Probability

The basic rules of probability

Complement, addition and multiplication rules, conditional probability and independence.

About 10 minutes

Probability assigns each event a number between 0 (impossible) and 1 (certain). When all outcomes are equally likely, the probability of an event is the number of favourable outcomes divided by the total number of outcomes. A handful of rules then lets you combine simple probabilities into answers to complicated questions.

The complement rule

An event either happens or it does not, so the two probabilities add to 1. “At least one” problems are almost always easiest through the complement: P(\text{at least one}) = 1 - P(\text{none}).

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

The addition rule

For “A or B”, add the probabilities and subtract the overlap, which would otherwise be counted twice. If the events are mutually exclusive — they cannot happen together — the overlap is zero.

P(A \cup B) = P(A) + P(B) - P(A \cap B)

Conditional probability and the multiplication rule

P(B \mid A) is the probability of B given that A has happened. For “A and B”, multiply the probability of A by the probability of B given A.

Events are independent when knowing that one happened does not change the probability of the other, so P(B \mid A) = P(B) and the rule becomes P(A \cap B) = P(A)\,P(B).

P(A \cap B) = P(A)\,P(B \mid A)

Worked examples

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

P(no six on one roll) = \frac56.
The rolls are independent: P(no six twice) = \frac56 \times \frac56 = \frac{25}{36}.
Complement: 1 - \frac{25}{36} = \frac{11}{36}.

Answer: \frac{11}{36} \approx 0.306

Two cards are drawn from a 52-card deck without replacement. What is the probability that both are aces?

P(first ace) = \frac{4}{52}.
Given that, P(second ace) = \frac{3}{51}.
Multiply: \frac{4}{52} \times \frac{3}{51} = \frac{1}{221}.

Answer: \frac{1}{221}

Common mistakes

Adding probabilities of events that can happen together without subtracting the overlap.
Multiplying probabilities of events that are not independent — drawing without replacement changes the second probability.
Confusing P(A \mid B) with P(B \mid A); Bayes’ theorem relates the two.
Believing that past independent outcomes affect future ones (the gambler’s fallacy).

Practice problems

1.P(drawing a heart from a full deck)
2.P(a head and a six) with a coin and a die
3.P(king or heart) from a full deck
4.P(at least one head in three coin flips)

Practise this topic with checked answers

Tools for this lesson

Probability CalculatorBasic, complement, conditional, independent and dependent event probabilities with the rule used shown for each result.Bayes' Theorem CalculatorUpdate a probability with Bayes’ theorem from a prior, a true-positive rate and a false-positive rate, with the full working.Permutations & Combinations CalculatorExact nPr, nCr and factorials for large numbers, with and without repetition, using big-integer arithmetic.Binomial Distribution CalculatorBinomial probabilities P(X = k), P(X ≤ k) and P(X ≥ k) with mean, variance and the full probability chart.

Next lesson: Introduction to derivatives

The complement rule

An event either happens or it does not, so the two probabilities add to 1. “At least one” problems are almost always easiest through the complement:

P(\text{at least one}) = 1 - P(\text{none})

P(\text{at least one}) = 1 - P(\text{none}).

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

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

The addition rule

P(A \cup B) = P(A) + P(B) - P(A \cap B)

P(A \cup B) = P(A) + P(B) - P(A \cap B)

Conditional probability and the multiplication rule

P(B \mid A)

P(B \mid A) is the probability of B given that A has happened. For “A and B”, multiply the probability of A by the probability of B given A.

Events are independent when knowing that one happened does not change the probability of the other, so

P(B \mid A) = P(B)

P(B \mid A) = P(B) and the rule becomes

P(A \cap B) = P(A)\,P(B)

P(A \cap B) = P(A)\,P(B).

P(A \cap B) = P(A)\,P(B \mid A)

P(A \cap B) = P(A)\,P(B \mid A)

Worked examples

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

P(no six on one roll) $= \frac56$ = \frac56.
The rolls are independent: P(no six twice) $= \frac56 \times \frac56 = \frac{25}{36}$ = \frac56 \times \frac56 = \frac{25}{36}.
Complement: $1 - \frac{25}{36} = \frac{11}{36}$ 1 - \frac{25}{36} = \frac{11}{36}.

Answer: $\frac{11}{36} \approx 0.306$ \frac{11}{36} \approx 0.306

Two cards are drawn from a 52-card deck without replacement. What is the probability that both are aces?

P(first ace) $= \frac{4}{52}$ = \frac{4}{52}.
Given that, P(second ace) $= \frac{3}{51}$ = \frac{3}{51}.
Multiply: $\frac{4}{52} \times \frac{3}{51} = \frac{1}{221}$ \frac{4}{52} \times \frac{3}{51} = \frac{1}{221}.

Answer: $\frac{1}{221}$ \frac{1}{221}

Common mistakes

Adding probabilities of events that can happen together without subtracting the overlap.

Multiplying probabilities of events that are not independent — drawing without replacement changes the second probability.

Confusing

P(A \mid B)

P(A \mid B) with

P(B \mid A)

P(B \mid A); Bayes’ theorem relates the two.

Believing that past independent outcomes affect future ones (the gambler’s fallacy).