The probability of a shooter hitting a target | vedclass

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Question

(C) Let the shooter fire $n$ times. These $n$ fires are $n$ Bernoulli trials. In each trial,the probability of hitting the target is $p = \frac{3}{4}$

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Let the shooter fire $n$ times. These $n$ fires are $n$ Bernoulli trials. In each trial,the probability of hitting the target is $p = \frac{3}{4}$ and the probability of not hitting the target is $q = 1 - p = \frac{1}{4}$. The probability of hitting the target at least once is given by $P(X \geq 1) = 1 - P(X = 0)$. Here,$P(X = 0)$ is the probability of not hitting the target in any of the $n$ trials,which is $q^n = (\frac{1}{4})^n$. We are given that $P(X \geq 1) > 0.99$. Therefore,$1 - (\frac{1}{4})^n > 0.99$. This simplifies to $1 - 0.99 > (\frac{1}{4})^n$,which means $0.01 > (\frac{1}{4})^n$. This is equivalent to $\frac{1}{100} > \frac{1}{4^n}$,or $4^n > 100$. Testing values for $n$: For $n = 3$,$4^3 = 64 For $n = 4$,$4^4 = 256 > 100$. Thus,the minimum number of times the shooter must fire is $4$.

The probability of a shooter hitting a target is $\frac{3}{4}$. What is the minimum number of times he/she must fire so that the probability of hitting the target at least once is greater than $0.99$?

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