If A and B are sets in the sample space C, then the probability of the union of sets (events) A and B is given by the addition rule.P\left(A\cup B\right)=P \left(A \right)+P\left(B\right)-P\left(A \cap B\right)This formula can be rearranged in terms of the probability of the intersection of sets A and B.

In some cases we need to find the probability an event does not occur. We can use the fact that the probabilities of all the possible events will sum to 1, because we are certain that something in the sample space will occur. With this, we can derive a formula for the probability of the complement of an event occuring.

We denote the complement of an event using the prime symbol. So event A \rq contains all the outcomes that are in the sample space S, but not in event A. Together, the events A and A\rq contain all the outcomes in the sample space. This means the probability of the events A and A\rq will sum to 1. So:

\begin{aligned} P(A) + P(A\rq) &=1 \\ P(A\rq) &=1-P(A) \end{aligned}

\displaystyle P(A\rq) =1-P(A)

\bm{A}

is an event

\bm{A\rq}

is the complement of event A

Sometimes we want to find the probability of events that are disjoint (mutually exclusive), meaning they don't have any elements in common.

Disjoint event

Two events are disjoint if their intersection is equal to the empty set.A\cap B=\emptyset

By the addition rule, the probability of the union of disjoint events is equal to the sum of probabilities of each event. P\left(A\cup B\right)=P \left(A \right)+P\left(B\right)

Worked examples

Example 1

In a school, there are 50 students in grade 12. There are 15 students that study physics, 10 that study both physics and biology, and 22 that study neither.

Find the probability that a randomly selected student studies physics and not biology.

Approach

We can simplify the problem by displaying our given information in a Venn diagram.

We can see that there are 15 students who study physics, and 10 of those students study both physics and biology. So we can calculate that there are 5 students who study only physics.

Solution

\displaystyle P(\text{Event})	\displaystyle =	\displaystyle \dfrac{\text{Number of outcomes satisfying the event}}{\text{Total number of possible outcomes}}	Probability formula
\displaystyle P(\text{Physics only})	\displaystyle =	\displaystyle \dfrac{\text{Students studying only physics}}{\text{Total number of students}}	Substitute in the event
	\displaystyle =	\displaystyle \dfrac{5}{50}	Substitute in the values
	\displaystyle =	\displaystyle \dfrac{1}{10}	Simplify

Find the probability that a randomly selected student studies biology.

Approach

In order to find the probability that a student studies biology, we want to find how many students study biology out of the whole grade using the Venn diagram.

Solution

Using the following Venn diagram, we can solve for the number of students studying biology.

A Venn diagram with two overlapping sets labeled Biology and Physics. The number inside the Biology set, but not in the Physics set, is the variable x. The number inside the Physics set, but not in the Biology set, is 5. The number inside the overlapping region of the two sets is 10. The number outside both sets, but in the Venn diagram, is 22.

From the Venn diagram we can see that there are x+10 students who study biology. We can solve for x using the information given:

\displaystyle 50	\displaystyle =	\displaystyle 22+x+10+5	Total number of students in the grade
\displaystyle 50	\displaystyle =	\displaystyle 37+x	Collect like terms
\displaystyle x	\displaystyle =	\displaystyle 13	Number of students studying biology only

So there are 13+10=23 students studying biology.

So the probability a student studies biology is given by:

\displaystyle P(\text{Event})	\displaystyle =	\displaystyle \dfrac{\text{Number of outcomes satisfying the event}}{\text{Total number of possible outcomes}}	Probability formula
\displaystyle P(\text{Biology})	\displaystyle =	\displaystyle \dfrac{\text{Students studying biology}}{\text{Total number of students}}	Substitute in the event
	\displaystyle =	\displaystyle \dfrac{23}{50}	Substitute in the values

Reflection

Consider how the probability would change if the student was randomly selected from those studying physics and/or biology, rather than the entire grade.

Example 2

A group of 280 people were asked if they had visited Radnor Lake or The Parthenon. The results showed that 66 people visited Radnor Lake, 84 people visited The Parthenon, and 12 people visited both.

Find the probability that a person has visited either Radnor lake or The Parthenon.

Approach

To avoid double counting the people have visited both locations, we can use the addition rule. To do this, we want to find the probability that a person has visited Radnor Lake, the probability that a person has visited The Parthenon, and the probability that a person has visited both locations.

Solution

Let A represent the subset of people that have visited Radnor Lake and B represent the subset that have visited the Parthenon. We can now find the probability of each subset by substituting the given values into the probability formula:P(\text{Event})=\dfrac{\text{Number of outcomes satisfying the event}}{\text{Total number of possible outcomes}}

This gives us: \begin{aligned} P(A)&=\dfrac{\text{People who have visited Radnor Lake}}{\text{Total number of people in the group}}&=\dfrac{66}{280} \\ \\ P(B)&=\dfrac{\text{People who have visited The Parthenon}}{\text{Total number of people in the group}}&=\dfrac{84}{280} \\ \\ P(A\cap B)&=\dfrac{\text{People who have visited both locations}}{\text{Total number of people in the group}}&=\dfrac{12}{280} \end{aligned}

We can now use the addition rule to find the probability that a person has visited either Radnor Lake or The Parthenon.

\displaystyle P\left(A \cup B \right)	\displaystyle =	\displaystyle P\left(A\right)+P\left(B\right)-P\left(A \cap B\right)	Addition rule
	\displaystyle =	\displaystyle \dfrac{66}{280}+\dfrac{84}{280}-\dfrac{12}{280}	Substitute in the values
	\displaystyle =	\displaystyle \dfrac{69}{140}	Evaluate

So, the probability that a person has visited Radnor lake or The Parthenon is\dfrac{69}{140} \approx 49\%.

Reflection

Think about why we had to subtract the intersection of A and B. We know that the probability of all events in a sample space sums to 1.

Try calculating the sum P\left(A\right)+P\left( B \right)+P(A \cup B)'. Is the sum greater, less than, or equal to 1?

Try subtracting P \left(A \cap B \right) from the total and compare your results.

Example 3

There are 40 students in a class, 20 people have brown eyes, 12 have blue eyes and 8 have neither. Find the probability a student has either brown or blue eyes.

Approach

We can display all of our information in a Venn diagram so that it's easier to visualise.

We can assume the events are disjoint since no student is reported to have multi-coloured eyes. This can also be confirmed by summing the number of elements in all the events in the Venn diagram and checking that it equals the number of students in the class. So the intersection must be empty.

Solution

We want to find P \left(\text{Brown} \cup \text{Blue} \right), so we want to focus on the following shaded subsets.

We can now find the probability for each shaded subset.

\displaystyle P \left(\text{Brown} \right)	\displaystyle =	\displaystyle \dfrac{\text{People with brown eyes}}{\text{Total people in the class}}	Probability that a student has brown eyes
	\displaystyle =	\displaystyle \dfrac{20}{40}	Substitute in the values
\displaystyle P \left(\text{Blue} \right)	\displaystyle =	\displaystyle \dfrac{\text{People with blue eyes}}{\text{Total people in the class}}	Probability that a student has blue eyes
	\displaystyle =	\displaystyle \dfrac{12}{40}	Substitute in the values

Since the subsets are disjoint, we can observe the probability can be calculated as

\displaystyle P \left(\text{Brown} \cup \text{Blue} \right)	\displaystyle =	\displaystyle P \left(\text{Brown} \right)+P \left(\text{Blue} \right)	Addition rule for disjoint events
	\displaystyle =	\displaystyle \dfrac{20}{40}+\dfrac{12}{40}	Substitute in the values
	\displaystyle =	\displaystyle \dfrac{4}{5}	Evaluate and simplify

Reflection

Think about why we didn't have to calculate the intersection for the addition rule in this example. Consider that the intersection of our two events, brown eyes and blue eyes, is empty.

Outcomes

M3.S.CP.C.6

Understand and apply the Addition Rule.*

M3.S.CP.C.6.A

Explain the Addition Rule, P(A or B) = P(A) + P(B) Ã¢â‚¬â€œ P(A and B) in terms of visual models (Venn diagrams, frequency tables, etc.).

M3.S.CP.C.6.B

Apply the Addition Rule to solve problems and interpret the answer in terms of the given context.

M3.MP1

Make sense of problems and persevere in solving them.

M3.MP2

Reason abstractly and quantitatively.

M3.MP4

Model with mathematics.

M3.MP5

Use appropriate tools strategically.

M3.MP6

Attend to precision.

M3.MP8

Look for and express regularity in repeated reasoning.

7.04 Probability

Concept summary

Worked examples

Example 1

Approach

Solution

Approach

Solution

Reflection

Example 2

Approach

Solution

Reflection

Example 3

Approach

Solution

Reflection

Outcomes

M3.S.CP.C.6

M3.S.CP.C.6.A

M3.S.CP.C.6.B

M3.MP1

M3.MP2

M3.MP4

M3.MP5

M3.MP6

M3.MP8

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