# Probability - Rule of Sum

The **probability rule of sum** gives the situations in which the probability of a union of events can be calculated by summing probabilities together. It is often used on **mutually exclusive** events, meaning events that cannot both happen at the same time.

Both the rule of sum and the rule of product are guidelines as to when these arithmetic operations yield a *meaningful* result, a result that is useful for problem-solving. In the case of the rule of sum, it tells you when adding probabilities together gives you *another* meaningful probability.

## Rule of Sum for Mutually Exclusive Events

Imagine rolling a six-sided fair die. It is possible to roll an odd number. If $O$ is the event that an odd number is rolled, then $O=\{1,3,5\}$. It is also possible that a composite number is rolled. If $C$ is the event that a composite number is rolled, then $C=\{4,6\}$.

Now think about whether or not it is possible to roll a number that is both odd *and* composite. As it turns out, it is not possible in the sample space of six-sided dice rolls to roll an odd composite number. The events $O$ and $C$ are called **mutually exclusive**, meaning they cannot both happen at the same time.

This distinction is very important for **the rule of sum for mutually exclusive events**:

Let $A$ and $B$ be mutually exclusive events. Then, the probability of the union of those events is

$P(A\cup B)=P(A)+P(B).$

"$\cup$" is the symbol for a union. Because events are sets, unions of events can be understood in much the same way as unions of sets. $P(A\cup B)$ is the probability of either event $A$ *or* event $B$ happening.

This rule can be intuitively understood with a Venn diagram showing the sample space which includes events $A$ and $B$:

Let $S$ be a sample space which includes

mutually exclusiveevents $A$ and $B$. The Venn diagram of this sample space is pictured below:

Note that there is no overlap between events $A$ and $B$. When events are mutually exclusive, it is not possible for both to happen at the same time.

The union of $A$ and $B$ is shown in blue. The probability of this union can be computed as follows:

$P(A\cup B)=\dfrac{|A|+|B|}{|S|}=\dfrac{|A|}{|S|}+\dfrac{|B|}{|S|}=P(A)+P(B).$

Mary has 2 green skirts, 3 red skirts, and 4 blue skirts to choose from for her outfit today. She chooses a skirt randomly, with each skirt equally likely to be chosen. What is the probability that Mary chooses a green

orblue skirt?

Define events $G$ and $B$ as follows:

- $G=$ Mary chooses a green skirt.
- $B=$ Mary chooses a blue skirt.
Mary cannot choose both a green skirt and a blue skirt, so the rule of sum applies:

$P(G\cup B)=P(G)+P(B)=\dfrac{2}{9}+\dfrac{4}{9}=\dfrac{2}{3}.\ _\square$

## Generalized Rule of Sum

The preceding examples and problems are somewhat limited because they require mutually exclusive events. The following is a **generalized rule of sum**:

Let $A$ and $B$ be events (not necessarily mutually exclusive). The probability of the union of these events is

$P(A\cup B)=P(A)+P(B)-P(A\cap B).$

This rule can be intuitively understood with a Venn diagram of events $A$ and $B$:

Let $S$ be a sample space which includes events $A$ and $B$. The Venn diagram of this sample space is pictured below:

In this example, events $A$ and $B$ are not mutually exclusive, and this can be seen by the overlapping section in the middle. This overlapping section is the event $A\cap B$. This event represents event $A$ happening at the same time as event $B$.

The union of $A$ and $B$ is shown in blue. If we want to calculate the probability of the union, then adding the probabilities of $A$ and $B$ results in double counting the section in the middle. The probability of this section must be subtracted to account for this:

$P(A\cup B)=\dfrac{|A|}{|S|}+\dfrac{|B|}{|S|}-\dfrac{|A\cap B|}{|S|}=P(A)+P(B)-P(A\cap B).$

If you are familiar with the principle of inclusion and exclusion, this concept is very similar.

In a bag, there are 34 marbles. There are 14 blue marbles in the bag, and 4 of those marbles are both blue and striped. 16 marbles are striped (of those marbles, the same 4 marbles are both blue and striped).

What is the probability that you draw a marble from the bag that is either blue or striped?

The problem mentions that you are calculating the probability of a marble being blue

orstriped. The use of the word "or" here is important. The "or" indicates that we are finding the probability of a union of events.Let $B$ be the event that a blue marble is drawn. Let $T$ be the event that a striped marble is drawn.

The problem mentions that marbles can be both blue and striped. This indicates that these events are

notmutually exclusive. $B\cap T$ is the event that a marble that is both blue and striped is drawn. $B \cup T$ is the event that either a blue marble or a striped marble is drawn.The goal is to find $P(B\cup T)$, the probability that either a blue marble is drawn or a striped marble is drawn.

By the above formula, $P(B\cup T)=P(B)+P(T)-P(B\cap T).$

This gives $P(B\cup T)=\frac{14}{34}+\frac{16}{34}-\frac{4}{34}=\frac{26}{34}=\frac{13}{17}.$

Therefore, the probability of drawing a blue or striped marble is $\frac{13}{17}.\ _\square$

## See Also

**Cite as:**Probability - Rule of Sum.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/probability-rule-of-sum/