The probability of a man hitting a target is | vedclass

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(B) Let $n$ be the number of shots fired.The probability of hitting the target in a single shot is $p = \frac{1}{10}$.The probability of missing the t

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Let $n$ be the number of shots fired.The probability of hitting the target in a single shot is $p = \frac{1}{10}$.The probability of missing the target in a single shot is $q = 1 - p = 1 - \frac{1}{10} = \frac{9}{10}$.The probability of missing the target in all $n$ shots is $q^n = \left(\frac{9}{10}ight)^n$.The probability of hitting the target at least once is $1 - P(	ext{missing all}) = 1 - \left(\frac{9}{10}ight)^n$.We are given that this probability is greater than $\frac{1}{4}$:$1 - \left(\frac{9}{10}ight)^n > \frac{1}{4}$$\Rightarrow \frac{3}{4} > \left(\frac{9}{10}ight)^n$.For $n = 1$: $\left(\frac{9}{10}ight)^1 = 0.9 > 0.75$ (False).For $n = 2$: $\left(\frac{9}{10}ight)^2 = 0.81 > 0.75$ (False).For $n = 3$: $\left(\frac{9}{10}ight)^3 = 0.729 Thus,the least number of shots required is $3$.

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

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