📚 The Geometric Distribution | 几何分布
The geometric distribution is a discrete probability distribution that models the number of independent Bernoulli trials needed to achieve the first success. It appears frequently in A-Level Statistics, particularly within the Edexcel specification, where it is used to solve problems involving waiting times, quality control, and games of chance.
几何分布是一种离散概率分布，用于描述在独立的伯努利试验中获得第一次成功所需的试验次数。它在A-Level统计中频繁出现，尤其符合Edexcel考试大纲的要求，常用于解决等待时间、质量控制和机会游戏等问题。
1. What is the Geometric Distribution? | 什么是几何分布？
A geometric distribution arises when we repeat independent trials, each with the same probability of success p, and count how many trials it takes until the first success occurs. The trials must be identical, independent, and have only two possible outcomes: success or failure.
当我们重复进行独立试验，每次试验的成功概率均为 p，并记录第一次成功出现时所进行的试验次数时，就产生了几何分布。这些试验必须相同、独立，并且只有两种可能的结果：成功或失败。
The random variable X denotes the trial number on which the first success is observed. Therefore, X can take values 1, 2, 3, … . Note that the Edexcel specification always uses this definition, where the count includes the successful trial.
随机变量 X 表示首次观察到成功时的试验编号。因此 X 可以取 1, 2, 3, … 这些值。值得注意的是，Edexcel 大纲始终使用这一定义，即计数包含成功的那一次试验。
2. Defining Conditions | 定义条件
For a random variable to follow a geometric distribution, four conditions must be satisfied:
要使一个随机变量服从几何分布，必须满足四个条件：
- Fixed probability of success: The probability of success, p, remains constant for every trial.
- 固定成功概率：每次试验的成功概率 p 保持不变。
- Independent trials: The outcome of any trial does not influence any other trial.
- 独立试验：任何一次试验的结果不会影响其他试验。
- Two outcomes: Each trial results in either 'success' or 'failure'.
- 两种结果：每次试验的结果只能是“成功”或“失败”。
- Count until first success: The variable measures the number of trials up to and including the first success.
- 计数至首次成功：变量测量的是直到并包含第一次成功在内的试验次数。
If these conditions hold, we can write X ~ Geo(p), where p is the probability of success on a single trial.
如果这些条件成立，我们可以记作 X ~ Geo(p)，其中 p 是单次试验的成功概率。
3. Notation and Parameters | 记号与参数
In the Edexcel A-Level course, the geometric distribution is written as X ~ fy 0 < p ≤ 1. When p = 1, the first trial is always a success, so X = 1 with certainty.
在 Edexcel A-Level 课程中，几何分布记作 X ~ Geo(p)。参数 p 表示每次独立试验的成功概率，必须满足 0 < p ≤ 1。当 p = 1 时，第一次试验总会成功，因此 X 必然等于 1。
o represent the probability of failure. This can simplify expressions such as P(X = x) = qx−1 p.
很多教材题目使用记号 q = 1 − p 表示失败的概率。这可以简化诸如 P(X = x) = qx−1 p 的表达式。
4. Probability Mass Function | 概率质量函数
The probability that the first success occurs on the x-th trial is given by the probability mass function (PMF):
第一次成功出现在第 x 次试验的概率由概率质量函数给出：
P(X = x) = (1 − p)x−1 p for x = 1, 2, 3, …
This formula arises because we must have (x − 1) consecutive failures, each with probability (1 − p), followed by one success with probability p.
这个公式之所以成立，是因为我们必须先经历 (x − 1) 次连续失败，每次失败的概率为 (1 − p)，然后出现一次成功，概率为 p。
For example, if p = 0.3, then P(X = 4) = (0.7)3 The distribution is strictly decreasing: larger values of x become less and less probable.
例如，若 p = 0.3，则 P(X = 4) = (0.7)3 × 0.3 = 0.1029。该分布严格递减：越大的 x 值出现的概率越小。
5. Cumulative Distribution Function | 累积分布函数
The cumulative distribution function (CDF) gives the probability that the first success occurs on or before trial x:
累积分布函数给出了第一次成功发生在第 x 次试验或之前的概率：
P(X ≤ x) = 1 − (1 − p)x
To see why, note that the event X ≤ x is the complement of 'no success in the first x trials'. The probability of x consecutive failures is (1 − p)x, so subtracting from 1 gives the desired result.
要理解这个公式，注意事件 X ≤ x 是“前 x 次试验中没有成功”这一事件的对立事件。连续 x 次失败的概率为 (1 − p)x，因此用 1 减去它就得到了所需的结果。
DF formula is extremely useful for calculating probabilities such as P(X > x) = (1 − p)x and P(a < X ≤ b) = P(X ≤ b) − P(X ≤ a).
这个累积分布公式对于计算诸如 P(X > x) = (1 − p)x 以及 P(a < X ≤ b) = P(X ≤ b) − P(X ≤ a) 的概率非常有用。
6. Expected Value (Mean) | 期望值（均值）
The expected number of trials to achieve the first success, E(X), is a measure of the long-run average. For X ~ Geo(p):
获得第一次成功所需的期望试验次数 E(X) 衡量的是长期平均结果。对于 X ~ Geo(p)：
E(X) = 1 ⁄ p
instance, if p = 0.2, we would on average need 1/0.2 = 5 trials to see the first success.
这在直观上是合理的：如果成功概率较低，我们预计会等待更长时间。例如，若 p = 0.2，我们平均需要 1/0.2 = 5 次试验才能看到第一次成功。
Exam questions often ask candidates to find the mean number of trials in context, such as the expected number of rolls of a fair die until a six appears, which would be 6.
考试题目经常要求考生根据背景求出平均试验次数，例如投掷一枚均匀骰子直到出现六点的期望次数为 6。
7. Variance and Standard Deviation | 方差与标准差
The variance of a geometric random variable measures the spread of the distribution:
几何随机变量的方差衡量分布的离散程度：
Var(X) = (1 − p) ⁄ p2
Consequently, the standard deviation is σ = √[(1 − p) / p2]. Notice that as p decreases, both the mean and the variance increase, reflecting greater uncertainty about when the first success will occur.
因此，标准差为 σ = √[(1 − p) / p2]。注意，当 p 减小时，均值和方差都会增大，这反映了第一次成功何时发生的不确定性更大。
In applied problems, you may need to use Var(X) = E(X2) − [E(X)]2 to find the variance from given summary statistics, but the formula above is the direct result for the geometric model.
在应用问题中，你可能需要使用 Var(X) = E(X2) − [E(X)]2 根据给定的汇总统计量计算方差，但上述公式是几何模型的直接结果。
8. Mode of the Geometric Distribution | 几何分布的众数
ying by (1 − p) reduces the probability, meaning P(X = 1) > P(X = 2) > P(X = 3) > … . Therefore, the mode is always at x = 1.
对于任何满足 0 < p < 1 的几何分布，其概率质量函数严格递减。因为 (1 − p) < 1，乘以 (1 − p) 会使概率减小，意味着 P(X = 1) > P(X = 2) > P(X = 3) > … 。因此，众数始终在 x = 1 处。
obabilities are zero, so the mode is still 1. This simple property can help check calculations: if you obtain a mode other than 1 for a standard geometric model, you have made a mistake.
唯一的例外是平凡情形 p = 1，此时 P(X = 1) = 1，其他概率均为 0，因此众数仍然是 1。这个简单的性质可以帮助检查计算：如果对标准几何模型得出的众数不是 1，那么你一定出错了。
9. Memoryless Property | 无记忆性
The geometric distribution possesses a unique property among discrete distributions: it is memoryless. Formally, for non-negative integers m and n:
几何分布在离散分布中具有一个独特的性质：无记忆性。形式上，对于非负整数 m 和 n：
P(X > m + n | X > m) = P(X > n)
This means that if we have already waited m trials without success, the probability of waiting a further n trials is the same as if we were starting from scratch. The past waiting time provides no information about future waiting time.
这意味着，如果我们已经等待了 m 次试验而没有成功，那么再等待 n 次试验的概率与从头开始等待 n 次的概率相同。过去的等待时间不提供关于未来等待时间的任何信息。
To prove this, recall that P(X > k) = (1 − p)k. Then P(X > m + n | X > m) = P(X > m + n) / P(X > m) = (1 − p)m+n / (1 − p)m = (1 − p)n = P(X > n).
为证明这一点，回顾 P(X > k) = (1 − p)k。那么 P(X > m + n | X > m) = P(X > m + n) / P(X > m) = (1 − p)m+n / (1 − p)m = (1 − p)n = P(X > n)。
This property is often tested by asking 'Given that a machine has not broken down in the first 10 days, what is the probability it will last at least 5 more days?' when the daily breakdown probability is p.
这一性质常被考查，例如问到：“已知一台机器在前 10 天没有发生故障，那么它至少还能再运行 5 天的概率是多少？”其中每日故障概率为 p。
10. Worked Examples | 例题解析
Example 1 (Basic PMF): repeatedly until a six appears. Let X be the number of rolls. Find P(X = 3) and P(X ≤ 3).
例1（基础概率质量函数）：反复投掷一枚骰子，直到出现六点为止。设 X 为投掷次数。求 P(X = 3) 和 P(X ≤ 3)。
Solution: Here p = 1/6. Using the PMF, P(X = 3) = (5/6)2(1/6) = 25/216 ≈ 0.1157. Using the CDF, P(X ≤ 3) = 1 − (5/6)3 = 1 − 125/216 = 91/216 ≈ 0.4213.
解：这里 p = 1/6。利用概率质量函数，P(X = 3) = (5/6)2(1/6) = 25/216 ≈ 0.1157。利用累积分布函数，P(X ≤ 3) = 1 − (5/6)3 = 1 − 125/216 = 91/216 ≈ 0.4213。
Example 2 (Mean and variance): In a board game, a player needs to roll a double to start. The probability of rolling a double with two dice is 1/6. How many rolls are expected before the player starts? What is the standard deviation?
例2（均值与方差）：在一个棋盘游戏中，玩家需要掷出对子才能出发。两个骰子掷出对子的概率为 1/6。在出发前预期需要掷多少次？标准差是多少？
Solution: X ~ Geo(1/6). Expected number of rolls E(X) = 1/(1/6) = 6. Variance Var(X) = (5/6) / (1/6)2 = (5/6) / (1/36) = 30, so standard deviation = √30 ≈ 5.48 rolls.
解：X ~ Geo(1/6)。期望掷骰次数 E(X) = 1/(1/6) = 6。方差 Var(X) = (5/6) / (1/6)2 = (5/6) / (1/36) = 30，因此标准差 = √30 ≈ 5.48 次。
11. Key Points and Exam Tips | 重点与考试技巧
When answering Edexcel exam questions on the geometric distribution, keep the following in mind:
在回答 Edexcel 考试中关于几何分布的题目时，请牢记以下几点：
- Always state the distribution clearly: Begin by writing 'Let X be the number of trials until the first success, so X ~ Geo(p)'. This clarifies your working and often earns the first mark.
- 明确写出分布：开始时写下“设 X 为直到第一次成功的试验次数，因此 X ~ Geo(p)”。这能清晰展示你的解题步骤，通常能拿到第一个分数。
- Check the definition: The Edexcel geometric distribution counts the trial on which the first success occurs. Do not use the alternative 'number of failures' model unless the question explicitly defines it that way.
- 确认定义：Edexcel 的几何分布计数的是第一次成功发生的那次试验。除非题目明确说明，否则不要使用“失败次数”的替代模型。
- Use the CDF for inequalities: For P(X ≤ k) or P(X > k), the formulae 1 − (1 − p)k and (1 − p)k are much faster than summing individual probabilities.
- 使用累积分布函数处理不等式：对于 P(X ≤ k) 或 P(X > k)，使用公式 1 − (1 − p)k 和 (1 − p)k 比逐个概率相加要快得多。
- Remember the memoryless property: If a conditional probability involves 'given that X > a', you can reset the count. E.g., P(X > a + b | X > a) = P(X > b).
- 记住无记忆性：如果条件概率中涉及“给定 X > a”，你可以重置计数。例如 P(X > a + b | X > a) = P(X > b)。
- Mean = 1/p, variance = (1 − p)/p2: Write these on your formula sheet if memorisation is a concern, and be ready to interpret them in context.
- 均值 = 1/p，方差 = (1 − p)/p2：如果担心记忆，可以在公式表中记下它们，并做好在情境中解释它们的准备。
- Mode is always 1: This can be used to spot errors if an answer suggests the most likely number of trials is 2 or higher.
- 众数始终为 1：如果答案暗示最可能的试验次数是 2 或更大，那么可以利用这一点发现错误。
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