Probability
Uncertainty becomes a single number between 0 and 1 — and a handful of diagrams turn even chained, dependent events into arithmetic you can trust.
The Probability of an Event
Probability turns "how likely?" into a number. When outcomes are equally likely, that number is just a count of the favourable outcomes over the total.
Equally likely outcomes
If every outcome has the same chance — a fair coin or a fair dice — the probability of an event is the fraction of outcomes that produce it: P(A) = number of favourable outcomes ÷ total number of equally likely outcomes. This theoretical probability needs no experiment, and is only valid when every outcome really is equally likely.
The complement: "not A"
Because some outcome is certain, all the probabilities add to 1. The chance an event does not happen is therefore one minus the chance it does, P(A′) = 1 − P(A) — often the quickest route to an answer.
Worked example: a fair spinner has 8 equal sectors: 3 red, 2 blue and 3 green. Find the probability that it lands on red, and the probability that it does not land on blue. Step 1: the 8 sectors are equally likely, so P(red) = 3 favourable ÷ 8 total = ⅜. Step 2: P(blue) = 2⁄8 = ¼, so by the complement P(not blue) = 1 − ¼. Step 3: both answers sit correctly between 0 and 1 — a quick sanity check that never costs time. P(red) = ⅜, P(not blue) = ¾.
Relative & Expected Frequency
Not every chance can be counted in advance. When a dice might be biased, a probability can only be estimated by running trials and counting how often the event happens.
Estimating from experiment
The relative frequency of an event is its share of the results so far — the best estimate the data allows: relative frequency = number of times the event occurs ÷ total number of trials. As trials grow, it settles towards the true probability.
Expected frequency
Turn the idea around and it predicts results: expected frequency = P(event) × number of trials — a prediction, not a guarantee.
Worked example: a biased dice is rolled 200 times and lands on a six on 56 of them. Estimate the probability of a six, then predict how many sixes to expect in 500 rolls. Step 1: relative frequency of a six = 56⁄200 = 0.28 — the best estimate of P(six). Step 2: expected sixes in 500 rolls = 0.28 × 500. Step 3: the dice is biased — a fair dice would give about ⅙ × 500 ≈ 83, far fewer. P(six) ≈ 0.28, expect 140 sixes.
Combined Events & Venn Diagrams
Most exam questions combine events: A or B, A and B. The trick is to organise the outcomes first — in a grid, or in a Venn diagram — and only then count. Two dice give a 6×6 grid of 36 equally likely outcomes, a sample space you can count directly.
The OR rule for exclusive events
If two events cannot happen together, the chance of one or the other is simply the sum of their chances: for mutually exclusive events, P(A ∪ B) = P(A) + P(B). On a Venn diagram this is the union ∪, and because there is no overlap, nothing is counted twice.
Reading a Venn diagram
A Venn diagram sorts a group into four regions: A only, B only, both (the intersection A ∩ B), and neither. Fill the overlap first, then work outwards so each region’s count is correct. Every probability is then a region total over the grand total.
Worked example: using Fig 8.3, a student is chosen at random. Find the probability that the student studies French or Spanish (or both). Step 1: the regions are French only 11, both 7, Spanish only 5 — total in the union = 11 + 7 + 5 = 23. Step 2: there are 30 students in all, so P(French or Spanish) = 23⁄30. Step 3: check — 18 + 12 − 7 = 23, subtracting the overlap counted in both, the same answer. P(French or Spanish) = 23⁄30.
Tree Diagrams
When events happen one after another, a tree diagram lays out every path. Following a path multiplies the branch probabilities; collecting the paths that satisfy the question adds them.
Multiply along the branches
For independent events, the chance of A and B is the product of their probabilities: P(A ∩ B) = P(A) × P(B). On a tree each complete path is one combined outcome, so its probability is the branches multiplied together.
With and without replacement
If the first item is put back, the second draw faces the same probabilities — the events are independent. If it is not replaced, both the total and the count change, so the second set of branch probabilities is different.
Worked example: a bag holds 5 red and 3 blue counters. Two are taken at random without replacement. Find the probability that both are red. Step 1: first counter red, P = 5⁄8 (5 of 8 counters are red). Step 2: one red is now gone, leaving 4 red of 7, so the second red has P = 4⁄7. Step 3: multiply along the top path, 5⁄8 × 4⁄7 = 20⁄56, which cancels to 5⁄14. P(both red) = 5⁄14.
Conditional Probability
Sometimes you are told part of what happened before you work out the rest. Conditional probability handles this by shrinking the picture: only the outcomes consistent with the given information are still in play. "Without replacement" in the previous section is conditional probability already — the second branch depends on the first.
Working from a restricted set
Knowing that event B has happened rules out everything outside B. The probability of A is then measured against B alone — the total becomes the number in B, not the number in the whole group. Venn diagrams, two-way tables and tree diagrams all make that smaller total easy to read off.
Reading it off the diagram
The count in the overlap is unchanged; what changes is the total you divide by. That single adjustment — a smaller denominator — is the whole of conditional probability at this level.
Worked example: using Fig 8.5, a person who owns a dog is chosen. Find the probability that they also own a cat. Step 1: restrict to dog owners — 15 own only a dog and 6 own both, so 21 in all. Step 2: of those 21, the ones who also own a cat are the 6 in the overlap. Step 3: so the probability is 6⁄21, which cancels to 2⁄7 — smaller than the unconditional P(cat) = 14⁄40, because the total has changed. P(cat, given dog) = 2⁄7.
Exam advice
Common mistakes
Model answer
Recall checklist
- State the probability scale and what 0 and 1 mean.
- Write and use the complement P(A′) = 1 − P(A).
- Calculate a relative frequency and an expected frequency.
- Distinguish mutually exclusive (add) from independent (multiply).
- Read ∩ and ∪ from a two-set Venn diagram.
- Complete a tree diagram and multiply along its branches.
- Adjust the probabilities for selection without replacement.
- Calculate a conditional probability from a restricted set.
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