The probability of a man hitting a target is 110The least number of shots required so that the probability of his hitting the target at least once is greater than 14 is

Solution:

Let X denote the number of shots hitting the target in n trials. Then, X follows binomial distribution with $p = \frac{1}{10}$

and $q = \frac{9}{10}$ .

It is given that

$\begin{array}{l} P (X = r) =^{n} C_{r} {(\frac{1}{10})}^{r} {(\frac{9}{10})}^{n - r}, r = 0, 1, 2, \dots, n \\ P (X \geq 1) > \frac{1}{4} \\ \Rightarrow 1 - P (X = 0) > \frac{1}{4} \\ \Rightarrow 1 -^{n} C_{0} {(\frac{9}{10})}^{n} > \frac{1}{4} \\ \Rightarrow {(\frac{9}{10})}^{n} < \frac{3}{4} \Rightarrow (0.9)^{n} < 0.75 \Rightarrow n = 3, 4, 5, \dots . \end{array}$

Hence n=3

The probability of a man hitting a target is $\frac{1}{10}$ The least number of shots required so that the probability of his hitting the target at least once is greater than $\frac{1}{4}$ is

Solution:

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