How to Use Sample Space to Calculate Probability (IGCSE Maths)
Understanding how to list a sample space is essential for solving probability questions in IGCSE Maths. This page explains how sample spaces are used to organise outcomes and calculate probabilities accurately, using exam style examples involving coins, dice and ordered pairs.
What is a Sample Space?
A sample space is the complete list of every possible outcome of an experiment. Drawing a sample space diagram (usually a table or grid) makes it easy to count outcomes and calculate probabilities.
Once you have the sample space, the probability of any event is:
The total of all probabilities in the sample space always equals 1.
Core Ideas
A sample space must include all possible outcomes. Missing even one will make every probability wrong.
When two things happen (e.g. two dice, or a coin and a die), draw a two-way table with one event along the top and the other down the side.
Count the outcomes that match what you want, then divide by the total number of outcomes. Always simplify the fraction.
Coin and Die
Flip a fair coin (Heads or Tails), then roll a fair six-sided die (1 to 6). There are \(2 \times 6 = 12\) equally likely outcomes.
| DieCoin | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| H | (H, 1) | (H, 2) | (H, 3) | (H, 4) | (H, 5) | (H, 6) |
| T | (T, 1) | (T, 2) | (T, 3) | (T, 4) | (T, 5) | (T, 6) |
💡 Example 1: Specific outcome
Find the probability of getting Heads and a 4.
There is exactly 1 outcome (H, 4) out of 12.
\[P(\text{H and 4}) = \frac{1}{12}\]💡 Example 2: Even number on the die
Find the probability that the die shows an even number.
The even numbers are 2, 4, 6. Each can appear with H or T, giving 6 outcomes.
\[P(\text{even}) = \frac{6}{12} = \frac{1}{2}\]Rolling Two Dice: Sum of Outcomes
Roll two fair dice. There are \(6 \times 6 = 36\) equally likely ordered pairs. The table below shows the sum of the two dice for each pair.
| Die 2Die 1 | 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 |
💡 Example 3: Sum of 7
Find the probability that the two dice add up to 7.
Count the 7s in the table: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). That is 6 outcomes.
\[P(\text{sum} = 7) = \frac{6}{36} = \frac{1}{6}\]💡 Example 4: Sum of 4 or less
Find \(P(\text{sum} \leq 4)\).
Sum = 2: one way (1,1).
Sum = 3: two ways (1,2), (2,1).
Sum = 4: three ways (1,3), (2,2), (3,1).
Total: \(1 + 2 + 3 = 6\) outcomes.
\[P(\text{sum} \leq 4) = \frac{6}{36} = \frac{1}{6}\]💡 Example 5: Least likely sum
Which sum is least likely, and what is its probability?
The sums 2 and 12 each appear only once in the table: (1,1) for 2, and (6,6) for 12.
\[P(\text{sum} = 2) = P(\text{sum} = 12) = \frac{1}{36}\]💡 Example 6: Expected frequency
If two dice are rolled 180 times, how many times would you expect a sum of 7?
\(P(\text{sum} = 7) = \frac{1}{6}\)
\[\text{Expected} = 180 \times \frac{1}{6} = 30\]You would expect a sum of 7 about 30 times.
Distinguishing Two Dice: Ordered Pairs
When two dice are different colours (e.g. one red and one blue), the outcomes are ordered pairs (red, blue). There are still 36 outcomes, but the order matters: (2, 5) is different from (5, 2).
| BlueRed | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| 1 | (1,1) | (1,2) | (1,3) | (1,4) | (1,5) | (1,6) |
| 2 | (2,1) | (2,2) | (2,3) | (2,4) | (2,5) | (2,6) |
| 3 | (3,1) | (3,2) | (3,3) | (3,4) | (3,5) | (3,6) |
| 4 | (4,1) | (4,2) | (4,3) | (4,4) | (4,5) | (4,6) |
| 5 | (5,1) | (5,2) | (5,3) | (5,4) | (5,5) | (5,6) |
| 6 | (6,1) | (6,2) | (6,3) | (6,4) | (6,5) | (6,6) |
💡 Example 7: Specific pair
Find the probability that the red die shows 3 and the blue die shows 5.
There is exactly 1 outcome: (3, 5).
\[P(\text{red 3 and blue 5}) = \frac{1}{36}\]💡 Example 8: Red greater than blue
Find the probability that the red die shows a higher number than the blue die.
Count the outcomes where red > blue by going through each row of the table:
Red 2: 1 outcome. Red 3: 2. Red 4: 3. Red 5: 4. Red 6: 5.
Total: \(1+2+3+4+5 = 15\)
\[P(\text{red} > \text{blue}) = \frac{15}{36} = \frac{5}{12}\]💡 Example 9: At least one 6
Find the probability that at least one of the dice shows a 6.
It is easier to use the complement here. "At least one 6" is the opposite of "no sixes at all".
Outcomes with no 6 on either die: each die has 5 non-six options, so \(5 \times 5 = 25\) outcomes.
\[\begin{aligned} P(\text{no sixes}) &= \frac{25}{36} \\[4pt] P(\text{at least one 6}) &= 1 - \frac{25}{36} = \frac{11}{36} \end{aligned}\]Using the complement avoids counting the 11 outcomes individually.
Other Ways to Combine Two Dice
A two-way table does not only add the two scores. The cell can hold the product (multiply the scores) or the difference (subtract the smaller score from the larger). The method never changes: fill the table, count the cells that match, then divide by 36.
💡 Example 10: Product of two dice
Two fair dice are rolled and the scores are multiplied. Find the probability that the product is 12.
The pairs that multiply to 12 are (2, 6), (3, 4), (4, 3) and (6, 2): 4 outcomes out of 36.
\[P(\text{product} = 12) = \frac{4}{36} = \frac{1}{9}\]💡 Example 11: Difference of two dice
Two fair dice are rolled. Find the probability that the difference between the scores is 1.
The pairs one apart are (1,2), (2,1), (2,3), (3,2), (3,4), (4,3), (4,5), (5,4), (5,6) and (6,5): 10 outcomes.
\[P(\text{difference} = 1) = \frac{10}{36} = \frac{5}{18}\]🔑 Key Points
- A sample space lists every possible outcome of an experiment.
- For two combined events, draw a two-way table (grid) with one event on each axis.
- The total number of outcomes = rows \(\times\) columns.
- Probability = favourable outcomes ÷ total outcomes. Always simplify the fraction.
- "At least one" questions are often easier using the complement: \(P(\text{at least one}) = 1 - P(\text{none})\).
- If two dice are distinguishable (different colours), (2, 5) and (5, 2) are different outcomes.
⚠️ Common Mistakes
- Missing outcomes: always draw the full table before answering. A missing row or column ruins every calculation.
- Treating (2, 5) and (5, 2) as the same: when dice are distinguishable, these are separate outcomes.
- Forgetting to simplify: \(\frac{6}{36}\) should be written as \(\frac{1}{6}\).
- Counting the wrong event: read the question carefully. "Sum equals 7" is different from "at least one die shows 7" (impossible on a standard die).
- Using the wrong total: two dice give 36 outcomes, not 12. A coin and a die give 12, not 6.