📚 Geometric Distribution | 几何分布
Imagine flipping a fair coin until you see heads for the first time. The number of tosses needed follows a geometric distribution. This discrete probability distribution models the waiting time until the first success in a sequence of independent Bernoulli trials. It is widely used in reliability engineering, quality control, and any scenario where we count trials until an event of interest occurs. Understanding the geometric distribution deepens your ability to analyse random phenomena with memoryless properties.
想象一下你不断抛一枚公平的硬币，直到第一次出现正面。所需的抛掷次数就服从几何分布。这种离散概率分布用于建模在一连串独立的伯努利试验中，等待第一次成功出现所需的试验次数。它被广泛应用于可靠性工程、质量控制和任何需要计算直到关注事件发生所需试验次数的场景。掌握几何分布能加深你对具有无记忆性质的随机现象的分析能力。
1. Introduction to Geometric Distribution | 几何分布导论
The geometric distribution arises from a sequence of independent trials, each having only two possible outcomes, usually called success and failure, with the probability of success constant at p. The random variable X represents the number of trials needed to obtain the first success. For example, if you repeatedly roll a fair die and wait for a six, the number of rolls until the first six follows a geometric distribution with p = 1/6. The distribution is discrete, with possible values 1, 2, 3, … . It is a fundamental building block of more complex waiting-time models.
几何分布产生于一系列独立试验，每次试验只有两种可能结果，通常称为成功和失败，且成功的概率恒为 p。随机变量 X 表示获得第一次成功所需的试验次数。例如，重复掷一枚公平骰子并等待六点出现，首次掷出六点所需的次数就服从 p = 1/6 的几何分布。该分布是离散的，可取值为 1, 2, 3, … 。它是更复杂的等待时间模型的基本组成部分。
2. Probability Mass Function (PMF) | 概率质量函数
The probability that the first success occurs on the kth trial is given by the product of k−1 failures and the final success. Hence, the probability mass function is:
第一次成功发生在第 k 次试验的概率等于 k−1 次失败与最后一次成功的乘积。因此，概率质量函数为：
P(X = k) = p(1 − p)ᵏ⁻¹, for k = 1, 2, 3, …
Here, p is the probability of success on any single trial, and (1 − p) is the probability of failure. The exponent k−1 the geometric series ∑ₖ₌₁∞ p(1−p)ᵏ⁻¹ = p × 1/(1 − (1−p)) = 1. This PMF is decreasing when p > 0, meaning that earlier successes are always more likely than later ones.
其中 p 是每次试验中成功的概率，(1 − p) 是失败的概率。指数 k−1 计入了成功之前的连续失败次数。所有概率之和为 1，因为等比级数 ∑ₖ₌₁∞ p(1−p)ᵏ⁻¹ = p × 1/(1 − (1−p)) = 1。当 p > 0 时，该概率质量函数是递减的，意味着较早出现成功总是比较晚出现更有可能。
3. Conditions for a Geometric Distribution | 几何分布的必要条件
To apply the geometric distribution correctly, the following conditions must be satisfied:
要正确应用几何分布，必须满足以下条件：
- Binomial trials: Each trial has exactly two outcomes – success or failure. | 二项试验：每次试验只有两个结果——成功或失败。
- Constant probability: The probability of success p remains the same for each trial. | 概率恒定：每次试验的成功概率 p 保持不变。
- Independence: The trials are independent of each other; the outcome of one trial does not affect another. | 独立性：试验相互独立；一次试验的结果不影响其他试验。
- Counting until first success: The random variable counts the number of trials needed to achieve the first success. | 计数至第一次成功：随机变量计入了获得第一次成功所需的试验次数。
If the trials continue after the first success or if failures are counted instead, we obtain a different distribution (such as the negative binomial). It is also essential to define whether the support starts at 0 (number of failures before the first success) or at 1 (trials including the success). In the IB and Cambridge (A Level) syllabuses, the definition with support {1,2,3,…} is standard.
如果试验在第一次成功后继续进行，或者改计失败次数，我们就会得到不同的分布（例如负二项分布）。同样重要的是要明确支撑集是从 0 开始（成功前的失败次数）还是从 1 开始（包含成功那一次）。在 IB 和 Cambridge（A Level）教学大纲中，支撑集为 {1,2,3,…} 的定义是标准形式。
4. Notation and Parameters | 符号与参数
We write X ~ Geo(p) to indicate that the random variable X follows a geometric distribution with success probability p. The parameter p must satisfy 0 < p ≤ 1. If p = 1, success is certain on the first trial and X is always 1; this degenerate case is not usually studied. Common notations used in exam questions include q for the failure probability, where q = 1 − p. Some textbooks denote the distribution as Geo(p) or Geom(p). Always check which variant is expected: number of trials including the success (as we do here) or number of failures before the first success.
我们写成 X ~ Geo(p) 来表示随机变量 X 服从成功概率为 p 的几何分布。参数 p 必须满足 0 < p ≤ 1。如果 p = 1，第一次试验必定成功，此时 X 恒为 1；这个退化情形通常不专门讨论。考题中常见的符号还包括用 q 表示失败概率，其中 q = 1 − p。有些教材将分布记为 Geo(p) 或 Geom(p)。请务必确认题目使用的是哪种定义：是包含成功那一次的试验次数（如本文所用），还是成功之前的失败次数。
5. Cumulative Distribution Function (CDF) | 累积分布函数
The cumulative distribution function gives the probability that the first success occurs on or before the kth trial. It can be derived from the sum of a geometric series:
累积分布函数给出了第一次成功发生在第 k 次试验或之前出现的概率。它可以从等比级数的求和推导出来：
P(X ≤ k) = 1 − (1 − p)ᵏ
This result is particularly useful for finding the probability that it takes more than k trials, because P(X > k) = (1 − p)ᵏ. For instance, if p xaminations, the CDF is often used to solve inequalities or to determine critical regions.
这个结果对于求需要超过 k 次试验的概率特别有用，因为 P(X > k) = (1 − p)ᵏ。例如，若 p = 0.2，需要多于 4 次试验的概率为 (0.8)⁴ ≈ 0.4096。累积分布函数还能快速计算区间概率：P(a < X ≤ b) = P(X ≤ b) − P(X ≤ a)。在考试中，累积分布函数常被用来求解不等式或确定临界区域。
6. Mean and Variance | 期望与方差
The expected value, or mean, of a geometric random variable represents the average number of trials required for the first success. Using the definition with support starting at 1, the mean is:
几何随机变量的期望值或均值代表了获得第一次成功所需的平均试验次数。对支撑集从 1 开始的定义，均值为：
E(X) = 1/p
The variance measures the spread of the distribution and is given by:
方差衡量分布的离散程度，由下式给出：
Var(X) = (1 − p) / p²
These formulas are derived using summation of infinite series. For example, if p = 0.25, then E(X) = 4 and Var(X) = 0.75 / 0.0625 = 12. The standard deviation is √12 reater uncertainty about the waiting time. In IB and Cambridge exams, you may be asked to compute E(X) and Var(X) directly from these formulas or to use them in problem-solving contexts involving the distribution of a sample mean.
这些公式借助无穷级数求和导出。例如，若 p = 0.25，则 E(X) = 4，Var(X) = 0.75 / 0.0625 = 12，标准差为 √12 ≈ 3.46。较大的方差意味着等待时间具有更大的不确定性。在 IB 和 Cambridge 考试中，可能会要求你直接运用这些公式计算期望和方差，或在涉及样本均值分布的问题中使用它们。
7. The Memoryless Property | 无记忆性
A unique feature of the geometric distribution is its memoryless property. This means that the probability of needing a certain number of additional trials, given that no success has occurred so far, does not depend on how many failures have already been observed. Formally:
几何分布的一个独特特征是其无记忆性。这意味着，在迄今为止尚未出现成功的条件下，还需要一定数量额外试验的概率，不依赖于已经观察到了多少次失败。其形式化表达为：
P(X > m + n | X > m) = P(X > n) = (1 − p)ⁿ
In words, the conditional probability that the remaining waiting time exceeds n trials after m consecutive failures is exactly the same as the original probability that the waiting time exceeds n trials from the start. Only the geometric and exponential distributions (its continuous analogue) possess this property among non-negative distributions. This concept is frequently examined: you might be asked to verify the property or to use it to simplify probability calculations involving conditional events.
通俗地说，在经历了 m 次连续失败之后，剩余等待时间超过 n 次试验的条件概率，恰好等于从一开始算起等待时间超过 n 次试验的概率。在所有非负分布中，只有几何分布及其连续对应——指数分布具有这一性质。这一概念经常出现在考题中：可能会要求你验证该性质，或利用它来简化涉及条件事件的概率计算。
8. Geometric Distribution vs Other Distributions | 几何分布与其他分布的比较
It is easy to confuse the geometric distribution with the binomial or negative binomial distributions. Here are the key differences to remember:
很容易将几何分布与二项分布或负二项分布混淆。以下是需要记住的关键区别：
- Binomial: The number of successes in a fixed number of trials, support {0,1,…,n}. | 二项分布：在固定试验次数中的成功次数，支撑集为 {0,1,…,n}。
- Geometric: The number of trials until the first success, support {1,2,3,…}. | 几何分布：直到第一次成功出现的试验次数，支撑集 {1,2,3,…}。
- Negative binomial: The number of trials until the rth success, a generalisation of the geometric distribution (r=1 gives the geometric). | 负二项分布：直到第 r 次成功出现的试验次数，它是几何分布的推广（r=1 即为几何分布）。
Also, the total number of failures before the first success (a shifted geometric) has support {0,1,2,…} and mean (1−p)/p. Always read the problem statement carefully to identify which variable is being counted. In Cambridge A Level and IB questions, the geometric distribution often appears in contexts such as 'number of attempts until a pass' or 'number of days until a fault occurs', making correct identification essential.
此外，第一次成功之前的失败总次数（一种平移后的几何分布）支撑集为 {0,1,2,…}，均值为 (1−p)/p。务必仔细阅读题目陈述，明确计数的是哪个变量。在 Cambridge A Level 和 IB 问题中，几何分布常出现在诸如“通过所需的尝试次数”或“出现故障前的天数”等语境中，因此正确识别分布类型至关重要。
9. Worked Examples (IB and Cambridge Style) | 例题解析
Example 1: A biased die shows a six with probability 0.15. Find the probability that the first six appears on exactly the 5th roll. | 例 1：一枚不均匀的骰子出现六点的概率为 0.15。求第一次出现六点恰好是在第 5 次抛掷时的概率。
: P(X = 5) = 0.15 × (0.85)⁴ ≈ 0.15 × 0.5220 ≈ 0.0783. Interpretation: there is about a 7.8% chance that the first six comes on the fifth toss.
使用概率质量函数：P(X = 5) = 0.15 × (0.85)⁴ ≈ 0.15 × 0.5220 ≈ 0.0783。解读：第一次出现六点恰好在第五次抛掷的概率大约为 7.8%。
Example 2: A manufacturer tests light bulbs and finds that 2% are defective. What is the expected number of bulbs tested until the first defective is found? | 例 2：某制造商测试灯泡，发现 2% 为次品。求检测出第一个次品所需的平均检测数量。
450, so the standard deviation is about 49.5. This shows high variability around the mean.
此处 p = 0.02，因此 E(X) = 1/0.02 = 50。平均需要检测 50 个灯泡。方差为 (0.98)/(0.0004) = 2450，标准差约为 49.5。这表明在均值附近有很高的变异性。
Example 3: Given X ~ Geo(0.25), find P(3 ≤ X ≤ 6). | 例 3：已知 X ~ Geo(0.25)，求 P(3 ≤ X ≤ 6)。
= 1 − 0.5625 = 0.4375. Therefore P(3 ≤ X ≤ 6) = 0.82202 − 0.4375 = 0.3845 (approximately).
使用累积分布函数：P(X ≤ 6) = 1 − 0.75⁶ ≈ 1 − 0.17798 = 0.82202；P(X ≤ 2) = 1 − 0.75² = 1 − 0.5625 = 0.4375。因此 P(3 ≤ X ≤ 6) = 0.82202 − 0.4375 = 0.3845（近似值）。
Example 4 (Memoryless): The probability of passing a test on any attempt is 0.3. Given that a student has already failed 4 times, what is the probability that at least 3 more attempts are needed? | 例 4（无记忆性）：每次尝试通过某项测试的概率为 0.3。已知一名学生已经失败了 4 次，求至少还需要 3 次尝试的概率。
By the memoryless property: P(X > 4 + 3 | ditional probability is exactly the same as starting fresh and needing more than 3 attempts. This property simplifies the calculation dramatically.
根据无记忆性：P(X > 4 + 3 | X > 4) = P(X > 3) = (0.7)³ = 0.343。这一条件概率与重新开始并需要超过 3 次尝试的概率完全相同。该性质极大地简化了计算。
10. Common Mistakes and Exam Tips | 常见错误与考试技巧
Students often miscount the exponent in the PMF, using k instead of k−1. Remember: k trials mean k−1 ntil first success. When reading questions, underline whether the waiting time is measured "before" or "until" rty; it only works for the geometric distribution (and the exponential in continuous time) – la you intend to use, as marks are often awarded for correct substitution even if a calculation error occurs. Practice recognising when the geometric distribution is appropriate: keywords like "first success", "first defective", "until a hit" are strong indicators.
学生们经常在概率质量函数的指数上出错，错误地使用 k 而不是 k−1。记住：第 k 次试验意味着需要 k−1 次失败。明确定义：X 是直到第一次成功为止的总试验次数。读题时，要圈出等待时间是“之前”还是“直到”的表述。另一个常见的陷阱是错误应用无记忆性；它只对几何分布（以及连续时间的指数分布）成立——不适用于二项分布或泊松分布。在考试中，务必先写下你打算使用的公式，因为即使计算有误，正确的代入步骤通常也能得分。多练习识别何时适合使用几何分布：诸如“第一次成功”、“首个次品”、“直到命中”等关键词都是强烈的信号。
ften underpins derivation questions. Moreover, learn to use the CDF to find probabilities of ranges efficiently – n (failures before success) that may appear in certain textbooks; ensure you convert correctly by checking the support and mean formula.
扎实掌握等比级数至关重要。无限级数的求和公式 ∑ arⁿ⁻¹ = a/(1−r) 常常是推导题的基础。此外，学会利用累积分布函数高效地求解区间概率——与逐项相加概率质量函数相比，这样更省时。最后，要注意某些教材中可能出现的另一种参数化形式（成功前的失败次数）；通过检查支撑集和均值公式，确保正确转换。
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