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Year 12 Maths Standard 2 (2027) Relative frequency and probability

Complementary Events

20 practice questions 0 video lessons Theory + worked examples

Learn complementary events for NSW Year 12 Mathematics Standard 2. The complement of an event \(A\) is “not \(A\)”, and because every probability lies between \(0\) and \(1\) and the two events cover all outcomes, they add to \(1\) — giving the rule \(P(\text{not }A)=1-P(A)\).

You will use the complement rule with fractions, decimals and percentages, and learn the key Standard 2 trick of solving “at least one” problems by finding the probability of “none” and subtracting from \(1\).

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Theory

Complementary events are an event and its opposite, “not \(A\)”. This Year 12 Standard 2 (NSW) guide explains why every probability lies between \(0\) and \(1\), how the rule \(P(\text{not }A)=1-P(A)\) works, and how to solve “at least one” problems quickly by taking the complement “none”.

The complement of an event \(A\) is the event “\(A\) does not happen”, written \(\overline{A}\) or “not \(A\)”. Every probability is a number between \(0\) (impossible) and \(1\) (certain), and the whole sample space has probability \(1\).

Because every trial gives either \(A\) or not \(A\), the two events cover all the outcomes, so their probabilities add to \(1\): \(P(A)+P(\overline{A})=1\). Rearranging gives the complementary event rule \(P(\overline{A})=1-P(A)\).

This Year 12 Standard 2 (NSW) shortcut is most useful when the “not” event is easier to count — in particular for “at least one” problems, whose complement is the single case “none”. It works with fractions, decimals or percentages.

Probability scale from 0 to 1The scale splits into P(A) and P(not A), which add to 1P(A)P(not A)00.51impossibleeven chancecertain
Probabilities run from \(0\) to \(1\); \(P(A)\) and \(P(\text{not }A)\) split the scale and add to \(1\).
Two-region probability barP(storm) and P(no storm) are complementary and fill the bar from 0 to 1P(storm)P(no storm)0.20.800.21
A \(20\%\) chance of a storm leaves an \(80\%\) chance of no storm — the two fill the bar.

An event and its complement cover the whole sample space, so their probabilities add to \(1\):

\[P(A)+P(\overline{A})=1\]
P(A)+P(A¯)=1

Rearranged, this is the complementary event rule:

\[P(\overline{A})=1-P(A)\]
P(A¯)=1P(A)

With percentages the two add to \(100\%\), so \(P(\overline{A})=100\%-P(A)\). For “at least one”, use the complement “none”:

\[P(\text{at least one})=1-P(\text{none})\]
P(at least one)=1P(none)
Between 0 and 1. A valid probability never falls below \(0\) or rises above \(1\); if a complement gives you something outside that range, recheck \(P(A)\).

How to use the complement

  1. Name the event \(A\) and its complement “not \(A\)”. For an “at least one” question, the complement is “none”.
  2. Find \(P(A)\) as a fraction, decimal or percentage using \(P=\dfrac{\text{favourable}}{\text{total}}\).
  3. Subtract from 1 (or from \(100\%\)): \(P(\overline{A})=1-P(A)\).
  4. Check the answer is between \(0\) and \(1\), and that \(P(A)\) and \(P(\overline{A})\) add back to \(1\).
Example 1 β€” A single die
A fair six-sided die is rolled once. Find the probability that the result is not greater than \(4\).
Solution

The scores greater than \(4\) are \(5\) and \(6\); take the complement.

\(n(S)\)\(=\)\(6\)
\(P(\text{greater than }4)\)\(=\)\(\dfrac{2}{6}=\dfrac{1}{3}\)
\(P(\text{not greater than }4)\)\(=\)\(1-\dfrac{1}{3}=\dfrac{2}{3}\)
113=23

\(P(\text{not greater than }4)=\dfrac{2}{3}\).

Example 2 β€” A percentage forecast
The forecast gives a \(20\%\) chance of a thunderstorm this afternoon. Find the probability that there is no thunderstorm.
Solution

The two percentages add to \(100\%\), so take the complement.

\(P(\text{storm})\)\(=\)\(20\%\)
\(P(\text{no storm})\)\(=\)\(100\%-20\%\)
\(\)\(=\)\(80\%\)
100%20%=80%

There is an \(80\%\) chance of no thunderstorm.

Example 3 β€” A pack of cards
One card is drawn at random from a standard deck of \(52\) playing cards. Find the probability that it is not an ace.
Solution

A deck holds \(4\) aces; find \(P(\text{ace})\), then take the complement.

\(P(\text{ace})\)\(=\)\(\dfrac{4}{52}=\dfrac{1}{13}\)
\(P(\text{not an ace})\)\(=\)\(1-\dfrac{1}{13}\)
\(\)\(=\)\(\dfrac{12}{13}\)
1113=1213

\(P(\text{not an ace})=\dfrac{12}{13}\).

Example 4 β€” At least one
Two fair six-sided dice are rolled. Find the probability of getting at least one six.
Solution

The complement of “at least one six” is “no six on either die”.

\(P(\text{no six on one die})\)\(=\)\(\dfrac{5}{6}\)
\(P(\text{no six on both})\)\(=\)\(\dfrac{5}{6}\times\dfrac{5}{6}=\dfrac{25}{36}\)
\(P(\text{at least one six})\)\(=\)\(1-\dfrac{25}{36}=\dfrac{11}{36}\)
12536=1136

\(P(\text{at least one six})=\dfrac{11}{36}\).

Common pitfalls

Stay between 0 and 1. A probability can never be negative or greater than \(1\); an answer like \(1.2\) means a mistake in \(P(A)\).
Subtract from 1, not from the count. The rule is \(1-P(A)\) (or \(100\%-P(A)\)) — not the number of outcomes minus one.
“At least one” pairs with “none”. Its complement is zero of the thing, not “exactly one”.

Frequently asked questions

What is a complementary event?

The complement of an event A is the event that A does not happen, written as not A. Together A and not A cover every possible outcome, so their probabilities add to 1.

What is the formula for a complementary event?

P(not A) equals 1 minus P(A). With percentages it is 100 percent minus P(A). This works because an event and its complement always add to 1, or 100 percent.

When should I use the complement rule?

Use it whenever the 'not' event is easier to count than the event itself. It is especially useful for 'at least one' problems, where the complement is the single case of getting none.

How do you find the probability of at least one?

Find the probability of the opposite, 'none', then subtract from 1. For example, the chance of at least one six when two dice are rolled is 1 minus the chance of no sixes, which is 1 minus 25 over 36, giving 11 over 36.

Can a probability be more than 1?

No. Every probability lies between 0 and 1, where 0 means impossible and 1 means certain. If a complement calculation gives a value outside this range, you have made an error in P(A).

Do P(A) and P(not A) always add to 1?

Yes. An event and its complement together include all the outcomes in the sample space, so P(A) plus P(not A) is always exactly 1, or 100 percent as a percentage.