Probability measures how likely an event is to happen, on a scale from 0 (impossible) to 1 (certain). In GCSE Maths you need to move fluently between fractions, decimals and percentages, calculate probabilities for single and combined events, and present your reasoning using sample space diagrams, tree diagrams and Venn diagrams. This topic rewards clear, organised working: most marks are earned by listing outcomes carefully, multiplying along branches and adding the right combinations. The higher-tier ideas of conditional probability and sampling without replacement build directly on these foundations, so it pays to make the basics automatic before tackling them.
The Probability Scale and Notation
Every probability is a number between 0 and 1 inclusive. A probability of 0 means the event is impossible, 1 means it is certain, and $0.5$ means it is as likely to happen as not. You can write a probability as a fraction, a decimal or a percentage, but a fraction is usually safest in exams because it avoids rounding. We write $P(A)$ to mean the probability that event $A$ happens. For equally likely outcomes the basic rule is $P(A)=\frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$. For example, rolling a fair six-sided dice, $P(\text{even})=\frac{3}{6}=\frac{1}{2}$. A key result is that the probabilities of all possible outcomes always add up to 1, which leads to the complement rule: the probability that $A$ does not happen is $P(\text{not }A)=1-P(A)$. So if $P(\text{rain})=0.3$ then $P(\text{no rain})=0.7$. Always check your answers are not negative and not greater than 1 - if they are, you have made a slip.
Theoretical vs Experimental Probability and Relative Frequency
Theoretical probability is what you expect to happen based on the assumption that outcomes are equally likely, such as $P(\text{head})=\frac{1}{2}$ for a fair coin. Experimental probability, also called relative frequency, is what you actually observe when you carry out trials: $\text{relative frequency}=\frac{\text{number of times the event happened}}{\text{total number of trials}}$. For example, if a drawing pin lands point-up 37 times in 50 throws, the relative frequency is $\frac{37}{50}=0.74$. The two values usually differ for small numbers of trials, but the more trials you carry out, the closer the relative frequency tends to get to the true probability - this is the law of large numbers. Relative frequency is the only way to estimate probability for a biased or unfair object, because you cannot assume equally likely outcomes. If experimental results differ a lot from the theoretical values, that is evidence the object may be biased.
Expected Frequency
Expected frequency tells you how many times you would expect an event to occur over a number of trials. The rule is $\text{expected frequency}=P(\text{event})\times\text{number of trials}$. For example, if a fair dice is rolled 300 times, the expected number of sixes is $\frac{1}{6}\times 300=50$. This works the other way too: you can use observed frequencies to estimate a probability, then scale up. Suppose a spinner landed on red 18 times in 60 spins, giving an estimated $P(\text{red})=\frac{18}{60}=0.3$; over 200 spins you would expect about $0.3\times 200=60$ reds. Remember that expected frequency is only an estimate of the most likely number - you will rarely get exactly that value in practice. Give your answer as a whole number when counting real objects or events, since you cannot have a fraction of a six.
Sample Space Diagrams
A sample space is the set of all possible outcomes of an experiment. A sample space diagram lays these out clearly so you can count favourable outcomes. For a single event it can just be a list, but for two combined events a two-way grid is ideal. For example, when rolling two fair dice and adding the scores, you draw a 6 by 6 grid; the 36 cells show every total from 2 to 12. From it you can read off that $P(\text{total of }7)=\frac{6}{36}=\frac{1}{6}$ because seven appears in six cells, and $P(\text{total} \gt 9)=\frac{6}{36}=\frac{1}{6}$. The power of a sample space diagram is that it guarantees you count every outcome and do not miss any. Always state the total number of outcomes clearly, then count the ones that satisfy the condition, and write the probability as a fraction over that total.
Mutually Exclusive and Exhaustive Events
Two events are mutually exclusive if they cannot both happen at the same time - for example, rolling a 2 and rolling a 5 on a single dice. For mutually exclusive events the addition rule simplifies to $P(A\text{ or }B)=P(A)+P(B)$. A set of events is exhaustive if together they cover every possible outcome, so their probabilities add up to 1. For instance, when rolling a dice the events odd and even are both mutually exclusive and exhaustive, and $P(\text{odd})+P(\text{even})=1$. These ideas let you find a missing probability: if a spinner can land on red, blue or green with $P(\text{red})=0.5$ and $P(\text{blue})=0.2$, then because the outcomes are exhaustive $P(\text{green})=1-0.5-0.2=0.3$. Watch out: not all events are mutually exclusive, so you cannot always simply add.
The Addition and Multiplication Rules
The general addition rule for the probability of A or B is $P(A\text{ or }B)=P(A)+P(B)-P(A\text{ and }B)$. The final term is subtracted to avoid double-counting outcomes that are in both events; when the events are mutually exclusive that overlap is zero and the rule reduces to simple addition. The multiplication rule deals with A and B both happening. For independent events it is $P(A\text{ and }B)=P(A)\times P(B)$. For example, the probability of getting heads on a coin and a six on a dice is $\frac{1}{2}\times\frac{1}{6}=\frac{1}{12}$. A useful way to remember the two rules: OR usually means add, AND usually means multiply. When you see the word and combining results along a path, multiply; when you see or combining separate possibilities, add the path results together at the end.
Independent and Dependent Events; Tree Diagrams
Events are independent when the outcome of one does not affect the other, such as two separate coin tosses. Events are dependent when one outcome changes the probabilities for the next, which is exactly what happens when you select items without replacement. Tree diagrams organise these combined events: each branch shows an outcome with its probability, you multiply along the branches to find the probability of a combined outcome, and you add down the diagram when several paths satisfy the condition. With replacement the probabilities stay the same on every set of branches. Without replacement they change because the total shrinks. For example, a bag has 3 red and 2 blue counters; drawing two without replacement, $P(\text{red then red})=\frac{3}{5}\times\frac{2}{4}=\frac{6}{20}=\frac{3}{10}$, where the second fraction uses 2 reds left out of 4 counters remaining. Note that the probabilities on each pair of branches still add to 1.
Venn Diagrams, Set Notation and Conditional Probability
Venn diagrams show how events overlap using overlapping circles inside a rectangle (the universal set). Set notation describes regions: $A\cap B$ means the intersection (A and B, the overlap), $A\cup B$ means the union (A or B, everything in either circle), and $A'$ means the complement (not A, everything outside A). For example, if 30 students study French (F) or Spanish (S), with 12 in F only, 8 in both, and 6 in S only, then $n(F\cap S)=8$ and $n(F\cup S)=26$, leaving $30-26=4$ outside both. Conditional probability is the probability of one event given another has already happened, written $P(A\mid B)$. Using the diagram, $P(\text{French}\mid\text{Spanish})=\frac{n(F\cap S)}{n(S)}=\frac{8}{14}=\frac{4}{7}$, because you restrict attention to the 14 Spanish students only. The key idea is that the condition reduces the sample space to just the outcomes where the given event is true.
Key terms
Probability scale
A number line from 0 (impossible) to 1 (certain) measuring how likely an event is.
Complement
The event not happening; $P(\text{not }A)=1-P(A)$.
Relative frequency
Experimental probability found from trials: number of successes divided by number of trials.
Expected frequency
The predicted number of occurrences, equal to probability times number of trials.
Sample space
The complete set of all possible outcomes of an experiment.
Mutually exclusive
Events that cannot both happen at once; their probabilities can be added directly.
Exhaustive
A set of events that covers every possible outcome, so their probabilities sum to 1.
Independent events
Events where one outcome does not affect the probability of the other.
Dependent events
Events where one outcome changes the probabilities of the next, as in selection without replacement.
Tree diagram
A branching diagram where you multiply along branches and add between paths.
Intersection and union
$A\cap B$ is outcomes in both A and B; $A\cup B$ is outcomes in A or B or both.
Conditional probability
The probability of A given B has happened, written $P(A\mid B)$, using a reduced sample space.
Exam technique
Always write probabilities as fractions, decimals or percentages between 0 and 1 - never as a ratio like 3:10, which loses marks.
Use the complement rule $P(\text{not }A)=1-P(A)$ to save work, especially for at least one type questions.
On tree diagrams, multiply along the branches and add the separate paths; check each pair of branches sums to 1.
For without replacement problems, reduce both the numerator and the denominator for the second draw.
Decide between the rules using the keywords: OR means add, AND means multiply.
For conditional probability, restrict to the given group: $P(A\mid B)=\frac{n(A\cap B)}{n(B)}$ using a Venn diagram.
Quick check
A bag contains 4 red and 6 green counters. Two counters are taken at random without replacement. What is the probability that both are red?
$\frac{4}{10}\times\frac{4}{10}=\frac{16}{100}$
$\frac{4}{10}\times\frac{3}{9}=\frac{2}{15}$
$\frac{4}{10}+\frac{3}{9}=\frac{11}{15}$
$\frac{4}{10}\times\frac{3}{10}=\frac{6}{50}$
Show answer
Answer: $\FRAC{4}{10}\TIMES\FRAC{3}{9}=\FRAC{2}{15}$. Without replacement, after taking one red there are 3 reds left out of 9 counters. Multiply the two probabilities (AND): $\frac{4}{10}\times\frac{3}{9}=\frac{12}{90}=\frac{2}{15}$.