📚 The Geometric Distribution | 几何分布
In Edexcel A Level Mathematics, the geometric distribution models the number of trials needed to achieve the first success in a sequence of independent Bernoulli trials. It is a key topic in the Statistics specification, especially in probability modelling and hypothesis testing. Understanding its conditions, probability mass function, cumulative probabilities, mean and variance is essential for exam success.
在 Edexcel A Level 数学中，几何分布用于建模在一系列独立伯努利试验中首次成功所需的试验次数。它是统计学课程中的一个重要主题，尤其是概率建模和假设检验。掌握其条件、概率质量函数、累积概率、均值与方差对于考试成功至关重要。
1. What is a Geometric Distribution? | 什么是几何分布？
The geometric distribution is a discrete probability distribution that describes the number of Bernoulli trials needed to get the first success. The random variable X counts the number of trials up to and including the first success. The possible values are 1, 2, 3, ... because you cannot succeed before the first trial.
几何分布是一种离散概率分布，描述获得第一次成功所需的伯努利试验次数。随机变量 X 计数到第一次成功为止并包括该次成功的试验次数。可能取值为 1、2、3……因为第一次试验之前不可能成功。
he roll on which the first 6 occurs, then X follows a geometric distribution with p = 1/6.
例如，如果你反复掷一枚公平骰子直到出现 6，并设 X 为首次出现 6 的那一次投掷，则 X 服从 p = 1/6 的几何分布。
2. Conditions for a Geometric Distribution | 几何分布的条件
A random variable X has a geometric distribution if there is a fixed probability of success p on each trial, the trials are independent, and each trial has only two outcomes: success or failure. The probability of success p remains constant from trial to trial.
如果每次试验有固定的成功概率 p，各次试验相互独立，并且每次试验只有成功或失败两种结果，则随机变量 X 服从几何分布。成功概率 p 在每次试验中保持不变。
The key difference from the binomial distribution is that a geometric experiment stops at the first success, while a binomial experiment has a fixed number of trials. In geometric settings, the number of successes is fixed at 1 and the number of trials is the random variable.
与二项分布的主要区别在于，几何实验在第一次成功时停止，而二项实验有固定的试验次数。在几何情境中，成功次数固定为 1，试验次数是随机变量。
3. Probability Mass Function | 概率质量函数
If X ~ Geo(p), then the probability that the first success occurs on the x-th trial is given by the probability mass function:
如果 X ~ Geo(p)，那么第一次成功发生在第 x 次试验的概率由概率质量函数给出：
P(X = x) = p(1 − p)ˣ⁻¹, x = 1, 2, 3, …
This formula arises because the first x − 1 trials must be failures, each with probability 1 − p, and the x-th trial must be a success with probability p.
这个公式的由来是：前 x − 1 次试验必须都是失败，每次失败概率为 1 − p，而第 x 次试验必须成功，概率为 p。
Since 0 < p < 1, the probabilities form a decreasing geometric sequence. That is why the distribution is named 'geometric'. The sequence of probabilities p, p(1 − p), p(1 − p)², ... has a common ratio of 1 − p.
由于 0 < p < 1，这些概率构成一个递减的等比数列，这就是该分布被称为 "几何分布" 的原因。概率序列 p、p(1 − p)、p(1 − p)²……的公比为 1 − p。
4. Cumulative Distribution Function | 累积分布函数
s the probability that the first success occurs on or before the x-th trial. For a geometric distribution, it can be calculated using:
累积概率 P(X ≤ x) 是第一次成功发生在第 x 次或之前试验的概率。对于几何分布，可以用以下公式计算：
P(X ≤ x) = 1 − (1 − p)ˣ
Similarly, the probability that more than x trials are needed to get the first success is:
类似地，需要超过 x 次试验才能获得第一次成功的概率为：
P(X > x) = (1 − p)ˣ
Remember that because the distribution is discrete, P(X < x) = P(X ≤ x − 1), not P(X ≤ x).
这些公式在考试题目中非常有用，尤其是求 P(X ≥ x) 或 P(a ≤ X ≤ b) 时。可以写成 P(a ≤ X ≤ b) = P(X ≤ b) − P(X ≤ a − 1)。请记住，由于分布是离散的，P(X < x) = P(X ≤ x − 1)，而不是 P(X ≤ x)。
5. Mean and Variance | 均值与方差
The expected value of a geometric random variable is the average number of trials needed to obtain the first success. It is given by:
几何随机变量的期望值是获得第一次成功所需的平均试验次数。其公式为：
E(X) = 1/p
The variance measures the spread of the number of trials and is given by:
方差衡量试验次数的离散程度，其公式为：
Var(X) = (1 − p)/p²
verage you need 5 trials, but the spread is large because the distribution is right-skewed.
例如，若 p = 0.2，则 E(X) = 1/0.2 = 5，Var(X) = 0.8/0.04 = 20。所以平均需要 5 次试验，但由于分布右偏，离散程度较大。
Notice that a smaller success probability p leads to a larger mean and variance. This makes sense: if success is rare, you expect to wait longer and the waiting time is more variable.
注意，成功概率 p 越小，均值和方差就越大。这是合理的：如果成功很罕见，你预期等待更久，而且等待时间变化更大。
6. Worked Example: Finding P(X = k) | 例题：求 P(X = k)
nding on heads. Let X be the number of tosses needed to get the first head. Find P(X = 4).
一枚有偏硬币正面朝上的概率为 0.3。设 X 为首次得到正面所需的投掷次数。求 P(X = 4)。
Here p = 0.3, so 1 − p = 0.7. Using the probability mass function:
这里 p = 0.3，因此 1 − p = 0.7。使用概率质量函数：
P(X = 4) = 0.3 × 0.7³ = 0.3 × 0.343 = 0.1029
n also interpret this as three tails followed by one head: 0.7 × 0.7 × 0.7 × 0.3 = 0.1029.
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