Probability

What probability measures

Probability is a number that tells you how likely an event is to happen. It runs on a scale from $0$ (impossible) to $1$ (certain), and every probability you ever calculate must land somewhere on that scale.

Key terms Event — an outcome or set of outcomes you are interested in (e.g. rolling an even number).

Outcome — a single possible result of a trial.

$P(A)$ — the probability that event $A$ happens. We write it as a fraction, decimal or percentage.

If all outcomes are equally likely, then

$$P(\text{event}) = \frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$$

For a fair six-sided die, $P(\text{prime}) = \frac{3}{6} = \frac{1}{2}$ because $2, 3, 5$ are prime.

Exam tip Always leave answers as exact fractions unless the question asks for a decimal or percentage. A probability greater than $1$ or less than $0$ is always wrong, so sanity-check your final value.

Theoretical vs experimental probability

Theoretical probability is worked out by reasoning about equally likely outcomes. Experimental probability (also called relative frequency) is found by actually doing trials:

$$\text{relative frequency} = \frac{\text{frequency of the outcome}}{\text{total number of trials}}$$

The more trials you carry out, the closer the relative frequency tends to get to the true theoretical probability. This is why a biased coin can only be detected by experiment.

Real world Insurance companies set premiums using relative frequency from millions of past claims. They cannot compute theoretical probabilities for car accidents, so huge amounts of data stand in for them.

Expected frequency

If an event has probability $p$ and you repeat the trial $n$ times, the expected frequency is

$$\text{expected frequency} = n \times p$$

If a spinner lands on red with probability $0.3$ and you spin it $200$ times, you expect red about $200 \times 0.3 = 60$ times. This is an estimate, not a guarantee.

Sample space diagrams

A sample space diagram is a grid or list of every possible outcome. They are ideal for two combined actions such as rolling two dice.

There are $36$ equally likely outcomes. A total of $7$ appears six times, so $P(7) = \frac{6}{36} = \frac{1}{6}$ — the most likely total.

Mutually exclusive events and the addition rule

Two events are mutually exclusive if they cannot both happen at the same time (e.g. a single card being both a King and a Queen). For mutually exclusive events:

$$P(A \text{ or } B) = P(A) + P(B)$$

Because one outcome must occur, the probabilities of all mutually exclusive outcomes that cover every possibility sum to $1$:

$$P(\text{not } A) = 1 - P(A)$$

Worked example A bag has red, blue and green counters. $P(\text{red}) = 0.45$ and $P(\text{blue}) = 0.2$. Find $P(\text{green})$.

The three colours are mutually exclusive and cover everything, so they total $1$:

Independent events and the multiplication rule

Two events are independent if one happening does not change the probability of the other (e.g. two separate coin flips). For independent events:

$$P(A \text{ and } B) = P(A) \times P(B)$$

The probability of throwing two sixes with two dice is $\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$.

Watch out Do not mix the rules. "Or" with mutually exclusive events means add; "and" with independent events means multiply. Reading the question for these words is half the battle.

Tree diagrams for combined events

A tree diagram shows the outcomes of two or more stages. Rules:

Probabilities on branches from the same point add to $1$.
To find the probability of a path, multiply along the branches.
To combine several paths, add the path probabilities.

Notice the second-stage denominators are $6$, not $7$: once a sweet is taken and not replaced, only $6$ remain.

Worked example A bag holds $4$ red and $3$ green sweets. Two are taken without replacement. Find $P(\text{one of each colour})$.

"One of each" means Red-then-Green or Green-then-Red. Multiply along each path, then add:

$P(\text{R, G}) = \frac{4}{7} \times \frac{3}{6} = \frac{12}{42}$

Watch out With replacement, the bag is restored so the second branch uses the same denominator $7$ and the events are independent. Without replacement the totals shrink and the events become dependent. Always check which one the question describes.

Conditional probability

Conditional probability is the probability of an event given that another has already happened, written $P(B \mid A)$. In the sweet example above, after a red is drawn, $P(\text{green second} \mid \text{red first}) = \frac{3}{6}$. The second branches of a "without replacement" tree are exactly these conditional probabilities.

At IGCSE you handle conditional probability through tree diagrams and restricted totals rather than a separate formula: just count from the situation after the condition is known.

Venn diagrams and set notation

A Venn diagram sorts outcomes into overlapping sets. You should know this notation:

$A \cup B$ — union, "$A$ or $B$ (or both)".
$A \cap B$ — intersection, "$A$ and $B$".

In the diagram, $11$ take only French, $8$ take only Spanish, $7$ take both, and $4$ take neither, totalling $30$.

Worked example A student is chosen at random from the $30$ above. Find $P(F \cap S)$ and $P(F \cup S)'$.

$F \cap S$ is the overlap: $7$ students, so $P(F \cap S) = \frac{7}{30}$.

Exam tip Fill a Venn diagram from the middle outwards: place the intersection first, then subtract it from each set total before filling the outer regions. This stops you double-counting the overlap.