Addition Rule

Counting Outcomes That Belong to Either Event

Suppose we ask everyone in a class how they traveled to school. We then choose one student at random and ask for the probability that the student came by bicycle OR motorcycle.

The word "OR" joins the two events. The addition rule calculates the probability that event $A$ occurs, event $B$ occurs, or both occur when an overlap is possible.

Two cases determine which formula to use:

1. The events cannot happen in the same trial. These are mutually exclusive events.
2. The events can happen in the same trial. These events are not mutually exclusive.

The difference is whether the two events have an overlap.

Mutually Exclusive Events

Mutually exclusive events, also called disjoint events, cannot occur together in a single trial.

If event $A$ occurs, event $B$ does not occur, and vice versa.

Examples:

1. One coin toss produces either "Heads" or "Tails," never both at once.
2. Zain traveled to school by motorcycle OR by bicycle, assuming each student used one mode of transport.
3. Rolling two dice: The event "the sum is $7$" OR the event "getting doubles (same number on both dice)".
   • Sum is $7$: $(1,6)$, $(2,5)$, , , ,

Formula for Mutually Exclusive Events

If events $A$ and $B$ are mutually exclusive, their intersection is empty. The probability of "A OR B" is therefore the sum of their individual probabilities.

Worked example:

For the two-dice experiment above:

• Probability of sum being $7$ ($P(S=7)$): there are $6$ pairs out of $36$ total, so the probability is $6/36$

The probability of obtaining a sum of $7$ OR doubles is:

Not Mutually Exclusive Events

Events that are not mutually exclusive can occur together in one trial, so their outcome sets overlap.

Examples:

1. Draw one card from a standard deck. The events "drawing an Ace" and "drawing a Heart ($\heartsuit$)" overlap at the Ace of Hearts ($A\heartsuit$).
2. Rolling two dice: The event "the sum is $8$" OR the event "getting doubles".
   • Sum is $8$: $(2,6)$, , , ,

Why simple addition is not enough

Adding $P(A) + P(B)$ would count the intersection twice: once as part of $A$ and once as part of $B$.

For the "sum is $8$ OR doubles" example:

• $P(S=8)$: there are $5$ pairs, so the probability is $5/36$
• $P(\text{Doubles})$ is $6$ pairs, namely

Direct addition gives $5/36 + 6/36 = 11/36$, but the pair $(4,4)$ appears in both groups. Subtracting its probability once removes the duplicate count.

Formula for Not Mutually Exclusive Events

For events $A$ and $B$ that are not mutually exclusive, the way to calculate the probability of "A OR B" is:

The term $P(A \text{ and } B)$ is the probability of the intersection, where $A$ and $B$ occur together. Subtracting it once corrects the double count.

Worked example:

Continue with the "sum is $8$ OR doubles" example:

• $P(S=8) = 5/36$
• $P(\text{Doubles}) = 6/36$
• The probability of the event "sum is $8$ AND doubles" () is just one outcome: the pair . So its probability is .

The probability of getting a sum of $8$ OR doubles is:

How mathematics uses "OR"

In mathematics, "$A$ or $B$" includes three possibilities: only $A$, only $B$, or both when an overlap exists. This is an inclusive use of "or," unlike everyday expressions that may require choosing exactly one alternative.

When to Use Which Formula

1. Check the overlap: Can events $A$ and $B$ occur in the same trial?
2. If they cannot occur together: Use $P(A \text{ or } B) = P(A) + P(B)$.

Exercise

In a class of $30$ students, $18$ belong to the mathematics club, $14$ belong to the science club, and $8$ belong to both clubs. One student is selected at random.

1. What is the probability that the selected student belongs to the mathematics club or the science club?
2. What is the probability that the student belongs to neither club?

Worked Solution

Let $M$ be the event “belongs to the mathematics club” and $S$ the event “belongs to the science club.” The $8$ students in both clubs are included in both counts, so adding $18+14$ would count them twice. The general addition rule removes that duplicate count:

There are $30-24=6$ students outside both clubs. Therefore,

The two probabilities add to $4/5+1/5=1$, which checks that the class has been divided into two complementary groups: students in at least one club and students in neither club.

The overlap must be subtracted exactly once because it was already included in both original counts. This complement check is a quick way to catch an omitted or duplicated group before accepting the result.