7.8 The Addition Rule for Probability

Mutual Exclusivity

Before we get to the key techniques of this section, we must first introduce some new terminology. Let’s say you’re drawing a card from a standard deck. We’ll consider 3 events: $H$ is the event “the card is a $♡$,” $T$ is the event “the card is a 10,” and $S$ is the event “the card is a $♠$.” If the card drawn is $\text{J}♠$, then $H$ and $T$ didn’t occur, but $S$ did. If the card drawn is instead $10♠$, then $H$ didn’t occur, but both $T$ and $S$ did.

We can see from these examples that, if we are interested in several possible events, more than one of them can occur simultaneously (both $T$ and $S$, for example). But, if you think about all the possible outcomes, you can see that $H$ and $S$ can never occur simultaneously; there are no cards in the deck that are both $♡$ and $♠$. Pairs of events that cannot both occur simultaneously are called mutually exclusive. Let’s go through an example to help us better understand this concept.

The Addition Rule for Mutually Exclusive Events

If two events are mutually exclusive, then we can use addition to find the probability that one or the other event occurs.

Why does this formula work? Let’s consider a basic example. Suppose we’re about to draw a Scrabble tile from a bag containing A, A, B, E, E, E, R, S, S, U. What is the probability of drawing an E or an S? Since 3 of the tiles are marked with E and 2 are marked with S, there are 5 tiles that satisfy the criteria. There are ten tiles in the bag, so the probability is $\frac{5}{10}=\frac{1}{2}$. Notice that the probability of drawing an E is $\frac{3}{10}$ and the probability of drawing an S is $\frac{2}{10}$; adding those together, we get $\frac{3}{10}+\frac{2}{10}=\frac{5}{10}$. Look at the numerators in the fractions involved in the sum: the 3 represents the number of E tiles and the 2 is the number of S tiles. This is why the Addition Rule works: The total number of outcomes in one event or the other is the sum of the numbers of outcomes in each of the individual events.

Finding Probabilities When Events Aren’t Mutually Exclusive

Let’s return to the example we used to explore the Addition Rule: We’re about to draw a Scrabble tile from a bag containing A, A, B, E, E, E, R, S, S, U. Consider these events: $J$ is “draw a vowel” and $K$ is “draw a letter that comes after L in the alphabet.” Since there are 6 vowels, $P(J)=\frac{6}{10}$. There are 4 tiles with letters that come after L alphabetically, so $P(K)=\frac{4}{10}$. What is $P(J\text{or}K)$? If we blindly apply the Addition Rule, we get $\frac{6}{10}+\frac{4}{10}=1$, which would mean that the compound event $J$ or $K$ is certain. However, it’s possible to draw a B, in which case neither $J$ nor $K$ happens. Where’s the error?

The events are not mutually exclusive: the outcome U belongs to both events, and so the Addition Rule doesn’t apply. However, there’s a way to extend the Addition Rule to allow us to find this probability anyway; it’s called the Inclusion/Exclusion Principle. In this example, if we just add the two probabilities together, the outcome U is included in the sum twice: It’s one of the 6 outcomes represented in the numerator of $\frac{6}{10}$, and it’s one of the 4 outcomes represented in the numerator of $\frac{4}{10}$. So, that particular outcome has been “double counted.” Since it has been included twice, we can get a true accounting by excluding it once: $\frac{6}{10}+\frac{4}{10}-\frac{1}{10}=\frac{9}{10}$. We can generalize this idea to a formula that we can apply to find the probability of any compound event built using “or.”

It’s worth noting that this formula is truly an extension of the Addition Rule. Remember that the Addition Rule requires that the events $E$ and $F$ are mutually exclusive. In that case, the compound event $(E\text{and}F)$ is impossible, and so $P(E\text{and}F)=0$. So, in cases where the events in question are mutually exclusive, the Inclusion/Exclusion Principle reduces to the Addition Rule.