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 occurs, or both occur when an overlap is possible.
B
Two cases determine which formula to use:
The events cannot happen in the same trial. These are mutually exclusive events.
The events can happen in the same trial. These events are not mutually exclusive.
The difference is whether the two events have an overlap.
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.
P(A orB)=P(A)+P(B)
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
Probability of getting doubles (P(Doubles)): 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:
P(S=7 or Doubles)=P(S=7)+P(Doubles)=366+366=3612=31
For events A and B that are not mutually exclusive, the way to calculate the probability of "A OR B" is:
P(A orB)=P(A)+P(B)−P(A andB)
The term P(A andB) 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(Doubles)=6/36
The probability of the event "sum is 8 AND doubles" (P(S=8 and Doubles)) is just one outcome: the pair (4,4). So its probability is 1/36.
The probability of getting a sum of 8 OR doubles is:
P(S=8 or Doubles)=P(S=8)+P(Doubles)−P(S=8 and Doubles)
=365+366−361=3610=185
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.
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.
What is the probability that the selected student belongs to the mathematics club or the science club?
What is the probability that the student belongs to neither club?
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:
n(M∪S)=n(M)+n(S)−n(M∩S)=18+14−8=24
P(M∪S)=3024=54
There are 30−24=6 students outside both clubs. Therefore,
P(neither club)=306=51
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.
