Year 8AC9M8P01

Complementary Events

Recognise that complementary events have a combined probability of one; use this relationship to calculate probabilities in applied contexts.

Everything that can happen adds to one. When the event you want is awkward to count, count the opposite instead.

Method

The complement is everything else

P(A) + P(A') = 1

$A'$ means 'not $A$'.

Rearrange when it is easier

P(A) = 1 - P(A')

Often far quicker than counting directly.

'At least one' is the classic signal

P(\text{at least one}) = 1 - P(\text{none})

Counting 'none' is almost always easier.

Probabilities never leave 0 to 1

0 \leq P \leq 1

An answer outside this range is wrong.

e.g.

\text{A bag holds 4 red, 3 blue and 5 green. Find } P(\text{not blue}).

Count the total. $4 + 3 + 5 = 12$
Find the probability of the event itself first. $P(\text{blue}) = \frac{3}{12} = \frac{1}{4}$
Use the complement rule. $P(\text{not blue}) = 1 - \frac{1}{4}$
Work it out. $= \frac{3}{4}$
Check by counting directly. $\frac{4+5}{12} = \frac{9}{12} = \frac{3}{4} \quad\checkmark$

If the event is awkward, find its complement and subtract from 1.

P(\text{rain}) = 0{,}35. \text{ Find } P(\text{no rain}).

Answer

0{,}65

\text{A die is rolled. Find } P(\text{not a six}).

Answer

\frac{5}{6}

\text{Find } P(\text{at least one head}) \text{ in two coin tosses.}

Answer

1 - \frac{1}{4} = \frac{3}{4}

P(A) = 0{,}8. \text{ Find } P(A').

Answer

0{,}2

\text{Why is 'at least one' easier by complement?}

Answer

\text{'None' is a single outcome; 'at least one' is many.}

\text{Can } P(A) + P(A') = 1{,}2?

Answer

\text{No. They must total exactly 1.}