13.2: Complements and Unions of Events

Complementary Events

Now let us examine the probability that an event does not happen. As in the previous section, consider the situation of rolling a six-sided die and first compute the probability of rolling a six: the answer is $P(\text{six})=\frac{1}{6}$. Now consider the probability that we do not roll a six: there are 5 outcomes that are not a six, so the answer is $P(\text{not a six})=\frac{5}{6}$. Notice that $P(\text{six})+P(\text{not a six})=\frac{1}{6}+\frac{5}{6}=\frac{6}{6}=1$

This is not a coincidence. Consider a generic situation with $n$ possible outcomes and an event $E$ that corresponds to m of these outcomes. Then the remaining $n - m$ outcomes correspond to $E$ not happening, thus $P(\text {not } E)=\frac{n-m}{n}=\frac{n}{n}-\frac{m}{n}=1-\frac{m}{n}=1-P(E)$

Complement of an Event

The complement of an event $E$ is the event “$E$ doesn’t happen”

The notation $\bar{E}$ is used for the complement of event $E$.

We can compute the probability of the complement using $P(\bar{E})=1-P(E)$

Notice also that $P(E)=1-P(\bar{E})$

Example 1

If you pull a random card from a deck of playing cards, what is the probability it is not a heart?

Solution

There are 13 hearts in the deck, so

$P(\text {heart})=\frac{13}{52}=\frac{1}{4}. \nonumber$

The probability of not drawing a heart is the complement:

$P(\text { not heart })=1-P(\text { heart })=1-\frac{1}{4}=\frac{3}{4} \nonumber$

Example 2

Suppose we flipped a coin and rolled a die, and wanted to know the probability of getting a head on the coin or a 6 on the die.

Solution

Here, there are still 12 possible outcomes:

$\{H1,H2,H3,H4,H5,H6,T1,T2,T3,T4,T5,T6\}$

By simply counting, we can see that 7 of the outcomes have a head on the coin or a 6 on the die or both – we use or inclusively here (these 7 outcomes are $H1, H2, H3, H4, H5, H6, T6$), so the probability is $\frac{7}{12}$. How could we have found this from the individual probabilities?

As we would expect, $\frac{1}{2}$ of these outcomes have a head, and $\frac{1}{6}$ of these outcomes have a 6 on the die. If we add these, $\frac{1}{2}+\frac{1}{6}=\frac{6}{12}+\frac{2}{12}=\frac{8}{12}$, which is not the correct probability. Looking at the outcomes we can see why: the outcome H6 would have been counted twice, since it contains both a head and a 6; the probability of both a head and rolling a 6 is $\frac{1}{12}$.

If we subtract out this double count, we have the correct probability: $\frac{8}{12}-\frac{1}{12}=\frac{7}{12}$.

$P(A\text{ or }B)$

The probability of either $A$ or $B$ occurring (or both) is $P(A \text { or } B)=P(A)+P(B)-P(A \text { and } B)$

Example 3

Suppose we draw one card from a standard deck. What is the probability that we get a Queen or a King?

Solution

There are 4 Queens and 4 Kings in the deck, hence 8 outcomes corresponding to a Queen or King out of 52 possible outcomes. Thus the probability of drawing a Queen or a King is:

$P(\text { King or Queen})=\frac{8}{52}$

Note that in this case, there are no cards that are both a Queen and a King, so $P(\mathrm{King} \text { and Queen})=0$. Using our probability rule, we could have said:

$P(\text { King or Queen})=P(\text { King })+P(\text { Queen})-P(\text { King and Queen})=\frac{4}{52}+\frac{4}{52}-0=\frac{8}{52}$

In the last example, the events were mutually exclusive, so $P(A \text { or } B)=P(A)+P(B)$.

Mutually Exclusive

Two events $A$ and $B$ are mutually exclusive if they have no outcomes in common.

Thus, $P(A \text { or } B)=P(A)+P(B)$.

Example 4

Suppose we draw one card from a standard deck. What is the probability that we get a red card or a King?

Solution

• Half the cards are red, so $P(\text{Red})=\frac{26}{52}$
• There are four kings, so $P(\text{King})= \frac{4}{52}$
• There are two red kings, so $P(\text {Red and King})= \frac{2}{52}$

We can then calculate

$$P(\text {Red or King})=P(\text {Red})+P(\text {King})-P(\text {Red and King})=\frac{26}{52}+\frac{4}{52}-\frac{2}{52}=\frac{28}{52}=\frac{7}{13}\nonumber$$

Try it Now 1

In your drawer you have 10 pairs of socks, 6 of which are white, and 7 tee shirts, 3 of which are white. If you reach in and randomly grab a pair of socks and a tee shirt, what the probability you pulled either a pair of white socks or a white tee shirt?

Answer

$P(\text { white sock })=\frac{6}{10}$

$P(\text { white tee })=\frac{3}{7}$

$P(\text { white sock and white tee })=\frac{6}{10} \cdot \frac{3}{7}=\frac{9}{35}$

$P(\text { white sock or white tee })=\frac{6}{10}+\frac{3}{7}-\frac{9}{35}=\frac{27}{35}$