📚 Outcome Tables in Probability | IB数学：结果表在概率中的应用
In the IB Mathematics curriculum, one of the most reliable techniques for solving probability problems is the construction of an outcome table. This grid or tabular display organises all the possible results of a random experiment, making it easier to count favourable outcomes and compute probabilities, especially when the sample space is finite and all outcomes are equally likely.
在国际文凭（IB）数学课程中，解决概率问题最可靠的方法之一就是构建结果表。这种网格或表格形式能够组织随机实验的所有可能结果，从而更容易地计算有利结果的数量并求出概率，尤其适用于样本空间有限且所有结果等可能的情形。
1. What Is an Outcome Table? | 什么是结果表？
An outcome table is a systematic arrangement of the possible results of a probability experiment. It is particularly useful when the experiment can be viewed as two or more stages, such as rolling two dice, tossing two coins, or drawing one item from each of two groups. Each row and column corresponds to a possible result of one stage, and the interior cells represent the combined outcome.
结果表是对概率实验的可能结果进行系统排列的表格。当实验可以被视为两个或更多阶段时（例如掷两枚骰子、抛两枚硬币、从两组中各抽取一个物品），这种方法特别有效。每一行和每一列对应某个阶段的可能结果，而内部单元格则表示组合后的结果。
For example, when tossing a fair coin twice, the outcome table would have two rows (Heads or Tails on the first toss) and two columns (Heads or Tails on the second toss), with four cells: HH, HT, TH, TT.
例如，掷一枚均匀硬币两次时，结果表有2行（第一次正面或反面）和2列（第二次正面或反面），共4个单元格：HH、HT、TH、TT。
The main benefit of an outcome table is clarity. Instead of listing outcomes in a set, the table shows the structure of the sample space. This helps students identify whether outcomes are equally likely and avoid double-counting.
结果表的主要优点是清晰。与用集合列出结果不同，表格展示了样本空间的结构。这有助于学生判断结果是否等可能，并避免重复计数。
2. The Sample Space Table for Two Dice | 两枚骰子的样本空间表
faced dice. There are 6 × 6 = 36 equally likely outcomes. The rows represent the first die's result from 1 to 6, and the columns represent the second die's result. Each cell contains the sum of the two numbers.
一个经典例子是掷两枚均匀六面骰子时所得点数之和的表格。样本空间共有6×6=36个等可能结果。行表示第一枚骰子的点数（1到6），列表示第二枚骰子的点数。每个单元格包含两个点数的和。
| + | 1 | 2 | 3 | 4 | 5 | 6 |
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| 6 | 7 | 8 | 9 | 10 | 11 | 12 |
to bottom left contains the value 7 exactly six times. Therefore, P(sum = 7) = 6/36 = 1/6.
借助这张表，我们可以快速计算得到点数之和为7的概率。从右上方到左下方的对角线中，数值7恰好出现6次。因此，P(和为7)=6/36=1/6。
3. Constructing an Outcome Table | 如何构建结果表
To build an effective outcome table, follow these steps:
要构建一个有效的结果表，请遵循以下步骤：
Identify the two (or more) independent stages of the experiment. For each stage, list all possible outcomes.
确定实验的两个（或更多）独立阶段。列出每个阶段的所有可能结果。
Create a grid with these outcomes as row and column headings.
使用这些结果作为行和列的标题，创建网格。
Fill every cell with the combined outcome according to the rule defined in the question (sum, product, difference, etc.).
按照题目中定义的规则（和、积、差等）填充每个单元格。
Count the total number of cells. This is the denominator of the probability.
计算单元格的总数。这就是概率的分母。
Shade or circle the cells that satisfy the required event, then count them as the numerator.
将符合条件的单元格着色或圈出，并数出它们的数量作为分子。
In some experiments, outcomes are not equally likely. For example, if a coin is biased, the table still lists the possible outcomes, but probabilities cannot be found simply by counting cells. In IB problems, unless stated otherwise, you may assume that dice and coins are fair.
在某些实验中，结果并不是等可能的。例如，如果一枚硬币有偏差，结果表仍然可以列出可能结果，但概率不能简单地通过计数单元格来求。在IB题目中，除非另有说明，否则通常假设骰子和硬币是均匀的。
4. Calculating Basic Probabilities from the Table | 从结果表计算基本概率
For equally likely outcomes, the probability of an event A is:
对于等可能结果，事件A的概率为：
P(A) = (number of outcomes in A) / (total number of outcomes)
s 10, 11, and 12. There are 3 + 2 + 1 = 6 such cells. Therefore, P(sum ≥ 10) = 6/36 = 1/6.
使用骰子求和表，我们可以求P(和≥10)。数一数和为10、11、12的单元格，分别是3+2+1=6个。因此，P(和≥10)=6/36=1/6。
onds to rows 1, 3, 5 and columns 1, 3, 5: 3 × 3 = 9 cells. So P(product odd) = 9/36 = 1/4.
另一个例子：掷两枚骰子时P(乘积为奇数)。只有当两个数都是奇数时乘积才是奇数。在表格中，这对应行1、3、5和列1、3、5：共3×3=9个单元格。因此P(乘积为奇数)=9/36=1/4。
5. Joint Events and 'And' Probabilities | 联合事件与“且”的概率
two conditions must hold simultaneously. For instance, from the two-dice table, the event "first die shows 2 AND the sum is 6" occurs only at the cell (2,4). This gives P = 1/36.
结果表天然地展示了联合事件，即两个条件必须同时成立。例如，在两枚骰子表格中，“第一枚骰子显示2且和为6”这一事件只发生在单元格(2,4)。因此P=1/36。
More generally, if events A and B are defined on the same sample space, the table helps identify the intersection A ∩ B. The cells that lie in the rows corresponding to A and also in the columns corresponding to B belong to A ∩ B.
更一般地，如果事件A和B定义在同一个样本空间上，表格有助于确定交集A∩B。位于A对应的行且同时位于B对应的列的单元格属于A∩B。
For independent events, the multiplication rule P(A∩ ample, when tossing two coins, P(Two heads) = 1/4 from the table, which matches 1/2 × 1/2.
对于独立事件，乘法规则P(A∩B)=P(A)×P(B)仍然适用，但表格提供了直接的视觉验证。例如，抛两枚硬币时，P(两个正面)=1/4，这与1/2×1/2一致。
6. Complementary Events and Table Symmetry | 对立事件与表格对称性
Every outcome table also simplifies finding complementary probabilities. The complement of an event A is the set of all cells not in A. Its probability is 1 − P(A).
每个结果表也可以简化对立事件的概率求解。事件A的对立事件是表格中所有不在A中的单元格。其概率为1−P(A)。
0 cells) or use the complement: P(sum = 7) = 6/36, so P(sum ≠ 7) = 1 − 6/36 = 30/36 = 5/6.
例如，要求P(和≠7)，我们可以数出不包含7的单元格（30个），也可以使用对立事件：P(和=7)=6/36，所以P(和≠7)=1−6/36=30/36=5/6。
Symmetry in the table can also reduce work. In a table of sums, the number of cells giving a sum of k is symmetric: the count for k equals the count for 14 − k. This is because rolling (a,b) and (b,a) are equally likely.
表格中的对称性也可以减少计算量。在求和表中，和为k的单元格数量关于k=7对称：和为k的数量等于和为14−k的数量。这是因为(a,b)和(b,a)等可能。
ymmetry, if we know P(sum = 4) = 3/36, then P(sum = 10) = 3/36 as well, since 4 + 10 = 14.
利用对称性，若已知P(和=4)=3/36，则P(和=10)也等于3/36，因为4+10=14。
7. Conditional Probability from an Outcome Table | 从结果表求条件概率
Conditional probability can be read directly from a table by restricting the sample space. The formula is:
条件概率可以直接从表格中读取，方法是限定样本空间。公式为：
P(A | B) = P(A ∩ B) / P(B)
When using a table, if we know that event B has occurred, we only consider the rows/columns that define B. The denominator is the number of outcomes in B, and the numerator is the number of outcomes in both A and B.
使用表格时，如果我们知道事件B已经发生，我们只考虑定义B的行/列。分母是B中的结果数，分子是同时在A和B中的结果数。
Example: From the two-dice sum table, find P(first die = 3 | Among these, only (3,5) has first die = 3. Therefore, the conditional probability is 1/5.
例子：在掷两枚骰子的求和表中，求P(第一枚=3 | 和=8)。已知和为8，可能的有序对为(2,6)、(3,5)、(4,4)、(5,3)、(6,2)，共5个结果。其中只有(3,5)满足第一枚=3。因此条件概率为1/5。
oning explicit, helping students avoid the common mistake of ignoring the given condition.
这与无条件概率P(第一枚=3)=6/36=1/6不同。表格使条件明确化，帮助学生避免忽略已知条件这一常见错误。
8. Union of Events and 'Or' Probabilities | 事件的并集与“或”的概率
The union of two events A or B can be counted from the table by marking all cells that belong to A or to B (or both). Care must be taken not to double-count the intersection.
两个事件的并集“A或B”可以通过在表格中标记所有属于A或B（或两者）的单元格来计数。注意不要重复计算交集。
The addition rule is: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). The table provides all three quantities directly.
加法规则为：P(A∪B)=P(A)+P(B)−P(A∩B)。表格直接提供了这三个量。
For example, in one roll of two dice, let A be "sum = 6" and B be "first die = 4". From the table, P(A) = 5/36, P(B) = 6/36. The intersection A ∩ B occurs at (4,2) only, so P(A∩B) = 1/36. Hence P(A∪B) = 5/36 + 6/36 − 1/36 = 10/36 = 5/18.
例如，掷两枚骰子一次，设A为“和为6”，B为“第一枚为4”。从表中可知P(A)=5/36，P(B)=6/36。交集A∩B仅发生在(4,2)，所以P(A∩B)=1/36。因此P(A∪B)=5/36+6/36−1/36=10/36=5/18。
A two-way table is especially helpful here because it visualises the overlap. If we simply added 5 and 6 without subtracting the overlap, we would incorrectly get 11/36.
这里二维表格特别有用，因为它直观地显示了重叠部分。如果简单地相加5+6而不减去重叠，就会错误地得到11/36。
9. Two-Way Tables for Contingency Data | 列联表（双向表）在分类数据中的应用
Outcome tables are not limited to dice and coins. In IB statistics, two-way tables (contingency tables) summarise categorical data from surveys or experiments. Each cell counts how many individuals fall into a particular combination of categories.
结果表不仅限于骰子和硬币。在IB统计学中，双向表（列联表）用于汇总来自调查或实验的分类数据。每个单元格统计属于特定类别组合的个体数量。
Consider a group of 80 students, 35 of whom study physics (P), 28 study economics (E), and 12 study both. The two-way table is:
考虑80名学生，其中35人学习物理（P），28人学习经济（E），12人两科都学。双向表如下：
| Studies E | Not E | Total | |
| Studies P | 12 | 23 | 35 |
| Not P | 16 | 29 | 45 |
| Total | 28 | 52 | 80 |
We can immediately read probabilities. P(P and E) = 12/80 = 3/20. P(P | = 12/28 = 3/7, because given the student studies E, only the first column is relevant. P(E | P) = 12/35, because given the student studies P, only the first row is relevant.
我们可以直接读取概率。P(学物理且学经济)=12/80=3/20。P(学物理 | 学经济)=12/28=3/7，因为已知该生学经济，只考虑第一列。P(学经济 | 学物理)=12/35，因为已知该生学物理，只考虑第一行。
tables also allow quick checks for independence. Two events A and B are independent if P(A∩B) = P(A)P(B). Here P(P)×P(E) = (35/80)×(28/80) = 980/6400 ≈ 0.1531, while P(P∩E) = 12/80 = 0.15. Since 0.15 ≠ 0.1531, the events are not independent.
双向表还可以快速检查独立性。事件A和B独立当且仅当P(A∩B)=P(A)P(B)。此处P(物理)×P(经济)=(35/80)×(28/80)=980/6400≈0.1531，而P(物理∩经济)=12/80=0.15。由于0.15≠0.1531，这两个事件并不独立。
10. Applications in Typical IB Exam Questions | 在典型IB考试题目中的应用
IB exam questions often combine outcome tables with tree diagrams, Bayes' theorem, and probability distribution functions. A small table can save time and reduce errors in multi-step questions.
IB考试题目通常将结果表与树状图、贝叶斯定理和概率分布函数结合。一个小表格可以在多步骤问题中节省时间并减少错误。
as 2, sum 5 has 4, sum 7 has 6, sum 11 has 2. Total = 1+2+4+6+2 = 15. So P = 15/36 = 5/12.
例子（不使用计算器）：掷两枚均匀骰子，求点数之和为质数的概率。可能的质数和为2、3、5、7、11。计数：和为2有1个，和为3有2个，和为5有4个，和为7有6个，和为11有2个。总计=1+2+4+6+2=15。因此P=15/36=5/12。
ment. In such cases, counts are still used, but the total is 52 × 51 = 2652 ordered pairs.
另一个常见的IB题：从一副标准扑克牌中不放回地抽取两张。可以针对第一张和第二张的颜色构建双向表，但要注意不放回时结果不是等可能的。此时仍然使用计数，但总数是52×51=2652个有序对。
When questions involve "at least one", tables become powerful. For example, if a fair coin is tossed 3 times, the table with 8 cells allows us to count "at least one head" as 7 cells. The only cell with no head is TTT.
当问题涉及“至少一次”时，表格变得非常强大。例如，均匀硬币抛3次，8个单元格的结果表可以让我们数出“至少一个正面”是7个单元格。唯一没有正面的单元格是TTT。
11. Common Pitfalls and How to Avoid Them | 常见误区及规避方法
Students often make mistakes when applying outcome tables. Here are the most important pitfalls:
学生在应用结果表时经常出错。以下是最重要的几个误区：
Forgetting that ordered outcomes (1,2) and (2,1) are distinct when rolling two dice. Always use ordered pairs in the table unless the question specifically says unordered.
忘记掷两枚骰子时(1,2)和(2,1)是有区别的。除非题目特别说明不区分顺序，否则在表格中应使用有序对。
Counting cells without checking that every cell is equally likely. For example, when drawing without replacement, each ordered pair is equally likely, but each unordered pair is not.
直接数单元格而未检查每个单元格是否等可能。例如在不放回抽取时，每个有序对等可能，但每个无序对不等可能。
Using the table for conditional probability but forgetting to restrict the denominator to the given condition.
求条件概率时使用表格，但忘记将分母限定为已知条件。
Confusing P(A|B) and P(B|A). In a two-way table, P(A|B) uses the column B total as denominator, while P(B|A) uses the row A total.
混淆P(A|B)与P(B|A)。在双向表中，P(A|B)用B列总数作为分母，而P(B|A)用A行总数。
Forgetting to subtract the intersection when computing P(A ∪ B) if A and B are not mutually exclusive.
在A与B不互斥时，计算P(A∪B)忘记减去交集。
A systematic habit of labelling rows, columns, and totals will eliminate most of these errors.
养成系统地标注行、列和总计的习惯将消除大部分这些错误。
12. Practice Question with Full Solution | 练习题与完整解答
Let us solve a typical IB-style question using an outcome table.
让我们用结果表解决一道典型的IB风格题目。
Question: A biased coin has probability 2/3 of showing Heads. The coin is tossed three times. Use a tree diagram or outcome table to find the probability of getting exactly two Heads in any order.
题目：一枚不均匀硬币正面朝上的概率为2/3。将这枚硬币抛掷三次。用树状图或结果表求恰好出现两次正面（顺序任意）的概率。
Solution: uences. The probability of each sequence is the product of the probabilities of each toss.
解答：由于硬币不均匀，简单计数不适用。我们使用带概率的结果表。设H=正面，T=反面。共有8种序列。每个序列的概率是各次抛掷概率的乘积。
| Outcome | 1st toss | 2nd toss | 3rd toss | Probability |
| HHH | H | H | H | (2/3)³ = 8/27 |
| HHT | H | H | T | (2/3)²(1/3) = 4/27 |
| HTH | H | T | H | 4/27 |
| THH | T | H | H | 4/27 |
| HTT | H | T | T | (2/3)(1/3)² = 2/27 |
| THT | T | H | T | 2/27 |
| TTH | T | T | H | 2/27 |
| TTT | T | T | T | (1/3)³ = 1/27 |
HH. Their probabilities are each 4/27. Thus P(exactly two Heads) = 3 × 4/27 = 12/27 = 4/9.
恰好出现两次正面的序列是HHT、HTH、THH。它们的概率均为4/27。因此P(恰好两次正面)=3×4/27=12/27=4/9。
This example shows that even when counting is invalid, an outcome table organises the computation of
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