3.5 The Addition Rule
LEARNING OBJECTIVES
- Calculate “or” probabilities using the addition rule.
- Determine if two events are mutually exclusive.
For two events [latex]A[/latex] and [latex]B[/latex] we might want to know the probability that at least one of the two events occurs. For example, we might want to find the probability of rolling a 2 or a 5 in a single roll of a die, or we might want to find the probability that someone has a smartphone or a tablet. In probability terms, we want to find [latex]P(A \mbox{ or }B)[/latex], the probability that either [latex]A[/latex] or [latex]B[/latex] occurs. In probability, “or” is always an inclusive “or,” which means that either [latex]A[/latex] occurs, or [latex]B[/latex] occurs, or both occur.
The Addition Rule for Or Probabilities
To find [latex]P(A \mbox{ or } B)[/latex], we start by adding the individual probabilities, [latex]P(A)[/latex] and [latex]P(B)[/latex]. But this means that the overlap between the two events [latex]A[/latex] and [latex]B[/latex] is counted twice: once by [latex]P(A)[/latex] and once by [latex]P(B)[/latex]. To correct for this double counting, we need to subtract [latex]P(A \mbox{ and }B)[/latex], the probability of both events occuring. This gives us the addition rule to find [latex]P(A \mbox{ or }B)[/latex]:
[latex]\begin{eqnarray*} P(A \mbox{ or }B) & = & P(A)+P(B)-P(A \mbox{ and }B) \\ \\ \end{eqnarray*}[/latex]
EXAMPLE
At a local language school, 40% of the students are learning Spanish, 20% of the students are learning German, and 8% of the students are learning both Spanish and German. What is the probability that a randomly selected student is learning Spanish or German?
Solution:
[latex]\begin{eqnarray*} P(\mbox{Spanish or German}) & = & P(\mbox{Spanish})+P(\mbox{German})-P(\mbox{Spanish and German}) \\ & = & 0.4+0.2-0.08 \\ & = & 0.52 \end{eqnarray*}[/latex]
EXAMPLE
There are 50 students enrolled in the second year of a business degree program. During this semester, the students have to take some elective courses. 18 students decide to take an elective in psychology, 27 students decide to take an elective in philosophy, and 10 students decide to take an elective in both psychology and philosophy. What is the probability that a student takes an elective in psychology or philosophy?
Solution:
[latex]\begin{eqnarray*} P(\mbox{psychology or philosophy}) & = & P(\mbox{psychology})+P(\mbox{philosophy})\\ & & -P(\mbox{psychology and philosophy}) \\ & = & \frac{18}{50}+\frac{27}{50}-\frac{10}{50} \\ & = & 0.7 \end{eqnarray*}[/latex]
TRY IT
At a local basketball game, 70% of the fans are cheering for the home team, 25% of the fans are wearing blue, and 12% of the fans are cheering for the home team and wearing blue. What is the probability that a randomly selected fan is cheering for the home team or wearing blue?
Click to see Solution
[latex]\begin{eqnarray*} P(\mbox{home team or blue}) & = & P(\mbox{home team})+P(\mbox{nlue})-P(\mbox{home team and blue}) \\ & = & 0.7+0.25-0.12 \\ & = & 0.83 \end{eqnarray*}[/latex]
EXAMPLE
Suppose a study of speeding violations and drivers who use cell phones produced the following fictional data:
| Speeding violation in the last year | No speeding violation in the last year | Total | |
| Cell phone user | 25 | 280 | 305 |
| Not a cell phone user | 45 | 405 | 450 |
| Total | 70 | 685 | 755 |
- What is the probability that a randomly selected person is a cell phone user or has no speeding violations in the last year?
- What is the probability that a randomly selected person had a speeding violation in the last year or does not use a cell phone?
Solution:
- [latex]\begin{eqnarray*} P(\mbox{cell phone or no violations}) & = & P(\mbox{cell phone})+P(\mbox{no violations}) \\ & & -P(\mbox{cell phone and no violations}) \\ & = & \frac{305}{755}+\frac{685}{755}-\frac{280}{755} \\ & = & \frac{710}{755} \end{eqnarray*}[/latex]
- [latex]\begin{eqnarray*} P(\mbox{violations or no cell phone}) & = & P(\mbox{violations})+P(\mbox{no cell phone}) \\ & & -P(\mbox{violations and no cell phone}) \\ & = & \frac{70}{755}+\frac{450}{755}-\frac{45}{755} \\ & = & \frac{475}{755} \end{eqnarray*}[/latex]
TRY IT
This table shows the number of athletes who stretch before exercising and how many had injuries within the past year.
| Injury in last year | No injury in last year | Total | |
| Stretches | 55 | 295 | 350 |
| Does not stretch | 231 | 219 | 450 |
| Total | 286 | 514 | 800 |
- What is the probability that a randomly selected athlete stretches before exercising or had an injury last year?
- What is the probability that a randomly selected athlete does not stretch before exercising or had no injuries in the last year?
Click to see Solution
- [latex]\displaystyle{\mbox{Probability}=\frac{350}{800}+\frac{286}{800}-\frac{55}{800}=0.72625}[/latex]
- [latex]\displaystyle{\mbox{Probability}=\frac{450}{800}+\frac{514}{800}-\frac{219}{800}=0.93125}[/latex]
Mutually Exclusive Events
Two events [latex]A[/latex] and [latex]B[/latex] are mutually exclusive if the two events cannot happen at the same time. That is, the events [latex]A[/latex] and [latex]B[/latex] do not share any outcomes and so [latex]P(A \mbox{ and }B)=0[/latex]. For example, in the experiment of flipping a coin, the events heads and tails are mutually exclusive because it is not possible to have both heads and tails on the top face. In the case of mutually exclusive events, the addition rule is [latex]\displaystyle{P(A \mbox{ or } B)=P(A)+P(B)}[/latex].
EXAMPLE
Suppose a bag contains 20 balls. 10 of the balls are white, 7 of the balls are red, and 3 of the balls are blue. Suppose one ball is selected at random from the bag.
- Are the events “selecting a white ball” and “selecting a red ball” mutually exclusive? Why?
- What is the probability of selecting a white or red ball?
Solution:
- The events “selecting a white ball” and “selecting a red ball” are mutually exclusive because the events cannot happen at the same time. It is not possible for the selected ball to be both white and red.
- [latex]\displaystyle{P(\mbox{white or red}) = P(\mbox{white})+P(\mbox{red}) = \frac{10}{20}+\frac{7}{20} = 0.85}[/latex]
NOTE
In the calculation of the probability in part 2, there is nothing to subtract. Because the events are mutually exclusive, [latex]\displaystyle{P(\mbox{white and red})=0}[/latex].
TRY IT
At a local college, 60% of the students are taking a math class, 50% of the students are taking a science class, and 30% of the students are taking both a math and a science class.
- Are the events “taking a math class” and taking a science class” mutually exclusive? Explain.
- What is the probability that a randomly selected student is taking a math class or a science class?
Click to see Solution
- The events “taking a math class” and “taking a science class” are not mutually exclusive because the events can happen at the same time (i.e. a student can be taking both a math class and a science class). As stated in the question, [latex]\displaystyle{P(\mbox{math and science})=0.3 \neq 0}[/latex].
- [latex]\displaystyle{P(\mbox{math or science}) = P(\mbox{math})+P(\mbox{science})-P(\mbox{math and science}) = 0.6+0.5-0.3 = 0.8}[/latex]
TRY IT
You roll a fair die one time.
- Are the events “rolling a 4” and “rolling an even number” mutually exclusive?
- Are the events “rolling a 4” and “rolling an odd number” mutually exclusive?
- What is the probability of rolling a 4 or rolling an odd number.
Click to see Solution
- The events “rolling a 4” and “rolling an even number” are not mutually exclusive because the events can happen at the same time (i.e. 4 is an even number).
- The events “rolling a 4” and “rolling an odd number” are mutually exclusive because the events cannot happen at the same time. It is not possible to roll a die and get a 4 (an even number) and an odd number on the top face at the same time
- [latex]\displaystyle{P(\mbox{4 or odd}) = P(\mbox{4})+P(\mbox{odd}) = \frac{1}{6}+\frac{3}{6}=\frac{4}{6}}[/latex]
Watch this video: Addition Rule for Probability Khan Academy [10:42] (transcript available).
Concept Review
To find the probability of events [latex]A[/latex] or [latex]B[/latex], use the addition rule: [latex]\displaystyle{P(A \mbox{ or }B)=P(A)+P(B)-P(A \mbox{ and }B)}[/latex]. Two events are mutually exclusive if the events cannot happen at the same time.
Attribution
“3.1 Terminology” , “3.2 Independent and Mutually Exclusive Events”, and “3.4 Contingency Tables” in Introductory Statistics by OpenStax is licensed under a Creative Commons Attribution 4.0 International License.
