In a binomial distribution B(n, p = frac{1}{4 | vedclass

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(A) The probability of at least one success is given by $P(X \geq 1) = 1 - P(X = 0)$.In a binomial distribution $B(n, p)$,$P(X = 0) = q^n$,where $q =

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$\frac{1}{\log_{10} 4 - \log_{10} 3}$

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(A) The probability of at least one success is given by $P(X \geq 1) = 1 - P(X = 0)$.In a binomial distribution $B(n, p)$,$P(X = 0) = q^n$,where $q = 1 - p$.Given $p = \frac{1}{4}$,we have $q = 1 - \frac{1}{4} = \frac{3}{4}$.The condition is $1 - (\frac{3}{4})^n \geq \frac{9}{10}$.Rearranging the inequality,we get $1 - \frac{9}{10} \geq (\frac{3}{4})^n$,which simplifies to $\frac{1}{10} \geq (\frac{3}{4})^n$.Taking the logarithm with base $10$ on both sides: $\log_{10}(\frac{1}{10}) \geq \log_{10}((\frac{3}{4})^n)$.$-1 \geq n(\log_{10} 3 - \log_{10} 4)$.Multiplying by $-1$ reverses the inequality sign: $1 \leq n(\log_{10} 4 - \log_{10} 3)$.Therefore,$n \geq \frac{1}{\log_{10} 4 - \log_{10} 3}$.

In a binomial distribution $B(n, p = \frac{1}{4})$,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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