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  • The probability of a man hitting a target is   . He fires at the target  <img

The probability of a man hitting a target is  \frac{2}{5} . He fires at the target  k times, ( k , a given number ) . Then the minimun k , so that the probability of hitting the target at least once is more than \frac{7}{10} , is : 

Option: 1

3


Option: 2

5


Option: 3

2


Option: 4

4


Answers (1)

best_answer

\\ \frac{2}{5}+\frac{3}{5} \times \frac{2}{5}+\left(\frac{3}{5}\right)^{2} \times \frac{2}{5}+\ldots . .+\left(\frac{3}{5}\right)^{k} \cdot \frac{2}{5}>\frac{7}{10} \\ \Rightarrow \frac{2}{5}\left[1+\frac{3}{5}+\left(\frac{3}{5}\right)^{2}+\ldots \ldots+\left(\frac{3}{5}\right)^{k}\right]>\frac{7}{10} \\ \Rightarrow \quad \frac{2}{5} \times \frac{1-\left(\frac{3}{5}\right)^{k}}{1-\frac{3}{5}}>\frac{7}{10} \\\Rightarrow 1-\left(\frac{3}{5}\right)^{k}>\frac{7}{10} \\ \Rightarrow\left(\frac{3}{5}\right)^{k}<\frac{3}{10} \Rightarrow k \geq 3

Hence, minimum value of k = 3.

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chirag

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