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5. Probability
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United States of America
Grades 9-12

5.04 Probability

Lesson
Practice

Introduction

In 7th grade, we were introduced to the concept of probability, and we found the probability of simple and compound events. We will continue building our understanding of probability by learning about different types of events and by finding the probability of the union and intersection of those events.

The addition rule

Probability is the mathematics of chance, and we use it to predict outcomes.

Probability

The measure of likelihood that an event occurs

Event

A set of outcomes of an experiment to which a probability is assigned. Any subset of a sample space.

Example:

The intersection, union, or complement of sets that exist within the sample space

A Venn diagram showing two circles A and B. Circle A is highlighted.

The Venn diagrams from the previous lesson represented sample spaces of compound events where set A is one event and set B is a second event.

To calculate the probability of an event occurring in general, divide the number of outcomes satisfying the event by the total number of outcomes.

P\left(\text{Event}\right)=\dfrac{\text{Number of outcomes satisfying the event}}{\text{Total number of possible outcomes}}

The probabilities of all the possible events in the sample space will sum to 1.

In some cases, we need to find the probability an event does not occur. In other words, we need to find the probability of the complement of an event. Together, the events A and its complement, 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.

P(A) + P(A\rq)=1

Solving this equation for P(A\rq), we find the probability that event A does not occur is:

\displaystyle P(A\rq) =1-P(A)
\bm{A}
is an event
\bm{A\rq}
is the complement of event A

To find probabilities involving the union of two events, we must first know whether or not the two events are related. If events are mutually exclusive, it means they cannot happen at the same time. Mutually exclusive events are also referred to as disjoint events.

Disjoint event

Two events are disjoint if their intersection is equal to the empty set

A\cap B=\emptyset

Some examples of experiments that involve mutually exclusive or disjoint events are:

  • Flipping heads and flipping tails with a single flip of a coin. You cannot flip heads and tails at the same time
  • Rolling an even number and rolling an odd number from a single roll of a die. We cannot roll any number which is both even and odd
  • Drawing a 7 and drawing a 10 when picking a single card from a deck. There is no card that is both a 7 and a 10

However, some events can happen at the same time, so they are not mutually exclusive. For example:

  • Drawing a club card and drawing a 7 when picking a single card from a deck. We could pick a card that is a club and a 7 because we could get the 7 of clubs
  • Rolling an even number and rolling a prime number from a single roll of a die. They have outcomes in common, namely, the number 2

Exploration

Consider the following two experiments using a standard deck of cards:

  • Experiment 1. With the following two events:

    A:\text{ Drawing a }7

    B:\text{ Drawing a }10

  • Experiment 2. With the following two events:

    A:\text{ Drawing a club}

    B:\text{Drawing a }10

Answer the following questions for Experiment 1 and then repeat for Experiment 2:

  1. How could we find P(A \cup B)?
  2. Are there any cards that appear in both events A and B?
  3. How will your answer to question 2 affect your answer to question 1?

By the addition rule, the probability of the union of mutually exclusive 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)

A Venn diagram showing two circles A and B. The entirety of both circles A and B is highlighted.

However, we cannot simply add the probabilities of A and B if events are not mutually exclusive because the shared elements in A and B would be counted twice.

P(A)
P(B)

Notice that the section in the middle is counted for both P(A) and P(B), so we need to account for that when calculating the probability.

Addition rule for non-mutually exclusive events

If two events A and B are not mutually exclusive, then

P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)

In fact, the relationship above holds for mutually exclusive events too. When events are mutually exclusive, P\left(A\cap B\right)=0 so the formula becomes P\left(A\cup B\right)=P\left(A\right)+P\left(B\right).

We can work backwards, given probabilities for the union or the intersection of events, to deduce if the events are mutually exclusive.

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.

a

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

Worked Solution
Create a strategy

To model the given information, we can construct a two-way table.

Study BiologyDo not study BiologyTotal
Study Physics1015
Do not study Physics22
Total50

We can use the total rows to fill in the rest of the table. Each row must add to the total on the right side, and the totals of both rows must add to 50.

Study BiologyDo not study BiologyTotal
Study Physics10515
Do not study Physics2235
Total50

Then, we can add the students in the Do not study Biology column to find that total, and we can use subtraction to fill in the first column.

Study BiologyDo not study BiologyTotal
Study Physics10515
Do not study Physics132235
Total232750
Apply the idea

Next, we need to use the table to find the probability that a student studies physics only.

\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

There is a 10\% chance that a randomly selected student studies Physics and not Biology.

Reflect and check

Alternatively, we could have represented the problem by displaying the given information in a Venn diagram.

We know that there are 15 students who study physics, and 10 of those students study both physics and biology. That means there are {15-10=5} students who study only physics.

Then, P(\text{Physics only})=\dfrac{5}{50}=\dfrac{1}{10} or 10\%

b

Find the probability that a randomly selected student studies biology.

Worked Solution
Create a strategy

We can use the Venn diagram to find the probability that a student studies biology. We will let x represent the number of students who only study biology. After solving for that, we can add the answer to the number students who study both biology and physics.

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.
Apply the idea

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

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

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
Reflect and check

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.

Worked Solution
Create a strategy

To avoid double counting the people that 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.

Apply the idea

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\%.

Reflect and check

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.

Worked Solution
Create a strategy

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.

20+12+8=40

So, the intersection must be empty.

Apply the idea

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
Reflect and check

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.

Idea summary

We can calculate the probability of an event using the following formula:

P\left(\text{Event}\right)=\dfrac{\text{Number of outcomes satisfying the event}}{\text{Total number of possible outcomes}}

We can find the probability of the complement of any event using the formula:

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

And we can find the union of compound events using the formula:

P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)

The multiplication rule

Independent events

Two events are independent if the occurrence of one event does not affect the likelihood of the occurrence of the other event

An independent event means that the chances of that event happening are not changed by what happened before. For example, when we flip a coin, there is always a \frac{1}{2} chance that it will land on heads. It doesn't matter whether you tossed heads or tails on any previous flips.

To find the probability of two independent events that occur in sequence, find the probability of each event occurring separately, then multiply the probabilities.

Multiplication rule for independent events

If two events A and B are independent, then

P\left(A\cap B\right)=P\left(A\right) \times P\left(B\right)

The converse is also true. This means you can also test for independence by verifying {P\left(A\cap B\right)=P\left(A\right) \times P\left(B\right)}.

Examples

Example 4

Answer the following questions to determine if the following events are independent.

a

A family has a baby girl. Determine if this will affect the chances of their next baby being a girl.

Worked Solution
Create a strategy

The question is wanting us to determine if the event of having a baby girl is independent from the event of having a second baby girl.

Apply the idea

We can logically reason that gender of the one child has no effect on the gender of the next child. There is the same chance that the baby will be a boy or a girl, regardless of the gender of the previous baby. Therefore, the events are independent.

Reflect and check

To validate this answer, you can conduct an experiment by gathering data on the gender of children in families you know. Then, use the multiplication rule to verify the independence of these events.

b

Valerie owns 9 red shirts and 9 blue shirts. She randomly chooses a shirt for the day, and she gets a red shirt. At the end of the day, she puts it into the washing basket. Will this affect the chances of choosing a red shirt tomorrow?

Worked Solution
Create a strategy

The question is asking if choosing a red shirt on the first day is independent from choosing a red shirt on the second day. We need to determine if the chance of choosing a red shirt on the second day was affected by choosing a red shirt on the first day.

Apply the idea

At first, there was a 50\% chance of choosing a red shirt. Now, there is one less red shirt because Valerie put one into the washing basket. This means there will be one less shirt from the total, and there is a higher chance of choosing a blue shirt. In other words, the chance of choosing a red shirt is different since we are now missing a red shirt.

Choosing a red shirt the first day and placing it in the washing basket does have an effect on the chances of choosing a red shirt the next day. The events are not independent.

Reflect and check

Will this affect the chances of choosing a blue shirt tomorrow?

Yes, it still affects the chances of choosing a blue shirt because there will still be one less shirt than there was the previous day.

Example 5

For each of the following scenarios, use probability to determine if the events are independent.

a

A color spinner has three equally sized sections labeled G, Y and R. Stella spins the spinner twice. The events are:

  • Y: The spinner lands on Y on the first spin
  • G: The spinner lands on G on the second spin
Worked Solution
Create a strategy

To find the probability, we will first need to find the sample space. The sample space of spinning the spinner twice can be modeled using a tree diagram.

A tree diagram with two levels. In the first level, there are three choices: G, Y and R. Each choice from the first level is connected by lines to three choices: G, Y and R.

There are 3 possibilities after spinning the spinner once. It can either land on G,Y, or R.

These are the same possibilities for the second spin, so we need to add 3 more branches from each of the first 3 possibilities.

From the tree diagram, we see there are 9 equally likely outcomes in the sample space: \{(G, G), (G, Y), (G, R), (Y, G), (Y, Y), (Y, R), (R, G), (R, Y), (R, R) \}

To test the independence of the two events, we can check if P(Y \cap G) = P(Y) \cdot P(G) is true.

Apply the idea

P(Y \cap G) is the probability of spinning Y on the first spin and G on the second spin. This is the event (Y,G) which only appears once in the set.

P(Y \cap G)=\dfrac{1}{9}

P(Y) is the probability of spinning Y on the first spin. Only looking at the first branch of the tree diagram, we see

P(Y)=\dfrac{1}{3}

P(G) is the probability of spinning G on the second spin. Looking at the second row of branches in the tree diagram, we can see that there are \dfrac{3}{9} ways this can happen.

P(G)=\dfrac{1}{3}

\displaystyle P(Y)\cdot P(G)\displaystyle =\displaystyle \dfrac{1}{3}\cdot \dfrac{1}{3}
\displaystyle =\displaystyle \dfrac{1}{9}

\therefore P(Y \cap G)=P(Y) \cdot P(G), so the two events are independent.

Reflect and check

We could have determined the events were independent intuitively because we know that spinning one color cannot affect the color of the second spin. However, using probability to show independence confirms our intuition.

b

30 dancers audition for a part. The judges decide that all the dancers can fit into at least one of two categories: the dancer is the right height, R, and the dancer is a good dancer, G. They decided 16 dancers have the right height and 20 dancers are good dancers.

Worked Solution
Create a strategy

We can model the number of people in each category using a Venn diagram. However, \\ n(R)+n(G) = 16+20 =36 which is greater than the number of dancers that audition. So, 6 dancers must be in R \cap G.

A Venn diagram with an outer rectangle and two intersecting circles. The circle on the left is labelled as Right height and has a 10 inside it and a 6 inside the part that intersects with the circle on the right. The circle on the right is labelled as Good dancer and has a 14 inside it in addition to the 6 inside the intersection with the other circle.

To test the independence of the two events we must determine if P(R \cap G) = P(R) \cdot P(G) is true.

Apply the idea

From the Venn diagram we see that: P(R)=\dfrac{16}{30}, \, P(G)=\dfrac{20}{30} and P(R \cap G)=\dfrac{6}{30}.

Determining if the two events are independent:

\displaystyle P(R \cap G)\displaystyle =\displaystyle \dfrac{1}{5}Simplified
\displaystyle P(R) \cdot P(G)\displaystyle =\displaystyle \dfrac{16}{30} \cdot \dfrac{20}{30}Substitute the probabilities
\displaystyle \text{}\displaystyle =\displaystyle \dfrac{16}{45}Evaluate
\displaystyle \therefore P(R \cap G)\displaystyle \neq\displaystyle P(R) \cdot P(G)

Since P(R \cap G) is not equal to the product of P(R) and P(G), the two events are dependent.

Reflect and check

In this problem, we found that the judges' opinions of "good dancers" was not independent from the dancers having the "right height".

Example 6

The following two-way table was made by surveying 100 students who were 9th and 10th graders. They were asked two questions, what their favorite type of music is and if they were in the 9th or 10th grade. The results are as follows:

PopCountryRockTotal
9th grade379652
10th grade3441048
Total711316100

Determine if the following events are independent:

Event A: Being in 10th grade

Event B: Favorite music is Pop

Worked Solution
Create a strategy

To find P(A), we need to look at the total column for the 10th grade row and divide it by the total number of students.

To find P(A), we need to look at the total column for pop music and divide it by the total number of students.

And to find P(A\cap B), we look at the cell for 10th graders who like pop music and divide it by the total number of students.

Apply the idea

P\left(A\right)=\dfrac{48}{100}=48\%

P\left(B\right)=\dfrac{71}{100}=71\%

P\left(A \cap B\right)=\dfrac{34}{100}=34\%

P\left(A\right)\times P\left(B\right)=(.71)(.48)=0.3408\approx 34\%

Because the probabilities are approximately equal, these events are independent.

Reflect and check

Interpreting the independence within the context, this means that the grade a student is in has no effect on which type of music they prefer.

Do you think the events would be independent if music preference responses from students in 9th grade about were compared to responses from students in their first year of college? What if 9th-graders were compared to 40-year-olds or 60-year-olds?

Idea summary

We can determine if two events are independent by using the multiplication rule:

P\left(A\cap B\right)=P\left(A\right) \times P\left(B\right)

Outcomes

S.CP.A.2

Understand that two events a and b are independent if the probability of aand b occurring together is the product of their probabilities, and use this characterization to determine if they are independent.

S.CP.A.4

Construct and interpret two-way frequency tables of data when two categories are associated with each object being classified. Use the two-way table as a sample space to decide if events are independent and to approximate conditional probabilities.

S.CP.B.7

Apply the Addition Rule, P(A or B) = P(A) + P(B) - P(A and B), and interpret the answer in terms of the model.

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