In a binomial distribution B(n, p = 1/4),if t | vedclass

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(C) The probability of at least one success is given by $P(X \geq 1) = 1 - P(X = 0)$.For a binomial distribution,$P(X = 0) = {}^nC_0 p^0 q^n = q^n$,wh

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

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(C) The probability of at least one success is given by $P(X \geq 1) = 1 - P(X = 0)$.For a binomial distribution,$P(X = 0) = {}^nC_0 p^0 q^n = q^n$,where $q = 1 - p = 1 - 1/4 = 3/4$.We are given $P(X \geq 1) \geq 9/10$,which implies $1 - (3/4)^n \geq 9/10$.Rearranging the inequality,we get $1 - 9/10 \geq (3/4)^n$,so $(3/4)^n \leq 1/10$.Taking the logarithm with base $10$ on both sides,we get $n \log_{10}(3/4) \leq \log_{10}(1/10)$.Since $\log_{10}(1/10) = -1$,we have $n(\log_{10} 3 - \log_{10} 4) \leq -1$.Multiplying by $-1$ reverses the inequality sign: $n(\log_{10} 4 - \log_{10} 3) \geq 1$.Thus,$n \geq \frac{1}{\log_{10} 4 - \log_{10} 3}$.

In a binomial distribution $B(n, p = 1/4)$,if the probability of at least one success is $\geq 9/10$,then $n \geq$ ?

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