GCSE数学 概率树图 乘法加法 不放回
概率树图是 GCSE 数学概率部分最常见的题型，在 Edexcel 1MA1、AQA 8300 与 OCR J560 的试卷里都会出现，Foundation 与 Higher 都要考。做法本身很固定：一层代表一步，枝上写这一步的概率，沿枝相乘，同层相加。真正拉开分数的不是这条规则，而是分母有没有跟着剩余总数走。下面按搭树的顺序拆成十二个环节，每一节都配一个能照着算的例子。
Tree diagrams appear in nearly every GCSE probability paper, on Foundation and Higher alike, and the method never changes: one layer per stage, a probability on each branch, multiply along a route and add between routes. What separates a full-mark answer from a half-mark one is whether the denominators follow the items actually left in the bag. The sections below follow the order you would build a tree, with a worked example at each step.
树状图要解决的是两步以上的试验
概率树图把一个试验拆成几层，每一层代表一步，枝上写这一步发生的概率。做题时先确定有几层，再决定每一层放哪些结果。学生常常一上来就写分数，写到第二层才发现自己漏了一种颜色。画树之前先把总数写在草稿纸角落，后面每一步都要回头看它。Edexcel 与 AQA 的 specification 都把树状图列在概率部分，用的场景多是连着取两次、连着转两次。树的作用不是好看，是让每一步的分母和分子有地方可放。这一步做对，后面的乘法只是例行计算。
A tree diagram splits one experiment into stages, and each branch carries the probability of that single step. Before drawing anything, decide how many stages the question has and which outcomes belong on each stage. In practice, examiners see a neat tree with the wrong denominator on the second set of branches, which wastes the whole diagram. Write the starting total in the corner of your page and keep referring back to it; the tree only works if every branch is fed by the correct count.
第一层分支要摆全所有结果
第一层分支要把第一步所有可能出现的结果都摆出来，彼此不重叠。取一个球，结果只有红和蓝两种，第一层就两条枝，两条枝的概率之和必须是一。题里是转盘，扇区有三块，第一层就是三条枝。漏掉一条枝是最常见的结构性错误，后面所有乘积都会少算一块。检查办法很直接：把第一层枝上的分数加起来，看差错出在哪里。很多学生以为第一层随便画画就行，实际上分母定错一次，整棵树的数字全要重写。
The first stage must list every possible outcome of the opening step, with no overlaps and no gaps. Two colours in a bag means two branches that sum to one, and a spinner with three sectors means three. A missing branch is a structural error rather than an arithmetic one, because every later product quietly drops a case. Count the branches, check the fractions sum to one, and only then move down the tree; that single check catches most lost marks before any multiplication happens.
沿枝相乘算两个结果同时发生
同一枝上前后两步都要发生，概率就相乘，这是树图的核心运算。掷骰子得到六点，再抛硬币得到正面，两个结果同时成立的概率是六分之一乘二分之一，等于十二分之一。乘法的顺序可以调换，每一层用的分母必须是那一层的总数。阅卷时经常看到学生把两层概率相加，那是把并且当成了或者。判断依据很简单：题目问的是两件事都发生，就沿一条路径把枝上的分数乘下去。数值小到几十分之一也不必回头改，概率本来就是小数字。
Moving along a branch means both steps happen, so the fractions multiply. A six on a die followed by a head on a coin gives one sixth times one half, so the answer is one twelfth. Students who add instead of multiply have usually read the word and as or, so check the wording again before writing anything. Multiplication is commutative, but each denominator has to come from the stage it sits on; mixing them produces answers that cannot be right.
同一层的枝相加处理互斥结果
树图末端的不同路径互斥，问某几种结果之一就把这些路径的概率相加。两次取球里问取到一红一蓝，先红后蓝和先蓝后红是两条不同的路径，两条都要算，再把分数加起来。相加之前先看分母，分母不同就先通分。常见错误是只算一条路径就交卷，或者把两条路径的分母又乘了一遍。题目里的顺序词是关键，先红后蓝与先蓝后红不是同一件事。答案写成分数要化成最简形式，除非题目明确要求保留某个分母。
Different routes at the end of the tree are mutually exclusive, so one of these outcomes means adding their probabilities. Taking one red and one blue covers two routes, red then blue and blue then red, and both belong in the sum. Convert to a common denominator first, and simplify the final fraction unless the question asks you to keep a particular denominator. Answers that stop after a single route are the ones that give away half the marks, and the order words in the question tell you how many routes there really are.
不放回时第二次分母要减一
不放回的意思是第一次取走的东西不再放回去，第二层的总数要减一，分子也可能跟着减。袋子里五个红球三个蓝球，第一次取红的概率是八分之五；如果第一次真的取走了红球，第二次取红的概率变成七分之四。画树时要把第二层两套枝都写出来，一套挂在先红下面，一套挂在先蓝下面，因为两种情况下的第二层数字不一样。很多学生以为第二层还能沿用八分之五，这个错误会让后面的乘积全部偏大。放回与不放回只差题干里几个字，做法却完全不同。
Without replacement, the item taken in the first step never goes back, so the second stage holds one fewer item in total and possibly one fewer of that colour. Five red and three blue give five eighths on the first draw; if a red has gone, the next red is four sevenths. That difference is why the second stage needs two separate sets of branches, one under each first-stage outcome. Reusing the original denominator is the single most common error in this topic, and it inflates every later product.
分数先摆原数再约分
把数量直接写成分数，比先算小数再转分数更安全，因为约分容易把中间步骤吃掉。八个球里五个红，第一层写八分之五与八分之三，第二层按先红或先蓝分开写。约分放在乘法之后做，不要一开始就把每个分数化到最简，那样第二步的分母容易看错。算完每一步回头看一眼分子分母是否还是正整数，出现大于一的概率就说明某一层写反了。草稿上保留原数，誊到卷面时再化简，中间过程不容易丢。
Writing counts straight as fractions is safer than converting to decimals, because decimals invite rounding and hide the structure of the problem. With eight balls, five of them red, the first stage is five eighths and three eighths, and the second stage then depends on which branch you are under. Cancel after multiplying rather than before, since pre-simplifying makes the second-stage denominators harder to read. Any branch showing a probability above one means a numerator and denominator have been swapped somewhere.
至少一次用一减去一次都没有
至少一次这类问法从反面算最快：先算一次都没有的概率，再用一减去它。转两次转盘，每次命中目标的概率是三分之一，两次都没命中的概率是三分之二乘三分之二，等于九分之四，至少命中一次就是九分之五。直接在树上把命中一次、命中两次的路径全加起来也可以，路径一多就容易漏一条。常见的错误是拿一减去命中一次的概率，那是把反面换错了。看到至少两个字，先在草稿边上写一次都没有，答案就有了方向。
At least once is almost always faster from the complement: find the probability of never, then subtract it from one. Two spins with a one third hit chance leave four ninths for missing both, so at least one hit is five ninths. Adding every route that contains a hit works too, but the routes multiply with each stage and students start skipping one. Subtracting the probability of exactly one hit answers a different question, and markers see that slip every summer.
独立与不独立看枝上的数字变不变
独立事件的树图两层数字完全一样，因为前一步的结果不影响后一步。抛两次硬币，第二次仍是二分之一，树的两层都写二分之一。不独立的事件，第二层的数字必须按第一层的结果重新算，前一步取走了什么就减什么。判断是不是独立要看题目描述，抽完放回是独立，抽完不放回是不独立。两种情形混在一起，多半是学生没注意题干里的放回两个字。写题时把这两个字圈出来，比事后检查整棵树省力得多。
Independent stages keep identical fractions on both layers, because the earlier result changes nothing. Two coin tosses give one half at each stage, so the tree looks the same at every depth. Dependent stages need the second layer recalculated from the outcome above, subtracting whatever has been removed. The words with replacement and without replacement decide which tree you are drawing, and skipping them is what produces a diagram that cannot score.
条件概率记法 P(A|B) 在 Higher 卷怎么用
Higher 卷会用到条件概率的记法，写成 P(A|B)，读作在 B 发生的条件下 A 发生的概率。它的算法是用 A 与 B 同时发生的概率除以 B 的概率，两个数字都能从树上直接读出来。红蓝球的例子里，已知第一次取到蓝球，第二次取到红球的概率是五十六分之十五除以八分之三，等于七分之五。树上那条枝已经给出分子，分母是整条第一层枝的概率。这个记法在 Edexcel 与 AQA 的 Higher 卷都会出现，Foundation 卷一般只要求读图算概率。写答案时把两个分数都摆出来，过程分比结果分好拿。
Higher tier papers use the conditional notation, written P(A|B) and read as the probability of A given B. Divide the probability of both events by the probability of the condition, and both numbers already sit on the tree. In the red and blue bag, given a blue first, the chance of a red second is fifteen fifty-sixths divided by three eighths, which is five sevenths. The branch supplies the numerator and the whole first-stage branch supplies the denominator, so no new arithmetic is needed.
三层以上的树分母连着变小
三层以上的树，分母会连着变小两次，每一步都要按剩余总数重算。七个球里取三个不放回，第三层的第一套枝分母是五，第二套来自六个球，第一层则是七。层数一多，枝的数量翻倍，画的时候把同类结果对齐写，别让路径交叉。题目通常只问某几条路径，不必每条都算完，先在草稿上圈出需要的那几条。数量跟着层数变，这一点比公式更容易出错。多一层就多一次核对，别把第二层的分母直接抄到第三层。
A three-stage tree shrinks the total twice, so recalculate at every layer instead of carrying one denominator down the page. Drawing three items without replacement from seven gives branches over seven, then six, then five. The number of routes doubles at each stage, so keep similar outcomes aligned and ignore routes the question never asks about. Circling the routes you actually need before multiplying saves both time and careless errors.
用总和等于一检查整棵树
一棵完整的树，所有末端路径的概率加起来必须正好是一。这个检查不需要额外计算，把已经算出的几个分数通分相加，看是不是一。不等于一通常意味着漏了一条路径，或者某一层的分母写错了。检查放在最后一步做，比重算一遍快得多。用一减去已算路径之和，还能顺手得到剩下那条路径的概率，题目问其他情况时正好用上。算完四分之一、四分之三这类常见分数，心里应该有个大概，答案明显偏大或偏小就回去看分母。
Every route on a finished tree must add up to one, and that check costs almost nothing. Convert the route probabilities to a common denominator, add them, and compare with one. A total above or below one points to a missing route or a denominator taken from the wrong stage. The same subtraction hands you the probability of the one route you have not computed, which is often exactly what the final part asks for.
卷面写法与最常见的扣分点
卷面上要把树画清楚，枝上的分数写成分子在上、分母在下，不要用斜杠挤在一行。每层旁边标一句这一步在做什么，比如第一次取球、第二次取球，阅卷时看得懂才有分。最后答案要写完整，问概率就写概率，必要时加一句文字说明。答完再读一遍题目问的是哪一个事件，问的是至少还是恰好，两者答案常常差一个分数。错得最多的是最后一问，因为前面的数已经算完，学生急着收笔。留一分钟回看分母，比多写一段解释有用。
Draw the tree large enough that each fraction sits as a numerator over a denominator rather than a slash squeezed into one line. Label each layer with what the step does, first draw and second draw, because a marker cannot award method marks for a diagram they cannot follow. Write the final answer as a probability with a short sentence, and reread whether the question asked for exactly or at least. The last part of a tree question is where the most marks disappear, usually because the earlier numbers look finished and the pen speeds up.
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