Sample Spaces and Two-Way Tables
A sample space lists every possible outcome. A careful list, grid or table lets you count the outcomes you want without missing any or counting the same outcome twice.
Foundation & Higher. Use Probability Basics and Rules first. For general counting strategies, see Systematic Listing and Counting.
Build a complete sample space
For a fair six-sided die, the sample space is {1, 2, 3, 4, 5, 6}. Curly brackets list the members of a set. For two stages, record both results in an agreed order.
Worked example — two coin tosses
For two independent fair coin tosses, list HH, HT, TH, TT. H means heads and T means tails; the first letter is the first toss.
Each ordered pair has probability 1/4. Exactly one head occurs in HT or TH, so its probability is 2/4 = 1/2.
The totals “0 heads, 1 head, 2 heads” are not equally likely: there are two routes to one head but only one route to each other total.
Work systematically: keep the first result fixed while listing every second result, then change the first. For the arrangements of distinct letters A, B and C, this gives ABC, ACB, BAC, BCA, CAB, CBA.
A grid for two spinners
Spinner A has two equal sectors labelled 1 and 2. Spinner B has three equal sectors labelled 1, 2 and 3. Spin both independently. There are six equally likely ordered pairs. The table records their sums.
| A / B | 1 | 2 | 3 |
|---|---|---|---|
| 1 | 2 | 3 | 4 |
| 2 | 3 | 4 | 5 |
Worked example — count favourable cells
A sum of 4 appears twice: (1,3) and (2,2). P(sum 4) = 2/6 = 1/3. A sum greater than 3 appears in three cells, so P(sum greater than 3) = 1/2.
The grid has equal-chance cells because the sectors on each spinner are equal and the spins are independent. Unequal sector sizes would require probabilities or weights rather than a simple favourable-cell count.
Two dice: outcomes are not the same as totals
Roll two independent fair six-sided dice, one designated first and one second. Each of the 36 ordered pairs is equally likely. Add the die scores in each cell.
Worked example — compare two sums
A sum of 7 comes from (1,6), (2,5), (3,4), (4,3), (5,2) or (6,1): P(sum 7) = 6/36 = 1/6.
Only (1,1) gives a sum of 2, so P(sum 2) = 1/36. It is wrong to assign each possible total from 2 to 12 a probability of 1/11.
Worked example — count without double counting
Equal scores occur in the six diagonal cells: (1,1) to (6,6). P(equal scores) = 1/6. The two ordered outcomes (2,5) and (5,2) are different, even though both sum to 7.
Read and complete a two-way table
A two-way table classifies the same group in two ways. Each person or trial belongs in one interior cell; the margins show totals. Here every student either cycles or walks, and is in Year 10 or Year 11.
| Year | Cycle | Walk | All |
|---|---|---|---|
| 10 | 9 | 11 | 20 |
| 11 | 6 | 14 | 20 |
| All | 15 | 25 | 40 |
If the Year 11 walking count were missing, subtract 6 from its row total: 20 − 6 = 14. Check the column too: 25 − 11 = 14. Do not add all the interior cells and then add the margins again.
Worked example — select from the whole group
One of the 40 students is chosen uniformly at random. P(cycles) = 15/40 = 3/8. P(Year 10 and cycles) = 9/40.
P(Year 10 or cycles) includes the 9 Year 10 cyclists, 11 Year 10 walkers and 6 Year 11 cyclists. This gives 26/40 = 13/20, counting each student once.
The denominator is 40 because the selection is from all 40 students. If you are told the selected student cycles, the relevant group changes; that is the Higher conditional probability lesson.
Choose a representation that fits
Use a list or grid for a manageable set of equally likely outcomes. Use a two-way table for counts classified by two categories. Use a tree for successive stages, especially when probabilities change after an earlier result.
A complete sample space still helps when some outcomes have different probabilities, but you must attach the correct probabilities instead of just counting cells. The general product rule for counting is covered in Number; here, listing and grids show why the count works.
Think Like a Mathematician
Why is “at least one head” more likely than “two heads” in two fair tosses? Mark HH, HT and TH for the first event, but only HH for the second. A description of an event can cover several equally likely outcomes.
Common Mistakes
Dropping the order: HT and TH are separate outcomes when the toss positions are distinguished.
Assuming totals are equally likely: count the original equally likely cells, not just distinct sums.
Double counting table margins: totals summarise the interior cells; they are not extra people.
Practice
1. List the outcomes of two independent fair coin tosses and find P(at least one head).
HH, HT, TH, TT. Three of the four outcomes have at least one head, so P = 3/4.
2. Spinner A has equal sectors 1 and 3; spinner B has equal sectors 2, 4 and 6. The spins are independent. List all six sums systematically.
For A = 1, the sums are 3, 5, 7. For A = 3, they are 5, 7, 9. These are six cells, although some sums repeat.
3. For those spinners, find P(sum 7) and P(sum 9).
Sum 7 occurs in two of six equally likely cells: 1/3. Sum 9 occurs in one cell: 1/6.
4. Two independent fair six-sided dice are rolled. Find P(sum 9).
The pairs are (3,6), (4,5), (5,4), (6,3): 4/36 = 1/9.
5. Using the same dice, find P(the first score is greater than the second).
There are 5 + 4 + 3 + 2 + 1 = 15 suitable cells, so 15/36 = 5/12. The six equal-score cells are excluded.
6. A two-way table contains 50 people. Group A has 18 people, including 7 walkers. Group B has 9 walkers. Complete the non-walker counts and find P(walker) for a random person.
A non-walkers = 18 − 7 = 11. B total = 50 − 18 = 32, so B non-walkers = 32 − 9 = 23. P(walker) = (7 + 9)/50 = 8/25.
7. In that table of 50 people, find P(group A and walker) and P(group A or walker).
AND: 7/50. OR: count all 18 in A plus the 9 walkers in B, giving 27/50.
8. A grid has six cells, but one spinner has unequal sector sizes. Is each cell necessarily probability 1/6?
No. The ordered outcomes need not be equally likely. Use the appropriate sector probabilities; merely counting cells is insufficient.
Frequently Asked Questions
When does order matter?
When outcomes are distinguished by stage or device, record them in that order. Both orders may later belong to the same event, such as getting one head and one tail.
Is a two-way table a probability tree?
No. A table cross-classifies counts; a tree shows stages or category splits. Both can represent related information, but their labels and totals must be interpreted correctly.
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