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In a binomial distribution $B\left(n, p=\frac{1}{4}\right)$, if the probability of at least one success is greater than or equal to $\frac{9}{10}$, then $n$ is greater than
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The correct answer is:
$\frac{1}{\log _{10}{ }^4-\log _{10}{ }^3}$
$\frac{1}{\log _{10}{ }^4-\log _{10}{ }^3}$
$$
1-q^n \geq \frac{9}{10} \Rightarrow\left(\frac{3}{4}\right)^n \leq \frac{1}{10} \Rightarrow n \geq-\log _{\frac{3}{4}} 10 \Rightarrow n \geq \frac{1}{\log _{10}{ }^4-\log _{10}{ }^3}
$$
1-q^n \geq \frac{9}{10} \Rightarrow\left(\frac{3}{4}\right)^n \leq \frac{1}{10} \Rightarrow n \geq-\log _{\frac{3}{4}} 10 \Rightarrow n \geq \frac{1}{\log _{10}{ }^4-\log _{10}{ }^3}
$$
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