A rifle man is firing at a distant target and | vedclass

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(A) Let $p$ be the probability of hitting the target in one shot,so $p = \frac{10}{100} = 0.1$.Let $q$ be the probability of missing the target in one

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(A) Let $p$ be the probability of hitting the target in one shot,so $p = \frac{10}{100} = 0.1$.Let $q$ be the probability of missing the target in one shot,so $q = 1 - p = 0.9$.Let $n$ be the number of rounds fired.The probability of hitting the target at least once in $n$ shots is given by $P(	ext{at least one hit}) = 1 - P(	ext{no hits}) = 1 - q^n$.We want this probability to be at least $50\%$,so $1 - (0.9)^n \ge 0.5$.This simplifies to $(0.9)^n \le 0.5$.Taking the logarithm on both sides: $n \log(0.9) \le \log(0.5)$.Since $\log(0.9)$ is negative,the inequality sign flips: $n \ge \frac{\log(0.5)}{\log(0.9)}$.Using $\log(0.5) \approx -0.3010$ and $\log(0.9) \approx -0.04576$,we get $n \ge \frac{-0.3010}{-0.04576} \approx 6.57$.Since $n$ must be an integer,the minimum number of rounds is $7$.

$A$ rifle man is firing at a distant target and has only $10\%$ chance of hitting it. The minimum number of rounds he must fire in order to have $50\%$ chance of hitting it at least once is

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