12.04 Complementary events

Complementary events

A complement of an event are all outcomes that are NOT the event.

The following are examples of events and their complements.

• Event is $\left(\text{Heads}\right)$(Heads), the complement is {not a head} which is {Tails}
• Event is rolling a die and getting $\left(2\right)$(2), then the complement is $\left\{1,3,4,5,6\right\}${1,3,4,5,6}
• Event is {winter}, the complement is {summer, autumn, spring}
• Event is {Diamonds}, the complement is {Hearts, Clubs, Spades}

Notation

We already know we can denote the probability of event $A$A as P( $A$A ).

Well, we denote the probability of its complement, not $A$A as $P$P ( $\overline{A}$A ) (see the little line above the $A$A - it means NOT)

The probabilities of complementary events always sum to $1$1.

$P$P( $A$A )  +  $P$P ( $\overline{A}$A )  =  $1$1

 

Remember: the sum of the probabilities of all possible outcomes is $1$1

For example, for a standard die $P$P ( $1$1 ) + $P$P ( $2$2 ) + $P$P ( $3$3 ) + $P$P ( $4$4 ) + $P$P ( $5$5 ) + $P$P ( $6$6 ) = $1$1

 

Using complementary events

Because of the fact that complementary events always sum to $1$1, we have a choice about how to solve problems involving complementary events.

Let's consider the event of rolling two dice.  What is the probability that you do not roll a double?

Method 1

Answering this question does not need to involve complementary events at all if we don't want to.  We could just list the sample space (the list of all possible outcomes from rolling $2$2 dice), and then count up how many are not doubles and calculate the probability.  The sample space for this event is pretty large, but we can still do it - 

$((1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),$((1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),

$(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),$(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),

$(5,1),(5,2),(5,3),(5,4),5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6))$(5,1),(5,2),(5,3),(5,4),5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6))

P(not a double) =$\frac{\text{total favourable outcomes }}{\text{total possible outcomes }}=\frac{30}{36}$total favourable outcomes total possible outcomes ​=3036​ = $\frac{5}{6}$56​

 

Method 2

Using complementary events we can see that 

P(not a double) =$1$1 - P(double)

this is easier to work out without having to list the entire sample space.

P(not a double) = $1-\frac{6}{36}=\frac{5}{6}$1−636​=56​

 

Practice questions

Question 1

Question 2