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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 that \frac{5}{6}, is:

  • Option 1)

    6

  • Option 2)

    5

  • Option 3)

    4

  • Option 4)

    3

Answers (1)

best_answer

 

Binomial Theorem on Probability -

If an experiment is repeated n times under similar conditions we say that n trials of the experiment have been made.

Let E be an event.

P= the Probability of occurrence of event E in one trial.

q = 1- p = probability of non occurrence of event E in one trial such that p+q = 1

x= number of successes.

-

 

 

Binomial Theorem on Probability -

Then

\dpi{100} P\left ( X= r \right )     or   P(r)

=\ ^{n}C_{\Upsilon }\cdot P^{r }\cdot q^{n-r}

-

From the concept

1-_{0}^{n}\textrm{C}(\frac{1}{3})^{0}(\frac{2}{3})^{n}>\frac{5}{6}

\frac{1}{6}>(\frac{2}{3})^{n}

=>0.16>(\frac{2}{3})^{n}

n_{min}=5

 

 


Option 1)

6

Option 2)

5

Option 3)

4

Option 4)

3

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