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

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Detailed Solution

Given, $p = \frac{1}{3}$ , so $q = 1 - \frac{1}{3} = \frac{2}{3}$

Let $n$ be the minimum number of shots required. As per given condition, we have

$1 -^{n} C_{0} (p)^{0} (q)^{n} > \frac{5}{6} \Rightarrow 1 - 1 {(\frac{1}{3})}^{0} {(\frac{2}{3})}^{n} > \frac{5}{6} \Rightarrow \frac{1}{6} > {(\frac{2}{3})}^{n}$

$\Rightarrow p .1666 > {(\frac{2}{3})}^{n},$ which is possible only when we take

$n = 5, 6, . . .$

$∴ Minimum value of n = 5$

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