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

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

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

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

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