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

# 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

1. A

5

2. B

6

3. C

3

4. D

4

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### 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

$1{-}^{n}{C}_{0}\left(p{\right)}^{0}\left(q{\right)}^{n}>\frac{5}{6}⇒1-1{\left(\frac{1}{3}\right)}^{0}{\left(\frac{2}{3}\right)}^{n}>\frac{5}{6}⇒\frac{1}{6}>{\left(\frac{2}{3}\right)}^{n}$

which is possible only when we take

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