# 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. Probabilities are numbers, and as such, arithmetic operations (addition, subtraction, multiplication, division, and exponentiation) can be performed on them.

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 dice. 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 shirt and a blue shirt, so the rule of sum applies.

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

What is the probability of getting a book that is not a math book?

## 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 \(P(B\cup T)=\dfrac{14}{34}+\dfrac{16}{34}-\dfrac{4}{34}=\dfrac{26}{34}=\dfrac{13}{17}\)

Therefore, the probability of drawing a blue or striped marble is \(\boxed{\dfrac{13}{17}}\)

## See Also

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

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