Resources For Teachers For Tutors For Students & Parents AI New Pricing
Year 10 Maths Core Probability

Complementary Events

20 practice questions 1 video lesson Theory + worked examples

Complementary Events in NSW Year 10 Core maths uses the fact that an event and its complement, everything else that could happen, have probabilities that add to 1 (NESA outcome MA4‑PRO‑C‑01). So \(P(\text{not } A) = 1 - P(A)\), which is often the quickest way to a probability.

You will find the probability of an event not happening, name the complement of an event, and work from percentages, ratios and playing cards. You will then use the complement in experiments with two or more steps (MA5‑PRO‑C‑01) to find the chance of at least one success, such as at least one 6 in three rolls, at least one boy in a family of four or a bus being late at least once. It is part of Year 10 Core Probability.

Practice 20 questions
Practice questions

Every question with a fully worked solution.

Start practising
Watch 1 video(s)
  • Complementary Events - Video - What Are Complementary Events In Probability Statistics, Definition And Formula Watch
Create a free accountTrack your progress and save your work as you go.
Create free account

Theory

This Year 10 maths page revises complementary events from earlier years: the complement of \(E\) is "not \(E\)", and \(P(\text{not } E) = 1 - P(E)\). It then uses the complement for the idea of "at least one" in experiments with two or more stages, with and without replacement.

The complement of an event \(E\) is the event "not \(E\)". It is written \(E'\), \(\overline{E}\) or \(E^c\). It contains every outcome in the sample space that is not in \(E\).

An event and its complement cannot both happen, and one of them must happen. So their probabilities add to \(1\).

The complement of "rolling a \(6\)" is "not rolling a \(6\)", which is "rolling \(1\), \(2\), \(3\), \(4\) or \(5\)". The complement of "scores" is "misses".

At least one means one or more. Its complement is none. Neither \(A\) nor \(B\) is the complement of "\(A\) or \(B\)".

Events in different stages are independent if one does not change the other's probability (such as two separate coin tosses). Then multiply their probabilities.

An event and its complementThe outcomes 1 to 6 of a die. The event less than 3 contains 1 and 2. Its complement, 3 or more, contains 3, 4, 5 and 6. Their probabilities, 2 sixths and 4 sixths, add to 1. 1 2 3 4 5 6 event E: less than 3 complement Eβ€²: 3 or more P(E) = 2/6 P(Eβ€²) = 4/6 2/6 + 4/6 = 1
The complement \(E'\) is every outcome not in \(E\). Together they cover the sample space, so \(P(E) + P(E') = 1\).
At least one head in two coin tossesThe four outcomes of tossing two coins: HH, HT, TH and TT. Every outcome except TT has at least one head, so the probability of at least one head is 1 minus one quarter, which is three quarters. HH HT TH TT at least one head no heads P(at least one H) = 1 βˆ’ P(no H) = 1 βˆ’ 1/4 = 3/4
"At least one" is the complement of "none". It is usually quicker to find \(P(\text{none})\) and subtract from \(1\).

The complement rule, and its most useful form:

\[P(E') = 1 - P(E) \qquad P(\text{at least one}) = 1 - P(\text{none})\]
P(E′)=1−P(E)
Percentages. \(P(\text{not } E) = 100\% - P(E)\). If \(88\%\) of parcels arrive on time, \(12\%\) are late.
Neither. If \(A\) and \(B\) cannot happen together, \(P(\text{neither}) = 1 - (P(A) + P(B))\). With \(\dfrac{1}{3}\) and \(\dfrac{1}{4}\): \(1 - \dfrac{7}{12} = \dfrac{5}{12}\).
Repeated independent trials. For \(n\) trials, \(P(\text{none}) = P(\text{not } E)^n\). For two dice, list the \(6 \times 6 = 36\) outcomes in a grid and count the ones you need.
Without replacement. After each draw there is one fewer item, and one fewer of the type drawn. Multiply along the path, then subtract from \(1\).

How to use the complement

  1. Decide whether the event or its complement is easier to count.
  2. Find the probability of the easier one.
  3. Subtract from \(1\) (or from \(100\%\)).

How to find P(at least one)

  1. Write the complement: "none" (every trial fails).
  2. Find the probability of failing once, then multiply for every trial. Adjust the numbers after each draw if there is no replacement.
  3. Subtract from \(1\), and simplify or round as asked.
Example 1 β€” Using the complement

(i) A card is drawn from a standard deck of \(52\) cards. Find the probability that it is NOT a picture card (jack, queen or king).
(ii) The chance of rain tomorrow is \(0.35\). What is the chance of no rain?
(iii) A die is rolled. Describe the complement of "rolling less than \(3\)".
Solution

(i) There are \(3\) picture cards in each of \(4\) suits: \(12\) picture cards.

\(P(\text{picture})\)\(=\)\(\dfrac{12}{52} = \dfrac{3}{13}\)
\(P(\text{not picture})\)\(=\)\(1 - \dfrac{3}{13} = \boldsymbol{\dfrac{10}{13}}\)
\((\text{ii})\quad P(\text{no rain})\)\(=\)\(1 - 0.35 = \boldsymbol{0.65}\)

(iii) The complement is every other outcome: rolling \(3\), \(4\), \(5\) or \(6\) ("rolling \(3\) or more").

P=1013
Example 2 β€” Count the smaller group
A whole number from \(1\) to \(50\) is chosen at random. Find the probability that it is NOT a multiple of \(3\).
Solution

Counting the multiples of \(3\) is easier than counting the rest.

\(\text{multiples of } 3\)\(=\)\(3, 6, 9, \ldots, 48\)
\(\text{how many}\)\(=\)\(48 \div 3 = 16\)
\(P(\text{multiple of } 3)\)\(=\)\(\dfrac{16}{50}\)
\(P(\text{not})\)\(=\)\(1 - \dfrac{16}{50} = \dfrac{34}{50} = \boldsymbol{\dfrac{17}{25}}\)
P=1725
Example 3 β€” At least one (independent)
A spinner has \(5\) equal sections and \(1\) of them is red. It is spun \(3\) times. Find the probability of landing on red at least once.
Solution

"At least once" is the complement of "never". The spins are independent, so multiply.

\(P(\text{not red})\)\(=\)\(\dfrac{4}{5}\)
\(P(\text{never red})\)\(=\)\(\dfrac{4}{5} \times \dfrac{4}{5} \times \dfrac{4}{5} = \dfrac{64}{125}\)
\(P(\text{at least once})\)\(=\)\(1 - \dfrac{64}{125}\)
\(\)\(=\)\(\boldsymbol{\dfrac{61}{125}}\)
P=61125

That is \(0.488\), just under half.

Example 4 β€” At least one (without replacement)
A bag has \(12\) lollies and \(4\) of them are sour. Two lollies are taken at random and eaten. Find the probability that at least one is sour.
Solution

The complement of "at least one sour" is "no sour": both are from the \(8\) that are not sour. The first lolly is eaten, so the second draw is from \(11\).

\(P(\text{no sour})\)\(=\)\(\dfrac{8}{12} \times \dfrac{7}{11}\)
\(\)\(=\)\(\dfrac{56}{132} = \dfrac{14}{33}\)
\(P(\text{at least one sour})\)\(=\)\(1 - \dfrac{14}{33}\)
\(\)\(=\)\(\boldsymbol{\dfrac{19}{33}}\)
P=1933

Common pitfalls

The complement includes every other outcome. The complement of "rolling a \(2\)" is not "rolling a \(5\)". It is rolling \(1\), \(3\), \(4\), \(5\) or \(6\).
At least one is not exactly one. "At least one head" includes two heads. Use \(1 - P(\text{no heads})\).
Do not add for repeated trials. The chance of at least one \(6\) in two rolls is not \(\dfrac{1}{6} + \dfrac{1}{6}\). It is \(1 - \left(\dfrac{5}{6}\right)^2 = \dfrac{11}{36}\).
Without replacement, change both numbers. After eating one of \(8\) plain lollies from \(12\), the next draw is \(\dfrac{7}{11}\), not \(\dfrac{8}{11}\) or \(\dfrac{7}{12}\).

Frequently asked questions

What is a complementary event?

The complement of an event is everything that is not that event. An event and its complement cannot both happen, and together they are certain.

What is the complement rule in probability?

\(P(\text{not } E) = 1 - P(E)\). For example, if the chance of rain is \(0.3\), the chance of no rain is \(0.7\).

How do you find the probability of at least one?

Find the probability of none, then subtract it from \(1\). For independent trials, the probability of none is the probability of failing once multiplied by itself for each trial.

Why use the complement instead of counting directly?

Often the complement has far fewer outcomes to count. "At least one" can happen in many ways, but "none" happens in only one way.

What is the complement of 'neither A nor B'?

It is "\(A\) or \(B\) or both", which means at least one of the two events happens.