Calculating Probabilities
Every probability calculation starts the same way — count what you want, count everything that's possible, and divide. The skill is in counting the right things.
Sample space and events
An experiment is any process with an uncertain result — rolling a die, spinning a spinner, drawing a card. Each possible result is an outcome, and the sample space, $\mathscr{E}$, is the set of every outcome the experiment could produce. An event is any subset of the sample space — a collection of outcomes you're interested in, usually described by a rule ("rolling an even number") rather than listed individually.
The probability of event $A$, written $P(A)$, is always a number between $0$ and $1$: $P(A)=0$ means the event is impossible, $P(A)=1$ means it's certain to happen.
Equally likely outcomes
When every outcome in the sample space is just as likely as every other — a fair die, a fair coin, a well-shuffled deck — probability becomes a counting problem:
This formula only applies when the outcomes really are equally likely. Rolling a fair die satisfies this; spinning a spinner that's been weighted to favour one side does not. Whenever a question describes a "fair" or "unbiased" device, or simply says outcomes are "equally likely", this shortcut is valid — otherwise, probabilities have to be worked out or given directly instead of counted.
Sample space diagrams
When an experiment combines two smaller experiments — two dice, a coin and a spinner — listing every outcome as a table (rows for one experiment, columns for the other) makes counting reliable. This is called a sample space diagram. Once the grid is filled in, $P(A)$ is just (cells satisfying $A$) $\div$ (total cells).
Probability from grouped frequency tables
When data has been collected into a grouped frequency table, a probability can be estimated directly: $P(A) \approx \dfrac{\text{frequency in }A}{\text{total frequency}}$. If the value needed falls exactly on a class boundary this is immediate — but if it falls inside a class, the class is assumed to be spread evenly across its width, and only the appropriate fraction of that class's frequency is included. This is the same interpolation idea used to estimate a median from a cumulative frequency diagram.
Worked example 1
Worked example 2
| A × B | B = 1 | B = 2 | B = 3 |
|---|---|---|---|
| A = 1 | 1 | 2 | 3 |
| A = 2 | 2 | 4 | 6 |
| A = 3 | 3 | 6 | 9 |
| A = 4 | 4 | 8 | 12 |
Filling in the whole grid before counting is far safer than trying to count multiples of 3 "in your head" — it's easy to miss a repeated value like the two 3s and two 6s here if you don't lay every cell out.
Worked example 3
| Time (min) | $30\le t<40$ | $40\le t<50$ | $50\le t<60$ | $60\le t<70$ | $70\le t<80$ |
|---|---|---|---|---|---|
| Frequency | 6 | 14 | 22 | 12 | 6 |
Estimate the probability that a runner selected at random finished in under $45$ minutes.
Grouped data hides exactly where, within each class, the individual values actually fall. Interpolation assumes an even spread across the class width — a reasonable working assumption, but not a fact the table itself confirms. That's why probabilities calculated this way are always described as estimates.
Worked example 4
With a fair spinner, "$4$ favourable outcomes out of $16$" and "add up the $4$ probabilities" give the same answer, because every probability is $\tfrac1{16}$. Once outcomes stop being equally likely, only the second method — working directly with each outcome's own probability — still gives a correct answer. This is exactly the reasoning tree diagrams formalise later in this unit.
Check yourself
A fair eight-sided die is labelled $1$ to $8$ and rolled once. Find $P(\text{the number is prime})$ and $P(\text{the number is not prime})$.
(Hint: list the primes between 1 and 8, then use the complement.)