What is probabilities outside 0 to 1?

Any probability must lie between 0 and 1; a value above 1 signals an error.

P(A) = 0.4 and P(B) = 0.25 for mutually exclusive events. Find P(A or B). [1 mark]

Two independent events each have probability 13. Find the probability that both occur. [2 marks]

SingaporeMathsSyllabus dot point

How do we calculate the probability of single events and of combined events?

Calculate the probability of single events, and of combined events using the addition and multiplication rules with tree diagrams

A focused answer to the O-Level E-Maths outcome on probability. The probability of a single event, mutually exclusive and independent events, the addition and multiplication rules, and tree diagrams.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
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What this dot point is asking

SEAB wants you to calculate the probability of a single event, and to combine probabilities of two or more events using the addition rule for mutually exclusive events and the multiplication rule for independent events, supported by tree diagrams. Probability measures how likely an outcome is, on a scale from $0$ to $1$ .

The answer

Probability of a single event

For equally likely outcomes, the probability of an event is the number of favourable outcomes over the total number of outcomes:

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

Every probability lies between $0$ (impossible) and $1$ (certain).

The complement

The probability that an event does not happen is one minus the probability that it does:

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

This is often the quickest route, especially for at least one type questions.

Mutually exclusive events and the addition rule

Two events are mutually exclusive if they cannot both happen at once. For such events, the probability of one or the other is the sum:

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

Independent events and the multiplication rule

Two events are independent if one does not affect the other. For independent events, the probability of both happening is the product:

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

On a tree diagram, multiply along the branches for and, and add the results of separate branches for or.

Worked example

A box has $4$ green and $6$ yellow counters. Two counters are drawn one after the other without replacement. Find the probability that both are green.

Step 1: Probability the first is green

There are $10$ counters, $4$ green, so $P(\text{first green}) = \dfrac{4}{10}$ .

Step 2: Update for the second draw

Without replacement, one green is gone: $9$ counters remain, $3$ green. So $P(\text{second green}) = \dfrac{3}{9}$ .

Step 3: Multiply along the branch

P(\text{both green}) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90}

Step 4: Simplify

\frac{12}{90} = \frac{2}{15}

The probability that both counters are green is $\dfrac{2}{15}$ .

Examples in context

Example 1. Weather and plans. If rain on two independent days has probability $0.3$ each, the chance of rain on both is $0.3 \times 0.3 = 0.09$ , while the chance of at least one rainy day is best found as $1 - (0.7 \times 0.7) = 0.51$ using the complement. Tree thinking organises the cases.

Example 2. Quality control. Drawing items without replacement to test for defects changes the probability at each draw, since each removed item alters the pool. Manufacturers model this dependence to estimate the chance of finding a faulty item in a sample.

Try this

Q1. A fair die is rolled. Find the probability of scoring a $5$ . [1 mark]

Cue. One favourable outcome out of six: $\dfrac{1}{6}$ .

Q2. $P(A) = 0.4$ and $P(B) = 0.25$ for mutually exclusive events. Find $P(A \text{ or } B)$ . [1 mark]

Cue. Add: $0.4 + 0.25 = 0.65$ .

Q3. Two independent events each have probability $\dfrac{1}{3}$ . Find the probability that both occur. [2 marks]

Cue. Multiply: $\dfrac{1}{3} \times \dfrac{1}{3} = \dfrac{1}{9}$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original3 marksA bag contains

5

red and

3

blue balls. One ball is drawn at random. Find the probability that it is (a) red and (b) not red.

Show worked answer →

There are $5 + 3 = 8$ balls in total.

(a) $P(\text{red}) = \dfrac{5}{8}$ .

(b) $P(\text{not red}) = 1 - P(\text{red}) = 1 - \dfrac{5}{8} = \dfrac{3}{8}$ .

Markers reward the probability as favourable over total for red, and using $1 - P(\text{red})$ (or the $3$ blue out of $8$ ) for not red.

Original4 marksA fair coin is tossed twice. Using a tree diagram, find the probability of getting (a) two heads and (b) exactly one head.

Show worked answer →

Each toss is independent with $P(\text{head}) = \dfrac{1}{2}$ and $P(\text{tail}) = \dfrac{1}{2}$ .

(a) Two heads: multiply along the branch $\dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}$ .

(b) Exactly one head happens via head then tail or tail then head: $\dfrac{1}{2} \times \dfrac{1}{2} + \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{1}{2}$ .

Markers reward multiplying along branches for two heads, and adding the two relevant branch outcomes for exactly one head.