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(A) Let $X$ be the number of correct answers. Since the student guesses,$X$ follows a binomial distribution $B(n=8, p=1/2)$.We want to find the smalle

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$5$

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(A) Let $X$ be the number of correct answers. Since the student guesses,$X$ follows a binomial distribution $B(n=8, p=1/2)$.We want to find the smallest $n$ such that $P(X \geq n) $P(X \geq n) = \sum_{r=n}^{8} {}^{8}C_{r} (\frac{1}{2})^{r} (\frac{1}{2})^{8-r} = \frac{1}{2^{8}} \sum_{r=n}^{8} {}^{8}C_{r} $\sum_{r=n}^{8} {}^{8}C_{r} We know that $\sum_{r=0}^{8} {}^{8}C_{r} = 2^{8} = 256$.So,$\sum_{r=n}^{8} {}^{8}C_{r} = 256 - \sum_{r=0}^{n-1} {}^{8}C_{r} $\sum_{r=0}^{n-1} {}^{8}C_{r} > 128$.For $n=5$,$\sum_{r=0}^{4} {}^{8}C_{r} = {}^{8}C_{0} + {}^{8}C_{1} + {}^{8}C_{2} + {}^{8}C_{3} + {}^{8}C_{4} = 1 + 8 + 28 + 56 + 70 = 163$.Since $163 > 128$,the condition holds for $n=5$.For $n=4$,$\sum_{r=0}^{3} {}^{8}C_{r} = 1 + 8 + 28 + 56 = 93$,which is not greater than $128$.Thus,the smallest value of $n$ is $5$.

$A$ student appeared in an examination consisting of $8$ true-false type questions. The student guesses the answers with equal probability. The smallest value of $n$,so that the probability of guessing at least $n$ correct answers is less than $\frac{1}{2}$,is:

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