GCSE数学 概率树 树形图 条件概率 全解析
概率树图（Probability Tree Diagram）是 GCSE Maths 概率单元中分值最高、最容易拿满分也最容易失分的题型。它把多步骤随机事件拆成一层一层的分支，让每个分支上的概率都看得见、算得清。无论是 AQA、Edexcel 还是 OCR，Paper 1 与 Paper 3 都反复考查树形图：读懂树的含义、沿着分支相乘、把互斥路径相加，最后处理无放回抽样带来的分母变化。很多同学能画出树，却在最后一问失分，原因通常不是概念不会，而是没有区分"有放回"和"无放回"、忘记把分数约分、或者在"至少一个"这类问题上硬算而不使用反面法。
A probability tree diagram is one of the highest-value topics in GCSE Maths. It breaks a multi-stage random experiment into layers of branches, so every probability is visible and checkable. AQA, Edexcel and OCR all test tree diagrams repeatedly across Papers 1 and 3: read the tree, multiply along the branches, add mutually exclusive paths, and handle the changing denominators that come with sampling without replacement. Many students can draw the tree but lose marks on the final part, usually because they confuse "with replacement" and "without replacement", forget to simplify fractions, or grind through "at least one" questions instead of using the complement rule.
一、概率树图是什么 / What a Probability Tree Diagram Is
概率树图是用分支表示多步骤试验的图形工具。第一层分支代表第一次试验的所有可能结果，每条分支上标注该结果的概率；第二层分支从第一层的每个结果继续延伸，代表第二次试验的条件概率。所有从同一节点出发的分支概率之和必须等于 1，这是最重要的自检规则。若某次试验有两个结果，概率分别为 0.6 和 0.4，那么树上同一点出发的两条分支加起来必须是 1；如果加起来不是 1，说明你漏抄了条件或算错了补数。树形图的末端结果数等于各层分支数的乘积，例如两层各两个结果，末端共有 2 乘 2 等于 4 个结果，对应 4 条完整路径。画树时还要养成一个习惯：在分支旁写下分数的同时，保留未约分的形式到最后一刻再约分，避免中间步骤反复化简导致错误。
A probability tree diagram is a visual tool for multi-stage experiments. The first set of branches shows the outcomes of the first stage, with the probability written on each branch. The second set extends from every first-stage outcome and shows conditional probabilities. The most important self-check is that the probabilities on branches leaving the same node must sum to 1. If a stage has two outcomes with probabilities 0.6 and 0.4, the two branches must add to 1; if they do not, you have misread a condition or miscalculated a complement. The number of final outcomes equals the product of the branch counts at each stage, so two stages with two outcomes each give 2 times 2, that is 4 complete paths. A useful habit is to keep fractions in a working form and simplify only at the end, which avoids repeated simplification errors.
二、乘法法则：沿分支相乘 / The Multiplication Rule: Multiply Along Branches
乘法法则（AND rule）用于求一条完整路径的概率：沿树的分支一路相乘。例如袋中有 5 个红球和 3 个蓝球，共 8 个球，无放回地取两个。第一次取到红球的概率是 5/8，取走后袋中只剩 7 个球、4 个红球，所以第二次仍取到红球的概率是 4/7。于是两个都是红球的概率为 5/8 乘 4/7，等于 20/56，约分为 5/14。要牢记分母会随抽样而变化：无放回时第二次的分母比第一次小 1，分子则按第一次取走的结果相应减少。如果是有放回抽样，袋中球数不变，第二次的概率与第一次完全相同，仍然是 5/8，此时两红概率为 25/64。考试中最常见的失分就是抄错第二层的分母：看到"without replacement"就必须把 8 改成 7，把分子中的那类球减 1。
The multiplication rule, often called the AND rule, gives the probability of one complete path: multiply the probabilities as you travel along the branches. Take a bag with 5 red and 3 blue balls, 8 in total, and pick two without replacement. The first pick is red with probability 5/8. After that ball is removed there are 7 balls left and 4 red ones, so the probability of a second red is 4/7. The probability of two reds is therefore 5/8 times 4/7, which equals 20/56, simplifying to 5/14. Remember that the denominator changes: without replacement the second denominator is one less than the first, and the numerator for the relevant colour drops by one. With replacement, the bag is unchanged, so the second probability is again 5/8 and the probability of two reds is 25/64. The most common exam error is copying the wrong denominator for the second layer: whenever you read "without replacement", you must change 8 to 7 and reduce the numerator of that colour by one.
三、加法法则：互斥路径相加 / The Addition Rule: Add Mutually Exclusive Paths
加法法则（OR rule）用于求多个合格路径中任意一个发生的概率：把各条路径的概率相加。继续用 5 红 3 蓝无放回取两球的例子。恰好取到一个红球和一个蓝球有两条路径：先红后蓝是 5/8 乘 3/7 等于 15/56，先蓝后红是 3/8 乘 5/7 也等于 15/56，两条路径互斥，所以总概率为 30/56，约分为 15/28。注意不要只算一条路径就停笔：题目问"one red and one blue"并不指定顺序，两条路径都需要计入，这也是评分标准明确给出的一个得分点（M1 认出一条路径，M1 认出第二条，A1 得到正确总和）。此外，树上所有末端路径的概率之和必须等于 1。用上面的结果自检：两红 20/56，两蓝 3/8 乘 2/7 等于 6/56，一红一蓝 30/56，三者相加 56/56，正好等于 1，说明计算无误。这个"总和为 1"的自检方法能在考场上快速定位错误。
The addition rule, or OR rule, gives the probability that any one of several qualifying paths happens: add the probabilities of those paths. Keeping the same bag of 5 red and 3 blue balls sampled without replacement, the event "exactly one red and one blue" has two paths. Red then blue is 5/8 times 3/7, which is 15/56. Blue then red is 3/8 times 5/7, also 15/56. These two paths are mutually exclusive, so the total probability is 30/56, that is 15/28. Do not stop after one path: the question "one red and one blue" does not fix an order, so both paths must be counted, and mark schemes usually award one mark for each path plus one for the correct total. In addition, all the final path probabilities must sum to 1. Check with the numbers above: two reds is 20/56, two blues is 3/8 times 2/7 which is 6/56, and one of each is 30/56. Together they give 56/56, exactly 1, so the working is correct. This sum-to-1 check is a fast way to locate mistakes in the exam.
四、条件概率与无放回抽样 / Conditional Probability and Sampling Without Replacement
条件概率是树形图第二层的本质：它表示"在第一个结果已经发生的条件下，第二个结果发生的概率"。用公式写作 P(A given B) 等于 P(A and B) 除以 P(B)。例如已知 P(A and B) 等于 0.3，P(B) 等于 0.5，那么 P(A given B) 等于 0.3 除以 0.5，等于 0.6。在树形图上，这个公式有两种用法。第一种是正向计算：题目给出的第二层概率就是条件概率，直接相乘即可。第二种是反向推断：题目只给出某个最终结果的概率，要求你反推第一层的某个概率。无放回抽样还会改变后续所有分支：取走的球越多，分母越小。若从 8 个球中取三个且不放回，第三次的分母是 6；若取到第二个红球之后，第三个仍是红球的概率分子还要再减 1。用一个具体例子验证条件概率的方向性：P(第二个是红球) 在无放回下仍然等于 5/8，因为对称性保证每一个位置的边际概率相同，但 P(第二个是红球 given 第一个是红球) 等于 4/7，两者不同，混淆这两个量是本单元最隐蔽的错误来源。
Conditional probability is the essence of the second layer of a tree: it is the probability that the second outcome occurs given that the first outcome has already occurred. In symbols, P(A given B) equals P(A and B) divided by P(B). If P(A and B) is 0.3 and P(B) is 0.5, then P(A given B) is 0.3 divided by 0.5, which is 0.6. On a tree diagram this formula works in two directions. In the forward direction, the probabilities printed on the second layer are conditional probabilities, so you simply multiply. In the reverse direction, the question may give the probability of a final outcome and ask you to recover a first-layer probability. Sampling without replacement changes every later layer: the more balls removed, the smaller the denominator. Drawing three balls from eight without replacement means the third denominator is 6, and after a second red has been taken, the numerator for red falls by one again. Test the direction with one comparison: without replacement P(second ball is red) is still 5/8 by symmetry, because each position has the same marginal probability, but P(second is red given first is red) is 4/7. These two quantities are different, and confusing them is the subtlest source of error in this topic.
五、逆向问题与至少一个 / Reverse Problems and At Least One
"至少一个"（at least one）类问题必须使用反面法（complement rule）：P(至少一个 A) 等于 1 减去 P(一个 A 都没有)。例如一枚不均匀硬币，正面概率 0.6，反面概率 0.4，抛两次。至少出现一次正面的概率是 1 减去两次都是反面的概率，即 1 减去 0.4 乘 0.4，等于 1 减去 0.16，等于 0.84。如果硬算"恰好一次正面"加"两次正面"，得到 0.6 乘 0.4 加 0.4 乘 0.6 加 0.36，等于 0.48 加 0.36，等于 0.84，答案一致但步骤更长、更易错。逆向问题的另一种形态是"已知最终结果，反推起始条件"。例如某地有雨的概率 0.3，下雨时迟到的概率 0.8，不下雨时迟到的概率 0.1。迟到的总概率为 0.3 乘 0.8 加 0.7 乘 0.1，等于 0.24 加 0.07，等于 0.31。若已知今天迟到，来自"下雨"这一支的概率是 0.24 除以 0.31，约等于 0.774。这类题目在 GCSE 高阶卷中属于 4 到 5 分的综合题，务必写出分式形式再计算，评分标准通常给"正确分子"和"正确分母"各一个得分点。
Questions that say "at least one" must be answered with the complement rule: P(at least one A) equals 1 minus P(no A at all). Take a biased coin with P(heads) equal to 0.6 and P(tails) equal to 0.4, tossed twice. The probability of at least one head is 1 minus the probability of two tails, which is 1 minus 0.4 times 0.4, that is 1 minus 0.16, which equals 0.84. Grinding through "exactly one head" plus "two heads" gives 0.6 times 0.4 plus 0.4 times 0.6 plus 0.36, which is 0.48 plus 0.36, also 0.84, but with more steps and more chance of error. Reverse problems form the other pattern: a final outcome is known and the question asks which cause produced it. Suppose the probability of rain is 0.3, the probability of being late given rain is 0.8, and the probability of being late given no rain is 0.1. The overall probability of being late is 0.3 times 0.8 plus 0.7 times 0.1, which is 0.24 plus 0.07, that is 0.31. Given that the person is late, the probability it was a rainy day is 0.24 divided by 0.31, roughly 0.774. Questions of this shape are worth four to five marks in the higher tier, so always write the fraction before evaluating, because mark schemes normally award one mark for the correct numerator and one for the correct denominator.
六、考试规范与常见失分点 / Exam Technique and Common Losses
树形图题目的评分标准高度结构化，得分点通常包括：正确画出层数与分支、分支上概率正确、沿分支相乘、把互斥路径相加、最终化简或按题目要求保留小数。常见的失分点有五类。第一，忘记标注分支概率，只画结构，直接失去 A 类分值。第二，无放回时第二层分母没有变化。第三，只计算一条路径而漏掉顺序相反的另一条。第四，把加法法则和乘法法则用反，导致结果远大于 1 或远小于合理范围。第五，分数未化简，或把 15/28 写成小数时精度不足。做题时建议按固定顺序书写：先写"路径一：... 路径二：..."，再写"总概率等于路径一加路径二"，最后计算。这样即使计算出现小错，过程分依然可以拿到。另外，遇到"given"字样时，务必判断是条件概率还是普通概率；遇到"without replacement"时，先把第二次的样本空间写下来，再动笔计算。
Mark schemes for tree diagram questions are highly structured. The marks usually cover drawing the correct number of stages and branches, labelling the probabilities correctly, multiplying along the branches, adding mutually exclusive paths, and simplifying the final answer or giving the accuracy the question demands. Five mistakes account for most lost marks. First, drawing the structure without labelling the branch probabilities, which immediately loses the accuracy marks. Second, leaving the second-layer denominator unchanged when sampling without replacement. Third, computing one path only and missing the reversed order. Fourth, swapping the addition and multiplication rules, which produces answers far above 1 or far below a sensible range. Fifth, failing to simplify a fraction, or giving insufficient precision when converting 15/28 to a decimal. In the exam, write in a fixed order: first "path one: ...", then "path two: ...", then "total probability equals path one plus path two", and only then evaluate. That way, even a small arithmetic slip still earns the method marks. Whenever you see the word "given", decide whether it means a conditional probability or an ordinary one, and whenever you see "without replacement", write down the new sample space before computing.
学习建议 / Study Tips
第一，做题时先画树再计算，不要凭直觉列式子；树本身就是解题结构。第二，每次画完树立刻检查同一节点的分支是否相加为 1。第三，把"有放回"与"无放回"各做五道对照题，专门比较第二层概率的差异，形成肌肉记忆。第四，整理一份错题清单，按失分类型归类，例如"分母未变""漏算一条路径""没有用反面法"，每周回看一次。第五，刻意练习分数运算，尤其是异分母相加与约分，树形图的失分有相当比例源自分数计算而非概率概念。第六，用真题限时训练：GCSE 概率综合题通常给 4 到 6 分钟，超时就说明树的结构还不熟练。第七，把每道题的最终答案带回"总和为 1"的自检，养成写完就验证的习惯。坚持两周，概率树图会从最容易失分的题型变成最稳定的得分点。
First, always draw the tree before computing; the diagram itself is the solution structure, so do not try to list expressions from intuition. Second, after drawing, immediately check that the branches leaving each node sum to 1. Third, work through five paired questions on "with replacement" and five on "without replacement" so the difference in the second layer becomes automatic. Fourth, keep an error log sorted by cause, for example "denominator unchanged", "missing path", or "complement rule not used", and review it weekly. Fifth, practise fraction arithmetic on purpose, especially adding fractions with different denominators and simplifying, because a large share of lost marks comes from arithmetic rather than from probability. Sixth, train with past papers under time pressure: comprehensive GCSE probability questions usually allow four to six minutes, and running over time means the tree structure is not yet fluent. Seventh, run the sum-to-1 check on every final answer until verification becomes a habit. Two weeks of this routine turns tree diagrams from a weak spot into one of the most reliable sources of marks on the paper.
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