Let X be the number of successes in n indepen | vedclass

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(A) We have $p = \frac{3}{4}$,so $q = 1 - p = \frac{1}{4}$.It is given that $P(X \ge 1) \ge 0.9375$.Using the complement rule,$P(X \ge 1) = 1 - P(X =

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$2$

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(A) We have $p = \frac{3}{4}$,so $q = 1 - p = \frac{1}{4}$.It is given that $P(X \ge 1) \ge 0.9375$.Using the complement rule,$P(X \ge 1) = 1 - P(X = 0)$.Thus,$1 - P(X = 0) \ge 0.9375$.$1 - ^nC_0 (p^0) (q)^n \ge 0.9375$.$1 - (\frac{1}{4})^n \ge 0.9375$.$1 - 0.9375 \ge (\frac{1}{4})^n$.$0.0625 \ge (\frac{1}{4})^n$.Since $0.0625 = \frac{625}{10000} = \frac{1}{16}$,we have $\frac{1}{16} \ge (\frac{1}{4})^n$.$(\frac{1}{4})^2 \ge (\frac{1}{4})^n$.Since the base is less than $1$,the inequality reverses for the exponents: $n \ge 2$.The least value of $n$ is $2$.

Let $X$ be the number of successes in $n$ independent Bernoulli trials with probability of success $p = \frac{3}{4}$. The least value of $n$ so that $P(X \ge 1) \ge 0.9375$ is . . . . . .

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