Access this content offline
Omniguides by NOX
Skip to the reading

3.2 Probability Rules

Section Overview

Probability Models

The idea of probability rests on the fact that chance behavior is predictable in the long run. Do we always need to repeat a chance process many times to determine the probability of a particular outcome? Fortunately, the answer is no.

A chance process is made up of two basic parts:

  1. The sample space SS, the set of all possible outcomes which can be seen in Interactive 3.2.1 for rolling two die.

  2. A probability for each outcome

Both of these comprise the probability model.

Interactive 3.2.1 Example 3.2.1

Captured portion of a two-die sample space. Six rows show blue die values 1 through 6, and four visible columns show red die values 2 through 5. The source callouts read die Roll (6, 2) at the top right and die Roll (2, 6) at the bottom. Outer columns and parts of the callouts are clipped in the original widget. Open full-size image S0218
Source gap (S0218): The original interactive is horizontally clipped. Its full 36-outcome display and expanded callout states are not captured.
Example 3.2.1

Imagine rolling two fair, six-sided die—one that’s blue and one that’s red.

Give a probability model for this chance process

Example 3.2.1 — recorded answer

There are 36 possible outcomes when we roll two dice. Interactive 3.2.1 displays these outcomes. They make up the sample space SS. If the dice are well-made (fair), all 36 outcomes will be equally likely. That is, each of the 36 outcomes will come up on 136\frac{1}{36} of all rolls in the long run. So each outcome has probability 136\frac{1}{36}.

A probability model does more than just assign a probability to each outcome. It allows us to find the probability of any collection of outcomes, which we call an event. Events are usually designated by capital letters, like A, B, C, and so on.

That brings up to several basic probability rules:

  1. For any event AA, 0≤P(A)≤10\le P(A)\le1

  2. If SS is the sample space in a probability model, P(S)=1P(S)=1

  3. P(A)=number outcomes in event Anumber of outcomes in sample space SP(A)=\frac{\text{number outcomes in event }A}{\text{number of outcomes in sample space }S}

We also have several slightly more advanced rules:

  1. Complement Rule: P(Ac)=1−P(A)P(A^c)=1-P(A)

  2. Addition Rule for Mutually Exclusive Events: If A and B are mutually exclusive, P(A or B)=P(A)+P(B)P(A\text{ or }B)=P(A)+P(B)

Two-Way Tables and Probability

When we’re trying to find probabilities involving two events, a two-way table can display the sample space in a way that makes probability calculations easier.

Example 3.2.2

Students wanted to find out how common it is for the younger generation to have tattoos. Data is displayed below:

GenderTattoo? YesTattoo? NoTotal
Male88997
Female157691
Total23165188

Find the probabilities that the student:

a) Has a tattoo

b) Is a male with a tattoo

c) Is a male or has a tattoo

Example 3.2.2 — recorded answer

We’ll define events A: is male and B: has a tattoo

a) Since each of the 188 students in the class is equally likely to be chosen, and there are 23 students with a tattoo, P(a tattoo)=P(B)=23188P(\text{a tattoo})=P(B)=\frac{23}{188}.

b) We want to find P(male and a tattoo)P(\text{male and a tattoo}), that is the intersection of the two events or P(A∩B)P(A\cap B). Looking at the intersection of the “Male” row and the “Yes” column, we see that there are 8 males with a tattoo. So P(male and a tattoo)=P(A and B)=8188P(\text{male and a tattoo})=P(A\text{ and }B)=\frac{8}{188}.

c) This time, we’re interested in P(male or a tattoo)P(\text{male or a tattoo}), that is the union of the two events or, P(A∪B)P(A\cup B). From the two-way table, we see that there are 97 males in the class, so P(A)=97188P(A)=\frac{97}{188}. Can we just add P(A)P(A) to P(B)P(B) to get the correct answer? No! These two events are not mutually exclusive, since there are 19 males with a tattoo.

23188+97178=120188≠1,\frac{23}{188}+\frac{97}{178}=\frac{120}{188}\ne1,

which is clearly wrong, since the probability is smaller than 1! From the two-way table, we see that there are 8+15+89=1128+15+89=112 students who are male or have a tattoo.

So P(A or B)=112188P(A\text{ or }B)=\frac{112}{188}.

We can also get this probability from subtracting the probability of the complement from 1. In other words:

1−76188=112188.1-\frac{76}{188}=\frac{112}{188}.
Source gap (S0220): Original source inconsistencies: this answer says “19 males with a tattoo” although its table and part b say 8; the displayed addition uses denominator 178; and it describes being smaller than 1 as the reason the probability is wrong. These statements and numbers are retained as recorded.

The previous example revealed that we can’t use the addition rule for mutually exclusive events unless the two events have no outcomes in common. If events A and B are not mutually exclusive, they can occur together. The probability that one or the other occurs is then less than the sum of their probabilities.

This intuitive result is known as the general addition rule. That is, if A and A are any two events result from some chance process, then

P(A∪B)=P(A)+P(B)−P(A∩B)P(A\cup B)=P(A)+P(B)-P(A\cap B)
Source gap (S0221): The original prose says “A and A”; the adjacent recorded formula uses A and B.

Venn Diagrams and Probability

Venn diagrams are a great tool for illustrating the events within the sample space of a chance process. Gallery 3.2.1 gives further clarification to the following concepts and notation:

The gallery notation, as recorded: Ac,A∪B,A∩B,A∩B=0A^c,\quad A\cup B,\quad A\cap B,\quad A\cap B=0.

Interactive 3.2.2 Example 3.2.2

Venn diagram for the tattoo example. Circle A is blue and contains 89 in its nonoverlapping portion. Circle B is yellow and contains 15 in its nonoverlapping portion. Their green overlap contains 8. Four callouts identify regions; the overlap is incorrectly labelled A-complement intersection B in the original. Open full-size image S0225
A ∩ Bᶜ — recorded annotation
A∩BcA\cap B^c

This represents males without a tattoo

Aᶜ ∩ B — recorded overlap annotation
Ac∩BA^c\cap B

This represents males with a tattoo.

Aᶜ ∩ B — recorded right-region annotation
Ac∩BA^c\cap B

This represents females with a tattoo

Aᶜ ∩ Bᶜ — recorded annotation
Ac∩BcA^c\cap B^c

This represents females with no a tattoo.

Source gap (S0227): The source labels the overlap Aᶜ ∩ B while describing males with a tattoo. Its duplicate notation for two different regions is retained. The area outside both circles is clipped; its count is not visible in this interactive.

What is Conditional Probability?

Suppose we know that event A has happened. Then the probability that event B happens given that event A has happened is denoted by P(B∣A)P(B\mid A). This is known as conditional probability.

For the pierced-ears scenario, however, P(B∣A)≠P(B)P(B\mid A)\ne P(B). When knowledge that one event has happened does not change the likelihood that another event will happen, we say that the two events are independent.

Source gap (S0225): The original paragraph refers to a “pierced-ears scenario” following the tattoo example; this reference is retained.

Tree Diagrams and Probability

If you toss a fair coin three times, what’s the probability that you get two heads? By the reasoning we’ve already established, it’s the proportion of times you would get “HH” if you tossed the coin three times over and over again. Since there is only one way to get two heads, and there are 22=42^2=4 possible outcomes, the probability is 14\frac{1}{4}.

Source gap (S0230): The original prose says three tosses, while its HH outcome and 2² = 4 count describe two tosses. No unrecorded replacement calculation has been added.

This idea of multiplying along the branches in a tree diagram leads to a general multiplication rule for finding the probability P(A∩B)P(A\cap B) that two events happen together:

P(A∩B)=P(A)⋅P(B∣A)P(A\cap B)=P(A)\cdot P(B\mid A)
Example Problem 3.2.3

57% of adult internet users are 18 to 30 years old, another 35% are 31 to 40 years old, and the remaining 8% are 41 and over. 50% of internet users aged 18 to 30 have played a first-person shooter, along with 51% of those aged 31 to 40 and 16% of those 41 or older. What percent of all internetrs have played a first person shooter?

Example Problem 3.2.3 — recorded answer

The percents of each group who play first person shooters are conditional probabilities. We want to find the unconditional probability P(shooter yes)P(\text{shooter yes}). Making a tree diagram should help.

To get the probabilities at the far right, we used the general multiplication rule and multiply the unconditional (original) probabilities with the conditional probabilities:

For instance, the probability that a randomly chosen internet user is an 18- to 30-year old who plays first person shooters is

P(18 to 30 and shooter yes)=P(18 to 30)⋅P(shooter yes∣18 to 30)=(0.57)(0.50)=0.285\begin{aligned}P(18\text{ to }30\text{ and shooter yes})&=P(18\text{ to }30)\cdot P(\text{shooter yes}\mid18\text{ to }30)\\&=(0.57)(0.50)=0.285\end{aligned}

This probability appears at the end of the topmost branch.

There are three disjoint paths to “shooter yes,” one for each of the three age groups. These paths are shown in Interactive 3.2.3. Because the three paths are disjoint, the probability that an adult internetr plays first person shooters is the sum of their probabilities:

P(video yes)=(0.57)(0.50)+(035)(0.51)+(0.08)(0.16)=0.4763=48%\begin{aligned}P(\text{video yes})&=(0.57)(0.50)+(035)(0.51)+(0.08)(0.16)\\&=0.4763=48\%\end{aligned}
Source gap (S0231): The original equation writes “035” without a decimal point and calls the event “video yes” after discussing “shooter yes”. Both are retained.

Interactive 3.2.3 Example 3.2.3

Captured tree diagram with branches to age groups 18–30, 31–40, and 41-plus with probabilities 0.57, 0.35, and 0.08. Each age branch divides into shooter YES and shooter NO, with conditional probabilities respectively 0.5 and 0.5; 0.51 and 0.49; 0.16 and 0.84. Both outer edges are clipped. Partly clipped source callouts label Unconditional Probability and Conditional Probability; another Conditional Probability callout points to the 18–30 age node. Open full-size image S0231

The branch probabilities visible in the original diagram:

Age groupOriginal probabilityShooter YESShooter NO
18–30 years0.570.50.5
31–40 years0.350.510.49
41+ years0.080.160.84
Source gap (S0231): The tree’s outer labels and endpoint probabilities are clipped in the original capture. Expanded annotation explanations are unavailable.

Pop Quiz 3.2.1

Question 1 of 5

Let event A=person wears glassesA=\text{person wears glasses} and event B=person is smartB=\text{person is smart}. Say P(A)=0.136P(A)=0.136 and P(B)=0.221P(B)=0.221 and P(A∪B)=0.295P(A\cup B)=0.295. What is P(A∩B)P(A\cap B)?

Question 2 of 5

Suppose that two drug tests operate independently of each other. Test 1 has probability 0.7 of being positive if drugs have been used. Test 2 has probability 0.6 of being positive if drugs have been used. What is the probability that both tests are negative if drugs have been used?

Question 3 of 5

Consider a pie chart, with 5 categories equality distributed and represented by the green, yellow, blue, red and white colors. Suppose that the pie chart has a centralized spinner moving. What is the probability the spinner will stop on a yellow or blue?

Question 4 of 5

From question 3, what is the probability the spinner will not stop on red or green?

Question 5 of 5

In a toy’s factory, 40% of the products come from Line A and 60% come from Line B. Line A has a defect rate of 8%, whereas Line B has a defect rate of 10%. If an item selected is defective, what is the probability that it came from Line A?

Source gap (S0236): Question 5’s last choice is the malformed original string “0.30.9088”. All five question prompts and choices are captured, but no selected or checked answer feedback is shown.
Glossary and deeper reading · 9 refreshers

These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.

  • Probability

    Probability quantifies how likely an event is under a specified chance model. It lies between 0 and 1, with larger values representing greater likelihood. For example, drawing a red card from a well-shuffled standard deck has probability one half.

    Read more: Probability in the captured Statistics reading →

  • Sample space

    The sample space is the set of all possible outcomes of a chance process. Outcomes must be described at the level relevant to the question being modeled. For example, the sample space for one draw from a red-blue bag can be red and blue.

  • Event in probability

    An event is a set of outcomes from a chance process’s sample space. The event occurs when the observed outcome belongs to that set. For example, rolling an even result is the event containing 2, 4, and 6 for a six-sided die.

  • Two-way table

    A two-way table organizes counts for combinations of two categorical variables. The cells show combinations, while row and column totals summarize each variable separately. For example, a table can classify students by grade level and whether they walk to school.

    Read more: Two-way table in the captured Statistics reading →

  • General addition rule

    The general addition rule finds the probability that at least one of two events occurs. It adds their probabilities and subtracts the probability that both occur, avoiding double counting. For example, the probability of a card being red or a king must subtract the red kings counted twice.

  • Mutually exclusive events

    Mutually exclusive events cannot occur together in the same trial. Their intersection is empty, so the probability that either occurs is the sum of their probabilities. For example, rolling exactly 1 and rolling exactly 4 are mutually exclusive outcomes of one die roll.

  • Venn diagram

    A Venn diagram represents sets or events with regions inside a sample-space boundary. Overlapping regions show outcomes shared by two events. For example, overlapping circles for red cards and kings identify the red kings.

  • Tree diagram

    A probability tree represents successive stages of a chance process using branches. Multiplying the relevant conditional probabilities along a path gives that path’s probability. For example, a two-draw tree can show how a bag’s second-draw probabilities change when the first item is not replaced.

  • General multiplication rule

    The general multiplication rule finds the probability that two events both occur. It multiplies the first event’s probability by the second event’s probability conditional on the first. For example, two red draws without replacement use the updated red proportion for the second draw.

Search lessons, terms and problems