Sample space and events
Probability: How likely something is: a number from 0 (impossible) to 1 (certain).
What is a sample space?: The sample space, written U, is the set of all possible outcomes of a trial.
For one coin toss: U = {heads, tails}. For one die roll: U = {1, 2, 3, 4, 5, 6}.
An event is a selection of those outcomes. For example, “roll an even number” means {2, 4, 6}.
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Trial and outcome: Picking one student is a trial. "Walks" is one possible outcome.
Check two things before you divide: 1. U holds every possible outcome: all 25 students, none left out.
2. Every outcome is equally likely: each student has the same chance of being picked.
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Calculating probabilities
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Exam-style question [2 marks]
A month of the year is chosen at random.
(a) Find the probability that its name starts with M.
Solution
- List U: the 12 months, each equally likely.
- A = {March, May}.
- Divide.
Final answer
1/6 ≈ 0.167
Part (b)
(b) Find the probability that its name does not start with M.
Solution
- Not A: subtract from 1.
Final answer
5/6 ≈ 0.833
Exam tip: Give the answer as a fraction (2/12 or 1/6), a decimal (0.167, to 3 significant figures) or a percentage (16.7%): all get full marks.
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Complementary events
Complementary: Events covering all: P(A) + P(A′) = 1.
Example: a flight is late, or it is not late.
Worked example
P(rain)=0.3.
Find P(no rain).
Solution
- Rain and no rain cover every day, and cannot both happen.
- P(no rain) = 1 − 0.3 = 0.7
Final answer
0.7
"Not A" in words: Exam questions often ask which wording means A′. A′ is everything that is not in A.
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Two dice: the 36-outcome grid
What is a sample space diagram?: A sample space diagram organises every possible outcome in a list or grid, with none missed or repeated.
For two experiments, put the outcomes of one down the rows and the outcomes of the other across the columns. Each cell records one possible pair.
If the outcomes are equally likely, find a probability by counting the matching cells and dividing by the total number of cells.
Start with a smaller grid: toss a fair coin and independently spin a fair spinner numbered 1, 2, 3. Put the coin result down the side and the spinner result across the top.
| Coin / spinner | 1 | 2 | 3 |
|---|---|---|---|
| Heads | (H, 1) | (H, 2) | (H, 3) |
| Tails | (T, 1) | (T, 2) | (T, 3) |
Read one cell, then count: (H, 2) means heads AND spinner 2. There are 2 × 3 = 6 equally likely pairs.
“Heads and an even number” matches only (H, 2), so its probability is 1/6. “Heads” matches the three cells in the Heads row, so its probability is 3/6 = 1/2.
Now use the same idea for two dice
For two independent fair six-sided dice, use 6 rows for the first die and 6 columns for the second die. There are 6 × 6 = 36 equally likely ordered pairs.
The cell at row 2, column 5 represents (2, 5). The cell at row 5, column 2 represents (5, 2). They are different outcomes, even though both sums are 7.
The grid below writes the sum inside each cell. Count the cells matching the event; do not count only the different sums.
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Two dice: counting pairs in the exam
No grid? Go through the first die: For one event, try each value of the first die and ask what the second die must be. It must be 1 to 6. This way no pair is missed, including the reversed one.
| First die | Second die needed (11 − first) | Possible? |
|---|---|---|
| 1 | 10 | no |
| 2 | 9 | no |
| 3 | 8 | no |
| 4 | 7 | no |
| 5 | 6 | yes: (5, 6) |
| 6 | 5 | yes: (6, 5) |
Another check: product = 12: Same method, different rule: the second die must be 12 ÷ first die, and it must be a whole number from 1 to 6.
| First die | Second die needed (12 ÷ first) | Possible? |
|---|---|---|
| 1 | 12 | no |
| 2 | 6 | yes: (2, 6) |
| 3 | 4 | yes: (3, 4) |
| 4 | 3 | yes: (4, 3) |
| 5 | 2.4 | no (not a whole number) |
| 6 | 2 | yes: (6, 2) |
P(product = 12): 4 pairs work: (2, 6), (3, 4), (4, 3), (6, 2). So P(product = 12) = 4/36 = 1/9.
One event, or several?: Only (5, 6) and (6, 5) work, so P(sum = 11) = 2/36.
One event: go through the first die like this. Several events in one question (e.g. P(sum = 7), P(sum = 8) and P(sum ≥ 10)): draw the 6 × 6 grid once and count each from it.
Exam-style question [2 marks]
Two fair six-sided dice are rolled.
(a) Find the probability that the sum is at least 10.
Solution
- U: 36 equally likely pairs.
- Count in the sum grid: 10, 11 or 12 sits in 6 cells, (4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6).
Final answer
1/6 ≈ 0.167
Part (b)
(b) Find the probability that the product of the two numbers is 6.
Solution
- Go through the first die: the second must be 6 ÷ first, a whole number from 1 to 6. 1 → 6 yes, 2 → 3 yes, 3 → 2 yes, 4 → 1.5 no, 5 → 1.2 no, 6 → 1 yes. So (1, 6), (2, 3), (3, 2), (6, 1): 4 pairs.
Final answer
1/9 ≈ 0.111
Exam tip: Count the pairs, not the totals. A sum of 7 can be made 6 ways: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1). A sum of 12 can be made only 1 way: (6, 6). So P(sum = 7) = 6/36 and P(sum = 12) = 1/36.