ProbabilityProbabilityHigh school

Probability Rules

Complements, unions, independence and conditional probability.

Probability measures how likely an event is on a scale from 0 (impossible) to 1 (certain). When outcomes are equally likely, $P(A) = \frac{\text{favourable outcomes}}{\text{total outcomes}}$ . The complement rule $P(\text{not } A) = 1 - P(A)$ is often the fastest route, especially for at-least-one questions.

The addition rule $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ avoids double counting overlapping outcomes. For independent events $P(A \cap B) = P(A)P(B)$ ; otherwise use conditional probability, $P(A \cap B) = P(A)P(B \mid A)$ , which captures how knowing one event changes the chance of the other.

Key terms

Conditional probability: Probability of A given that B occurred.
$P(A\mid B) = \frac{P(A\cap B)}{P(B)}$

Key ideas

Mutually exclusive events cannot happen together, so $P(A \cap B) = 0$ .
Independent means $P(B \mid A) = P(B)$ ; it is not the same as mutually exclusive.
At least one: $1 - P(\text{none})$ .
Tree diagrams and two-way tables organise conditional probabilities.

Worked example

A fair die is rolled 3 times. What is the probability of at least one six?

$P(\text{no six on one roll}) = \frac{5}{6}$ .
Independent rolls: $P(\text{no six in 3}) = \left(\frac{5}{6}\right)^3 = \frac{125}{216}$ .
Complement: $1 - \frac{125}{216} = \frac{91}{216}$ .
About $0.421$ .

Answer: $\frac{91}{216} \approx 0.421$

Common mistakes

Adding probabilities of overlapping events without subtracting the overlap.
Treating mutually exclusive events as independent.
Computing at-least-one by adding $\frac{1}{6}$ three times.

Practice questions

Try each one, then open the answer. Numerical answers on this page are verified by an automated calculation check.

1.Two coins are tossed. What is the probability of two heads?
Show answer
$\frac{1}{4}$
2. $P(A) = 0.5$ , $P(B) = 0.4$ , $P(A \cap B) = 0.2$ . Find $P(A \cup B)$ .
Show answer
$0.7$
3.Two cards are drawn without replacement from a standard deck. Find the probability both are aces, to four decimals.
Show answer
$\frac{1}{221} \approx 0.0045$

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Calculators

Probability Calculator