Every probability sits on a scale from 0 (impossible) to 1 (certain), with $0.5$ representing an even chance. If a calculation ever produces a value outside $0$–$1$, that's an immediate sign something has gone wrong — the probability scale doubles as a built-in error check.

An event and its opposite ("not happening") always cover every possibility between them:

This is often the fastest route to an answer — rather than listing every way an event can fail to happen, calculate the probability it does happen and subtract from 1.

The more trials conducted, the closer this estimate tends to get to the true theoretical probability. Multiplying a relative frequency by a new number of trials gives the expected frequency — a predicted count, not a probability: $\frac16\times300=50$ expected occurrences out of 300 trials.

This is the single biggest source of errors in this topic. Ask what the question is really asking for:

• "OR" — mutually exclusive events (can't both happen in one trial): add the probabilities. $P(A \text{ or } B)=P(A)+P(B)$.
• "AND" — independent events (one doesn't affect the other): multiply the probabilities. $P(A \text{ and } B)=P(A)\times P(B)$.

Each branch is labelled with its own probability, and the probabilities leaving any single point must always sum to 1 (a built-in check). Multiply along a path of branches to find the probability of one specific sequence; add across different paths' endpoints when more than one sequence leads to the outcome you want. The diagram makes the correct rule visually obvious at each stage, which is exactly why it's required technique at this level.

Always write out the actual fractions being combined (e.g. $\frac16\times\frac12$), not just the final simplified result — examiners award method marks for seeing which rule was applied, and a bare final answer with no visible working risks losing them even when it's correct.

A bag contains 5 red counters and 3 blue counters. A counter is picked at random, replaced, then a second counter is picked. Find the probability both counters are the same colour.

1. $P(\text{red})=\frac58$, $P(\text{blue})=\frac38$ (with replacement, these don't change between picks).
2. $P(\text{RR}) = \frac58\times\frac58=\frac{25}{64}$.
3. $P(\text{BB}) = \frac38\times\frac38=\frac{9}{64}$.
4. These are the only two ways to get "same colour," and they're mutually exclusive, so add: $\frac{25}{64}+\frac{9}{64}=\frac{34}{64}=\frac{17}{32}$.