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Year 7 Maths Probability

Complementary Events

20 practice questions 1 video lesson Theory + worked examples

An event’s complement is everything in the sample space that is not in the event. This NSW Year 7 maths subtopic (NESA outcome MA4‑PRO‑C‑01) shows that any event and its complement always have probabilities that add to 1.

You will use the complement rule, \(P(\text{not } A) = 1 - P(A)\), with fractions, decimals and percentages: a \(0.08\) chance of rain in Alice Springs means a \(0.92\) chance of no rain. You will also find a missing probability when several outcomes are known, using bags of marbles, playing cards and spinners. This subtopic is part of Year 7 Probability.

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Theory

The complement of an event \(A\) is “not \(A\)”: every outcome in the sample space that is not in \(A\). An event and its complement together make up all the outcomes, so \(P(A) + P(\text{not } A) = 1\). This Year 7 maths page shows how to describe complements and use the complement rule.

The complement of an event \(A\) is “not \(A\)”. List the sample space, then take away the outcomes in \(A\). For one roll of a die, the complement of “an even number” \(\{2, 4, 6\}\) is “not an even number” \(\{1, 3, 5\}\). The complement of \(E\) is sometimes written \(\overline{E}\).

An event and its complement have no outcomes in common, and together they make up the whole sample space. They are complementary events, and their probabilities add to \(1\).

The complement rule: \(P(\text{not } A) = 1 - P(A)\). It works in every form: \(1 - \dfrac{4}{9} = \dfrac{9}{9} - \dfrac{4}{9} = \dfrac{5}{9}\); \(1 - 0.3 = 0.7\); \(100\% - 22\% = 78\%\).

For “neither \(A\) nor \(B\)”, first count “\(A\) or \(B\)” (each outcome once), then take it away from the total.

The probabilities of all the outcomes add to \(1\). Use this to find a missing probability.

Cards 1 to 10 split into prime and not prime A box for the sample space of cards 1 to 10. The shaded left part holds the primes 2, 3, 5, 7; the right part holds 1, 4, 6, 8, 9, 10. Sprimenot prime 2 3 5 7 1 4 6 8 9 10
Ten cards numbered \(1\) to \(10\). Every card is either prime or not prime, never both: \(P(\text{prime}) = \dfrac{4}{10}\) and \(P(\text{not prime}) = \dfrac{6}{10}\).
Bar from 0 to 1 split into an event and its complement A bar for probability 1 split into 4 tenths (prime) and 6 tenths (not prime). 1 whole 4/10 prime 6/10 not prime
\(\dfrac{4}{10} + \dfrac{6}{10} = 1\): together the event and its complement fill the whole bar.

The complement rule:

\[ P(A) + P(\text{not } A) = 1 \]
P ( A )+ P ( not A )=1
\[ P(\text{not } A) = 1 - P(A) \]
P ( not A )=1− P ( A )
Write 1 as a fraction first. \(1 - \dfrac{3}{8} = \dfrac{8}{8} - \dfrac{3}{8} = \dfrac{5}{8}\). Check: \(\dfrac{3}{8} + \dfrac{5}{8} = 1\).
“Not red” is every other colour. With red, blue and green counters, not red means blue or green.

How to find the probability of a complement

  1. Find \(P(A)\), or read it from the question.
  2. Take it away from \(1\) (or from \(100\%\)).
  3. Check: the two probabilities add to \(1\).

How to find “neither”

  1. Count the outcomes in “\(A\) or \(B\)”, each once.
  2. Take them away from the total, then divide by the total.
Example 1 — A five-sector spinner
A spinner has \(5\) equal sectors numbered \(1\) to \(5\). It is spun once.
(i) List the sample space.
(ii) Find \(P(4)\).
(iii) List the outcomes in the complement of “spinning a \(4\)”.
(iv) Find \(P(\text{not } 4)\).
Solution
\(1 - \dfrac{1}{5}\)\(=\)\(\dfrac{5 - 1}{5} = \dfrac{4}{5}\)
1−1/5=(5−1)/5=4/5

(i) \(\{1, 2, 3, 4, 5\}\)
(ii) \(\dfrac{1}{5}\) (\(1\) chance in \(5\))
(iii) \(\{1, 2, 3, 5\}\)
(iv) \(\dfrac{4}{5}\)

Example 2 — Describing a complement
A card is chosen at random from eight cards numbered \(1\) to \(8\).
(i) Describe the complement of “a number greater than \(6\)”.
(ii) Find the probability of the event and of its complement.
Solution
\(\dfrac{2}{8}\)\(=\)\(\dfrac{1}{4}\)
\(\dfrac{6}{8}\)\(=\)\(\dfrac{3}{4}\)
\(\dfrac{1}{4} + \dfrac{3}{4}\)\(=\)\(1\)
2/8=1/4,6/8=3/4,1/4+3/4=1

(i) The event is \(\{7, 8\}\); its complement is “\(6\) or less”: \(\{1, 2, 3, 4, 5, 6\}\).
(ii) \(\dfrac{1}{4}\) and \(\dfrac{3}{4}\)

Example 3 — Three forms
(i) The probability that a train is late is \(0.15\). Find the probability that it is not late.
(ii) There is a \(40\%\) chance of a storm today. What is the chance of no storm?
(iii) \(P(\text{green}) = \dfrac{3}{10}\). Find \(P(\text{not green})\).
Solution
\(1 - 0.15\)\(=\)\(0.85\)
\(100 - 40\)\(=\)\(60\)
\(1 - \dfrac{3}{10}\)\(=\)\(\dfrac{7}{10}\)
1−0.15=0.85,100−40=60,1−3/10=7/10

(i) \(0.85\)
(ii) \(60\%\)
(iii) \(\dfrac{7}{10}\)

Example 4 — Neither one nor the other
A box of \(36\) muffins has \(15\) blueberry, \(12\) chocolate and the rest banana. One is chosen at random. Find
(i) \(P(\text{not chocolate})\)
(ii) \(P(\text{neither blueberry nor chocolate})\)
Solution
\(1 - \dfrac{12}{36}\)\(=\)\(\dfrac{2}{3}\)
\(36 - 15 - 12\)\(=\)\(9\)
\(\dfrac{9}{36}\)\(=\)\(\dfrac{1}{4}\)
1−12/36=2/336−15−12=9,9/36=1/4

(i) \(\dfrac{2}{3}\)
(ii) \(\dfrac{1}{4}\). Check: \(1 - \dfrac{27}{36} = \dfrac{1}{4}\).

Example 5 — Repeated letters
The letters of BANANA are written on \(6\) cards. One card is chosen at random. Find \(P(\text{not A})\).
Solution
\(1 - \dfrac{3}{6}\)\(=\)\(\dfrac{1}{2}\)
1−3/6=1/2

There are \(3\) A cards out of \(6\), so \(P(\text{not A}) = \dfrac{1}{2}\).

Example 6 — A missing probability
A jar holds \(60\) jelly beans that are red, green or white. There are \(15\) green ones, and \(P(\text{red}) = \dfrac{2}{5}\).
(i) How many red jelly beans are there?
(ii) Find \(P(\text{white})\).
(iii) Find \(P(\text{not white})\).
Solution
\(\dfrac{2}{5} \times 60\)\(=\)\(24\)
\(60 - 24 - 15\)\(=\)\(21\)
\(\dfrac{21}{60}\)\(=\)\(\dfrac{7}{20}\)
\(1 - \dfrac{7}{20}\)\(=\)\(\dfrac{13}{20}\)
2/5×60=24,60−24−15=21,21/60=7/20,1−7/20=13/20

(i) \(24\)
(ii) \(\dfrac{7}{20}\)
(iii) \(\dfrac{13}{20}\)

Example 7 — Fractions that add to 1
A bag has red, blue and green counters only. \(P(\text{red}) = \dfrac{1}{5}\) and \(P(\text{blue}) = \dfrac{3}{10}\).
(i) Find \(P(\text{green})\).
(ii) There are \(40\) counters. How many are green?
Solution
\(1 - \dfrac{2}{10} - \dfrac{3}{10}\)\(=\)\(\dfrac{5}{10}\)
\(\dfrac{1}{2} \times 40\)\(=\)\(20\)
1−2/10−3/10=5/10=1/2,1/2×40=20

(i) \(\dfrac{1}{2}\)
(ii) \(20\)

Common pitfalls

Missing the middle. On a die, the complement of “less than \(5\)” is “\(5\) or more” \(\{5, 6\}\), not “greater than \(5\)”.
Counting twice for “neither”. On cards \(1\) to \(10\), “even or more than \(7\)” is \(\{2, 4, 6, 8, 9, 10\}\): \(6\) cards, not \(5 + 3 = 8\). So “neither” is \(10 - 6 = 4\) cards.
Subtracting from 100 for a fraction. For fractions and decimals take away from \(1\); for percentages take away from \(100\%\).
Not every pair is complementary. Two events are complementary only if they never overlap and together cover every outcome.

Frequently asked questions

What is a complementary event?

All the outcomes that are not in the event. Together the event and its complement make up the whole sample space.

What do complementary probabilities add to?

\(1\) (or \(100\%\)).

How do you find the probability that something does not happen?

Take its probability away from \(1\): \(P(\text{not } A) = 1 - P(A)\).

If there is a 22% chance of rain, what is the chance of no rain?

\(100\% - 22\% = 78\%\).

Are “less than 3” and “more than 3” complementary on a die?

No. The \(3\) is in neither event, so together they do not cover every outcome.