The probability of a man hitting a target is | vedclass

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Question

(B) Let $n$ be the number of times he fires at the target. Let $X$ be the number of times he hits the target. Since hitting the target is a Bernoulli

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Let $n$ be the number of times he fires at the target. Let $X$ be the number of times he hits the target. Since hitting the target is a Bernoulli trial,$X$ follows a binomial distribution. Here,$p = \frac{2}{3}$ (probability of hitting) and $q = 1 - p = \frac{1}{3}$ (probability of missing). The probability of hitting the target at least once is given by $P(X \geq 1) = 1 - P(X = 0)$. We are given that $P(X \geq 1) > 90 \%$,which means $P(X \geq 1) > 0.9$. Therefore,$1 - P(X = 0) > 0.9$. Since $P(X = 0) = {}^nC_0 q^n p^0 = (\frac{1}{3})^n$,we have: $1 - (\frac{1}{3})^n > 0.9$ $0.1 > (\frac{1}{3})^n$ $(\frac{1}{3})^n $3^n > 10$. For $n = 1$,$3^1 = 3 For $n = 2$,$3^2 = 9 For $n = 3$,$3^3 = 27 > 10$. Thus,the minimum number of times he must fire is $3$.

The probability of a man hitting a target is $\frac{2}{3}$. What is the minimum number of times he must fire so that the probability of hitting the target at least once is more than $90 \%$?

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