Probability Basics

AS · S1 · 3 min

Probability measures how likely an event is, on a scale from $0$ (impossible) to $1$ (certain). For equally likely outcomes it is a ratio of counts; for sequences of events it is built by multiplying along a path and adding across alternatives. Sample spaces, tree diagrams and the counting techniques from permutations and combinations do the work.

Equally likely outcomes

Key result

P(A) = \frac{\text{number of outcomes in } A}{\text{total number of equally likely outcomes}}

$P(A') = 1 - P(A)$ . Probabilities of all outcomes in a sample space add to $1$ .

Two dice

Two fair dice are thrown. Find the probability that the total is $8$ , and the probability that the total is at least $10$ .

Solution

36 equally likely pairs. Total $8$ : $(2,6), (3,5), (4,4), (5,3), (6,2)$ , five pairs: $\tfrac{5}{36}$ .

Total $\geq 10$ : $(4,6), (5,5), (6,4), (5,6), (6,5), (6,6)$ , six pairs: $\tfrac{6}{36} = \tfrac{1}{6}$ .

Selection without replacement using combinations

A bag holds 5 red and 4 blue balls. Three are taken at random. Find the probability that exactly 2 are red.

Solution

$\dfrac{\binom{5}{2}\binom{4}{1}}{\binom{9}{3}} = \dfrac{10 \times 4}{84} = \tfrac{10}{21}$ .

Alternatively with a tree: $P(RRB) + P(RBR) + P(BRR) = 3 \times \tfrac{5}{9} \cdot \tfrac{4}{8} \cdot \tfrac{4}{7} = \tfrac{10}{21}$ .

Addition and multiplication

Key result

Multiplication: $P(A \text{ and then } B) = P(A) \times P(B \mid A)$ . For independent events, $P(A \cap B) = P(A)P(B)$ .
Addition: for mutually exclusive events, $P(A \text{ or } B) = P(A) + P(B)$ . In general, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ .

Independent events

A biased coin shows heads with probability $0.6$ . It is tossed three times. Find the probability of (a) three heads; (b) exactly two heads; (c) at least one head.

Solution

(a) $0.6^3 = 0.216$ .

(b) Three orders (HHT, HTH, THH), each $0.6^2 \times 0.4 = 0.144$ : $0.432$ .

A tree diagram with replacement

A box has 3 white and 2 black counters. One is taken, its colour noted, and it is replaced; then a second is taken. Find the probability the two are different colours.

Solution

$P(WB) + P(BW) = \tfrac{3}{5} \cdot \tfrac{2}{5} + \tfrac{2}{5} \cdot \tfrac{3}{5} = \tfrac{12}{25}$ .

Without replacement

The same box, but the first counter is not replaced. Find the probability the two are different colours, and the probability that the second is white.

Solution

$P(WB) + P(BW) = \tfrac{3}{5} \cdot \tfrac{2}{4} + \tfrac{2}{5} \cdot \tfrac{3}{4} = \tfrac{12}{20} = \tfrac{3}{5}$ .

$P(\text{2nd white}) = \tfrac{3}{5} \cdot \tfrac{2}{4} + \tfrac{2}{5} \cdot \tfrac{3}{4} = \tfrac{3}{5}$ , the same as $P(\text{1st white})$ .

Sample spaces for games

When outcomes are not equally likely, or when the experiment has stages, list them in a table or tree with their probabilities, then add the ones you want.

A game with unequal probabilities

A spinner gives $1$ with probability $0.5$ , $2$ with probability $0.3$ and $3$ with probability $0.2$ . It is spun twice. Find the probability that the total is $4$ .

Solution

Total $4$ from $(1,3), (2,2), (3,1)$ : $0.5 \times 0.2 + 0.3 \times 0.3 + 0.2 \times 0.5 = 0.1 + 0.09 + 0.1 = 0.29$ .

Watch out

Probabilities multiply along a branch only when the second probability is the conditional probability given the first outcome. Without replacement, the denominator changes; with replacement, it does not.

Exam tip

"At least one" is almost always $1 - P(\text{none})$ . Computing every case separately is slow and error-prone; the complement is one line.

Practice

Question

A card is drawn from a pack of 52. Find the probability it is a king or a heart.
Two dice are thrown. Find the probability that the product is even.
A bag has 6 red and 4 green sweets. Two are eaten at random. Find the probability that they are the same colour.
Three independent components each work with probability $0.9$ . Find the probability that at least two work.

Answers

$\tfrac{4}{52} + \tfrac{13}{52} - \tfrac{1}{52} = \tfrac{16}{52} = \tfrac{4}{13}$ .
$1 - P(\text{both odd}) = 1 - \tfrac{1}{4} = \tfrac{3}{4}$ .
$\tfrac{6}{10} \cdot \tfrac{5}{9} + \tfrac{4}{10} \cdot \tfrac{3}{9} = \tfrac{42}{90} = \tfrac{7}{15}$ .
$3(0.9^2)(0.1) + 0.9^3 = 0.243 + 0.729 = 0.972$ .

Equally likely outcomes

Addition and multiplication

Sample spaces for games

Practice

Console