Sample spaces
The sample space is simply the list of every possible outcome of an experiment. Once you have it, the probability of an event is:
For two events happening together, a sample space diagram (a grid) is the clearest way to list everything. Rolling two fair dice and adding the scores gives \(6 \times 6 = 36\) equally likely outcomes.
- Compound event
- An event made of more than one part, such as "roll two dice" or "draw two counters". The sample space grows quickly, so a diagram keeps it organised.
Two fair dice are rolled and the scores added. Find the probability the total is 7.
Venn diagrams
A Venn diagram sorts outcomes into overlapping circles, one per event. The overlap holds outcomes in both events. Venn diagrams make the addition rule visible:
You subtract the overlap \(P(A \cap B)\) because adding \(P(A)\) and \(P(B)\) counts it twice. The symbol \(\cup\) means "or" (union) and \(\cap\) means "and" (intersection).
In a class of 30, 18 play football (F), 12 play basketball (B), and 5 play both. Find \(P(F \cup B)\) and the number who play neither.
Fill the middle first
Always write the "both" number into the overlap before anything else, then subtract it from each total. Filling the circle totals directly is the quickest way to double-count.
Tree diagrams: multiplying along branches
When events happen in sequence, a tree diagram is the tool. Each branch carries the probability of that step, and there are two golden rules:
- Multiply along the branches (left to right) to find the probability of a full path.
- Add between separate paths when more than one path satisfies the event.
The probabilities on each pair of branches from a point must add to \(1\), which is a handy self-check.
A bag holds 5 red and 3 blue counters. Two are drawn without replacement. Find \(P(\text{both red})\).
With and without replacement
The single most important question in these problems is whether the first item is put back.
| With replacement | Without replacement | |
|---|---|---|
| Total each draw | Stays the same | Drops by one |
| Second-draw probability | Unchanged (independent) | Changes (dependent) |
| Bag of 5 red, 3 blue | \(P(\text{red}) = \tfrac{5}{8}\) both times | \(\tfrac{5}{8}\) then \(\tfrac{4}{7}\) |
Continuing the bag above without replacement, the full set of outcomes checks out to \(1\): \(P(\text{RR}) = \tfrac{5}{8}\times\tfrac{4}{7} = \tfrac{20}{56}\); \(P(\text{one of each}) = \tfrac{5}{8}\times\tfrac{3}{7} + \tfrac{3}{8}\times\tfrac{5}{7} = \tfrac{30}{56}\); \(P(\text{BB}) = \tfrac{3}{8}\times\tfrac{2}{7} = \tfrac{6}{56}\). Total \(= \tfrac{20+30+6}{56} = \tfrac{56}{56} = 1\).
Where this is assessed
Choosing the right diagram and computing correctly is Criterion A. A clearly labelled tree with probabilities on every branch is Criterion C. Investigating "is this game fair?" by comparing probabilities pulls in Criterion B and D.
Do not simplify too early
Keep fractions over the same denominator (like \(56\) above) while you add paths. Cancelling each branch to lowest terms first makes the adding step far more error-prone.
Check yourself
Work each through before opening the answer.
1. Two fair dice are rolled. What is the probability of a total of 5? +
Out of \(36\) outcomes, the pairs totalling 5 are \((1,4), (2,3), (3,2), (4,1)\), which is \(4\) outcomes. So \(P = \dfrac{4}{36} = \) \(\dfrac{1}{9}\).
2. Of 40 people, 22 like tea, 18 like coffee, 10 like both. How many like neither? +
Tea only \(= 22 - 10 = 12\); coffee only \(= 18 - 10 = 8\). At least one \(= 12 + 10 + 8 = 30\). Neither \(= 40 - 30 = \) \(10\).
3. From the bag of 5 red and 3 blue, two are drawn without replacement. Find \(P(\text{both blue})\). +
First blue: \(\dfrac{3}{8}\). Then \(2\) blue remain of \(7\): \(\dfrac{2}{7}\). Multiply: \(\dfrac{3}{8} \times \dfrac{2}{7} = \dfrac{6}{56} = \) \(\dfrac{3}{28}\).
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