区间 0.5–0.55

设 $X_1,\\ldots,X_n\\stackrel{iid}{\\sim}U(0,1)$。命中区间 $[0.5,0.55]$ 的概率为 $0.05$。\n\n命中至少一次的概率为 $1-0.95^n$，要求\n\n$$\n1-0.95^n\\ge 0.95\n\\Rightarrow 0.95^n\\le 0.05\n\\Rightarrow n\\ge \\frac{\\ln 0.05}{\\ln 0.95}\\approx 58.4.\n$$\n\n因此最小整数为\n\n$$\n\\boxed{n=59}.\n$$

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英文解析

Let $X_{1}, X_{2}, \ldots , X_{n}$ be independent random variables uniformly distributed on [0, 1]. Denote by $A_{n}$ the event that at least one of $X_{1}, X_{2}, \ldots , X_{n}$ belongs to [0.5, 0.55]. Then, the complement $A_{n}^{C}$ of $A$ is

$$A_{n}^{C} = \bigcap_{i = 1}^{n} \{X_{i} \notin [0.5, 0.55]\}$$

Note that

$$\mathbb{P}\left(A_{n}^{C}\right) = \prod_{i = 1}^{n} \mathbb{P}\left(X_{i} \notin [0.5, 0.55]\right) = 0.95^{n}$$

since $X_{1}, X_{2}, \ldots , X_{n}$ are independent. Then,

$$\mathbb{P}\left(A_{n}\right) = 1 - 0.95^{n}$$

and therefore

$$\begin{array}{c}\mathbb{P}\left(A_{n}\right) \geq 0.95 \Longleftrightarrow 0.05 \geq 0.95^{n} \\ \Longleftrightarrow n \geq \frac{\ln 0.05}{\ln 0.95} \approx 58.40 \end{array}$$

We conclude that at least 59 independent uniform random variables have to be generated in order to ensure with probability $95\%$ that at least one of them is between 0.5 and 0.55.