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(C) Let $n$ be the number of independent shots.The probability of hitting the target in a single shot is $p = \frac{1}{3}$.The probability of missing

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$5$

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(C) Let $n$ be the number of independent shots.The probability of hitting the target in a single shot is $p = \frac{1}{3}$.The probability of missing the target in a single shot is $q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$.The probability of missing the target in all $n$ shots is $q^n = \left(\frac{2}{3}ight)^n$.The probability of hitting the target at least once is $1 - P(	ext{missing all}) = 1 - \left(\frac{2}{3}ight)^n$.We are given that this probability must be greater than $\frac{5}{6}$:$1 - \left(\frac{2}{3}ight)^n > \frac{5}{6}$$\left(\frac{2}{3}ight)^n $\left(\frac{2}{3}ight)^n Now,we test values for $n$:For $n = 3$: $\left(\frac{2}{3}ight)^3 = \frac{8}{27} \approx 0.296 > 0.166$For $n = 4$: $\left(\frac{2}{3}ight)^4 = \frac{16}{81} \approx 0.197 > 0.166$For $n = 5$: $\left(\frac{2}{3}ight)^5 = \frac{32}{243} \approx 0.131 Thus,the minimum number of shots required is $n = 5$.

If the probability of hitting a target by a shooter,in any shot,is $\frac{1}{3}$,then the minimum number of independent shots at the target required by him so that the probability of hitting the target at least once is greater than $\frac{5}{6}$,is

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