The probability that he misses the target is \( q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3} \).
The probability that he hits the target for the first time on his \( n \)th attempt can be expressed as:
\[ P(\text{hits first time on } n\text{th attempt}) = q^{n-1} \cdot p \]
Now substituting \( q \) and \( p \):
\[ P(\text{hits first time on } n\text{th attempt}) = \left( \frac{2}{3} \right)^{n-1} \cdot \frac{1}{3} \]
This must equal \( \frac{64}{2187} \):
\[ \left( \frac{2}{3} \right)^{n-1} \cdot \frac{1}{3} = \frac{64}{2187} \]
To isolate \( \left( \frac{2}{3} \right)^{n-1} \), we multiply both sides by \( 3 \):
\[ \left( \frac{2}{3} \right)^{n-1} = \frac{64 \cdot 3}{2187} \]
Calculating \( \frac{64 \cdot 3}{2187} \):
\[ \frac{192}{2187} \]
Next, we can express \( 2187 \) as \( 3^7 \) since \( 2187 = 3^7 \):
\[ 192 = 64 \cdot 3 = 2^6 \cdot 3^1 \]
Thus, we have:
\[ \left( \frac{2}{3} \right)^{n-1} = \frac{2^6 \cdot 3^1}{3^7} = \frac{2^6}{3^6} = \left( \frac{2}{3} \right)^6 \]
From this, we can set up the equation:
\[ \frac{2}{3}^{n-1} = \frac{2}{3}^{6} \]
By comparing exponents:
\[ n - 1 = 6 \implies n = 7 \]
Thus, the value of \( n \) is:
\[ \boxed{7} \]
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