3.1: Basics of Probability
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)We see probabilities almost every day in our lives. When you pick up the newspaper or read the news on the internet, you most likely encounter probability. There is a 60% chance of rain today, or a poll shows that 52% of voters approve of the President's job performance. Probabilities are essential in sports, gaming, and gambling establishments, but probabilities are also used to make business decisions, figure out insurance premiums, and determine the price of raffle tickets. In its most general sense, probability provides us a way to measure the chance or likelihood that something will happen.
Terminology Used in Probability
Before discussing how to find probabilities, we need to familiarize ourselves with some basic terminology. When studying probability, we consider an random experiment to be an activity or an operation that gives a result that can be observed but not predicted ahead of time. If we roll a pair of dice, pick a card from a deck of playing cards, spin a spinner, or randomly select a person and observe their hair color, we are executing an experiment and observing the result.
Any possible result of conducting an experiment is called an outcome. For the experiment of flipping a coin, there are only two outcomes: head or tail. For the experiment of rolling a single die, there are 6 outcomes: 1, 2, 3, 4, 5, or 6. For an experiment, this collection of all possible outcomes is called the sample space.
An event is a collection of outcomes from an experiment. In some instances events contain only one outcome while at other times an event may contain more than one outcome. Consider the event of rolling a single dice. The event "rolling a 3" contains only the outcome \(\{3\}\) while the event "rolling an even number" contains the outcomes \(\{2,4,6\}\).
- A random experiment is an activity or operation with a result that cannot be predicted ahead of time.
- Any result from conducting an experiment is called an outcome.
- The sample space of an experiment is the set of all its possible outcomes.
- An event is a subset of the sample space and describes a collection of outcomes.
These terms are best illustrated with some examples.
Consider an experiment of rolling a single die. When we roll it, only one outcome will occur, but we are unsure which outcome. There are six possible outcomes and so the sample space is \(S=\{1, 2, 3, 4, 5, 6\}\).
'"Rolling a 2" is an event that contains only one outcome \(\{2\}\).
"Rolling a number greater than 2" is a another event that contains multiple outcomes \(\{3, 4, 5, 6\}\).
Two pennies are tossed at the same time. Both pennies may land heads up (which we write as HH), or the first penny might land heads up and the second one tails up (which we write as HT), and so on. Write the sample space for the experiment and list the outcomes in the event "getting at least one heads."
Solution
The sample space for this experiment is \(S = \{HH, HT, TH, TT\}\).
If we define event \(A\) as "getting at least one heads," the outcomes in event \(A\) can be written as \(A = \{HH, HT, TH\}\).
Gabe performs an experiment of flipping a coin and then rolling a regular six-sided die.
- Give the sample space for how the coin and die could land.
- Give the outcomes in event \(A\): rolls an odd number.
- Give the outcomes in the event \(B\): gets tails and rolls an even number.
- Answer
-
- \(\{H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6 \}\)
- \(\{H1, H3, H5, T1, T3, T5\}\)
- \(\{T2, T4, T6 \}\)
Probability is one way to measure the chance or the likelihood that an event will occur. Probability is usually denoted in function notation by \(P\), and the event is denoted by a capital letter such as \(A, B, C,\) etc. The mathematical notation that indicates the probability that event A happens is \(P(A)\).
Probability is a numerical measure of the chance or the likelihood that an event will occur.
We will examine two types of probability in this chapter: empirical probability and theoretical probability. But, we will begin with theoretical probability and discuss the connection and differences between the two types later in this section.
Theoretical Probability
We turn to the concept of theoretical probability and relate it to the sample space, outcomes, and events. We define theoretical probability of an event as the ratio of the number of outcomes in the event to the number of outcomes in the sample space of an experiment. For this definition of probability, it is very important that the outcomes in the sample space must be equally likely to occur.
A theoretical probability is based on a mathematical model where the number of outcomes in the event is compared with the number of outcomes in the sample space of an experiment. A formula for the theoretical probability of event \(E\) is
\(P(A) = \dfrac{\text{number of outcomes* in event E}}{\text{number of outcomes* in the sample space }S}\)
*where the outcomes are equally likely.
Let's apply this formula in some relatively simple examples.
Consider the experiment of rolling a regular six-sided die. Find the probability of each event:
- rolling a 5
- rolling an even number
- rolling a number greater than 4
- rolling a 7
- rolling a number less than 7
Solution
There are 6 possible equally likely outcomes in the sample space for this experiment: \(S=\{1, 2, 3, 4, 5, 6\}\).
- There is only 1 outcome in the event "rolling a 5": \(\{5\}\)
\(P(\text{rolling a 5}) = \frac{\text{number of ways to roll a 5}}{\text{number of outcomes in the sample space}} = \frac{1}{6}\)
- There are 3 outcomes in the event "rolling an even number": \(\{2, 4, 6\}\).
\(P(\text{rolling an even number}) = \frac{\text{number of ways to roll an even number}}{\text{number of outcomes in the sample space}} = \frac{3}{6}=\frac{1}{2}\)
- There are 2 outcomes in the event "rolling a number greater than 4": \(\{5, 6\}\).
\(P(\text{rolling a number greater than 4}) = \frac{\text{number of ways to roll a number greater than 4}}{\text{number of outcomes in the sample space}} = \frac{2}{6}=\frac{1}{3}\)
- There are no outcomes in the event "rolling a 7": \(\{ \; \; \}\).
\(P(\text{rolling a 7}) = \frac{\text{number of ways to roll a 7}}{\text{number of outcomes in the sample space}} = \frac{0}{6}=0\)
- There are 6 outcomes in the event "rolling a number less than 7": \(\{1, 2, 3, 4, 5, 6\}\).
\(P(\text{rolling a number less than 7}) = \frac{\text{number of ways to roll a number less than 7}}{\text{number of outcomes in the sample space}} = \frac{6}{6}=1\)
The previous example illustrates some important properties about values that can be legitimate probabilities.
- The number of outcomes in an event can never be lower than 0. So, the smallest a probability can be is 0. If the probability of an event is 0 (such as the probability of rolling a 7 in Example 3), we say that event is impossible.
- Furthermore, the number of outcomes in an event can never be more than the number of outcomes in the sample space. Therefore, the largest a probability can be is 1. If the probability of an event is 1 (such as the probability of rolling a number less than 7 in Example 3), we say that event is certain.
- The previous two facts tell us that the probability of any event must always fall between 0 and 1, inclusive. In the course of this chapter, if you compute a probability and get an answer that is negative or greater than 1, you have made a mistake and should recheck your work.
An event that cannot occur has a probability of 0. This event is impossible.
An event that must occur has a probability of 1. This event is certain.
The probability of any event must be between 0 and 1, inclusive. That is, \(0 \leq P(E) \leq 1\).
Recall the experiment in Example 2: Two pennies are tossed simultaneously and how they land is recorded. Find the probability getting each result:
- exactly two heads
- exactly one head
- at least one head
- more than 2 heads
Solution
Recall the the sample space for this experiment is \(S = \{HH, HT, TH, TT\}\).
- The event "exactly two heads" occurs in only 1 outcome \(\{HH\}\): \(P(\text{exactly two heads}) = \frac{1}{4}\)
- The event "exactly one head" occurs in 2 outcomes, \(\{HT,TH\}\): \(P(\text{exactly one head}) = \frac{2}{4}= \frac{1}{2}\)
- The event "at least one head" occurs in 3 outcomes, \(\{HH,HT,TH\}\): \(P(\text{at least 1 head}) = \frac{3}{4}\)
- The event "more than 2 heads" occurs in 0 outcomes, \(\{ \; \;\}\): \(P(\text{more than 2 heads}) = \frac{0}{4}= 0\)
Gabe performs an experiment of flipping a coin and then rolling a regular six-sided die. (See Try it Now 1.) Find the probability of each event:
- Gabe rolls an odd number.
- Gabe gets tails and rolls an even number.
- Gabe rolls a number than is less than 10.
- Answer
-
- \(\frac{6}{12}=\frac{1}{2}\)
- \(\frac{3}{12}=\frac{1}{4}\)
- \(\frac{12}{12}=1\)
A small bookcase contains five math books, three English books, and seven science books - all the same size. One book is chosen at random. What is the probability that a math book is chosen?
Solution
There are 15 possible equally likely books that could be selected, so the number of possible outcomes in the sample space is 15. Of these 15 outcomes, 5 are in the event "math book," so the probability that the book will be a math book is \(\frac{5}{15}=\frac{1}{3}\).
Let's say you have a container of 20 cherries, 14 sweet and 6 sour. If you pick a cherry at random, what is the probability that it will be sweet?
- Answer
-
There are 20 possible cherries that could be picked, so the number of possible outcomes in the sample space is 20. Of these 20 possible outcomes, 14 are in the event "sweet," so the probability that the cherry will be sweet is \(\frac{14}{20}=\frac{7}{10}\).
Find the probability of drawing each of these cards on the first draw from a standard deck of playing cards:
- a red card
- a club
- a King
- a King of clubs
Solution
To assist in analyzing the outcomes and sample space, here is a diagram of a deck of cards. Counting the cards, there are 52 equally likely outcomes in the sample space.
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[1] ""English pattern" playing cards deck laid out, in SVG vector format"(opens in new window) by Dmitry Fomin(opens in new window) is in the Public Domain, CC0(opens in new window).
- There are 26 outcomes in the event "drawing a red card." (13 hearts, 13 diamonds). \(P(\text{red card} = \frac{26}{52} = \frac{1}{2}\).
- There are 13 outcomes in the event "drawing a club." \(P(\text{club}= \frac{13}{52} = \frac{1}{4}\).
- There are 4 outcomes in the event "drawing a King." \(P(\text{King}) = \frac{4}{52}= \frac{1}{12}\).
- There is only 1 outcome in the event "drawing a King of clubs." \(P(\text{King of clubs}) = \frac{1}{52}\)
Find the probability of drawing each of these cards on the first draw from a standard deck of playing cards:
- a 3 card
- a red 3 card
- a 3 of diamonds
- a red face card (Jack, Queen, or King)
- Answer
-
- \(\frac{4}{52}=\frac{1}{13}\)
- \(\frac{2}{52}=\frac{1}{26}\)
- \(\frac{1}{52}\)
- \(\frac{12}{52}=\frac{3}{13}\)
Empirical Probability
Up to this point, we have been finding probability theoretically by counting the equally likely outcomes in the sample space and in the event. For example, the theoretical probability of a coin ‘landing heads up’ is \(\frac{1}{2}\) because there are two equally likely outcomes in the sample space and only one of those outcomes is in the event ‘landing heads up.’
However, if you flip a coin 10 times, you may very well observe that the coin lands heads up 40%, 60%, or even 70% of the time! This discrepancy is perfectly natural and expected when conducting experiments, and it is important to recognize it. This brings us to the difference between theoretical probability and empirical probability.
- Theoretical probability is the calculated chance that an event occurs if the same experiment were completed an infinite number of times. We sometimes use our knowledge of sets and set operations when finding theoretical probabilities.
- Empirical probability is the calculated chance that an event occurs based on how often the outcome has occurred in the past over a limited number of trials. Generally speaking, when we determine the chance of something happening in the future by studying past results we are using empirical probability. Because empirical probability is determined by actual observations of an experiment it is also known as experimental probability.
- A theoretical probability is based on a mathematical model where all outcomes are equally likely to occur.
- An empirical probability is based on collected data and is the relative frequency of the event occurring.
Lawrence is playing with a standard 52-card deck and wants to find the probability of selecting a Queen from the deck.
The theoretical probability that Lawrence pulls a single Queen is \(\frac{4}{52} = \frac{1}{13} \approx 0.0769\) or about \(7.69\%\).
If Lawrence decides to try it out 25 times and pulls a Queen at random 3 times in 25 trials of “pull a card, record it, put it back,” the empirical probability of pulling a Queen is \(\frac{3}{25} = 0.12\), or about 12%.
A game uses a spinner like the one shown below. When spun, the arrow lands on either a white, yellow, blue, or red region. Let’s describe how we could use empirical probability to investigate the chance of landing on the white region.
The regions do not appear to be the same size, so the four color outcomes are not equally likely. We cannot apply our theoretical probability formula here unless we are told the areas of the regions or the measures of the central angles in the circular spinner. Instead, we can spin the spinner many times and record the results. Assume these hypothetical results have occurred:
| Color | Number of Spins |
|---|---|
| White | 18 |
| Red | 12 |
| Blue | 6 |
| Yellow | 4 |
Using the data in the table, the spinner landed on 'white' 18 times out of a total of \(18+12+6+4=40\) times. The empirical probability that the spinner lands on white is \( \frac{18}{40} = \frac{9}{20} = 0.45 = 45\% \). This seems reasonable as the white region makes up slightly less than half the spinner.
Often times, we calculate empirical probabilities from data that have already been collected and organized for us.
In a given week, a veterinarian recorded how many times she treated each type of animal. What is the probability that the next animal she treats is a cat?
| Animal | Number Treated |
|---|---|
| Dog | 40 |
| Cat | 32 |
| Bird | 9 |
| Rabbit | 7 |
| Iguana | 2 |
Solution
The total number of trials (animals treated) is \(40 + 32+9+7+2=90\). Of these, 32 were cats. The fraction of trials that resulted in the event "cat" gives the empirical probability: \(P(\text{cat})=\frac{32}{90}=\frac{16}{45} \approx 0.3556 \text{ or } 35.56\%\).
A survey was conducted to determine the programs that students enrolled in MAT 1130 were taking. The results are shown in the table. What is the probability that a student in MAT 1130 is in the Computer Information Technology program?
| Major | English | Art | Nursing | Culinary Arts & Hospitality | Computer Information Technology |
|---|---|---|---|---|---|
| Number of Students | 3 | 8 | 2 | 12 | 10 |
- Answer
-
\(\frac{10}{35}=\frac{2}{7} \approx 0.2857 \text{ or } 28.57\%\)
The table shows the results of a survey where each person was asked 1) Do you have cable TV? and 2) Did you go on vacation in the past year?.
| Took a Vacation | No Vacation | Total | |
|---|---|---|---|
| Has Cable TV | 97 | 38 | 135 |
| Doesn't Have Cable TV | 14 | 17 | 31 |
| Total | 111 | 55 | 166 |
Use the table to find the probability of each event:
- \(P( \text{a person takes a vacation})\)
- \(P( \text{a person does not have cable TV})\)
- \(P( \text{a person takes a vacation and has cable TV})\)
- \(P( \text{a person takes a vacation or has cable TV })\)
Solution
There are 166 people surveyed so the number of trials of this experiment is 166. We need to determine the number of outcomes in each event and then find its relative frequency.
- The total in the first column shows 111 people took a vacation last year:
\(P(\text{a person takes a vacation}) = \frac{111}{166} \approx 0.6687 \text{ or } 66.87\%\).
- The total in the second row shows 31 people do not have cable TV:
\(P(\text{a person does not have cable TV}) =\frac{31}{166} \approx 0.1867 \text{ or } 18.67\%\).
- There are 97 people who did both of these things:
\(P(\text{a person takes a vacation and has cable TV}) = \frac{97}{166} \approx 0.5843 \text{ or } 58.43\%\).
- There are \(97+38+14=149\) people who did one or the other or both of these things:
\(P(\text{a person takes a vacation or has cable TV}) = \frac{149}{166} \approx 0.8976 \text{ or } 89.76\%\).
Nate recorded the topping and type of crust for all the pizzas he sold yesterday.
| Pepperoni | Sausage | Veggie | Total | |
|---|---|---|---|---|
| Thick Crust | 10 | 15 | 5 | 30 |
| Thin Crust | 24 | 16 | 10 | 50 |
| Total | 34 | 31 | 15 | 80 |
Use the table to find the probability of each event the next time Nate sells a pizza.
- \(P( \text{sausage topping})\)
- \(P( \text{sausage topping or thick crust})\)
- \(P( \text{veggie topping and thin crust})\)
- \(P( \text{topping contains meat })\)
- Answer
-
- \(\frac{31}{80} = 0.3875 \text{ or } 38.75\%\)
- \(\frac{46}{80} = 0.575 \text{ or } 57.5\%\)
- \(\frac{10}{80} = 0.125 \text{ or } 12.5\%\)
- \(\frac{65}{80} = 0.8125 \text{ or } 81.25\%\)
The Law of Large Numbers
Flipping a coin is often used to randomly make a decision when there are only two choices. For example, you may flip a coin to decide whether you have steak or fish for dinner. Or, a referee uses a coin flip to decide which football team receives the ball prior to kickoff. The reason why a coin flip seems fair in these circumstances is that most of us agree that the probability of getting heads (and tails) on a coin is \(\frac{1}{2} = 0.5 = 50\%\). But what does this mean in practice?
Does that mean if we flip a coin twice we will get heads exactly once? If a coin is tossed 10 times, will we necessarily get heads five times? Most of intuitively know the answer is no. Indeed, if we flip a coin 10 times we might find that it lands on heads 7 times. So what does it mean to say that the probability of heads on a fair coin is \(\frac{1}{2}\)?
To investigate this question, consider the table showing results that may happen when a coin is toss several times. The top row shows the number times the coin has been tossed. The next row shows the number of heads that have occurred. The bottom row shows the empirical probability which is the ratio of the number of heads observed to the number trials.
| Number of Trials | 10 | 20 | 30 | 40 | 50 |
|---|---|---|---|---|---|
| Number of Heads Observed | 7 | 13 | 17 | 22 | 26 |
| Empirical Probability of Heads | \(\frac{7}{10}=0.7\) | \(\frac{13}{20}=0.65\) | \(\frac{17}{30} \approx 0.57\) | \(\frac{22}{40}=0.55\) | \(\frac{26}{50}=0.52\) |
Notice that as the number of trials increases, the empirical probability gets closer to 0.5, which is what we expect to happen theoretically. In fact, if we kept increasing the number of trials, we would find that the empirical probability would eventually be very, very close to \(\frac{1}{2}= 0.5\).
This relationship between empirical probability and theoretical probability can be summarized by the Law of Large Numbers. The probability of an event applies to a large number of trials, not a single or a few trials. We should not be surprised that the empirical probability calculated from only a few trials is different from the theoretical probability. It is only empirical probability calculated over the long-run that gives an accurate probability.
The Law of Large Numbers states that we can only expect the empirical probability of an event to approximate its true probability when the number of trials of the experiment is large.
Stanley picks a card from a standard deck of cards and gets heart. He returns the card to the deck, picks a second time, and gets another heart. Stanley repeats this process a total of five times and gets a heart in four of those trials. Should Stanley conclude that the probability of selecting a heart is \(\frac{4}{5} = 0.80 = 80\%\)?
Solution
No. Stanley's reasoning is incorrect. He should not make a conclusion about the true probability of selecting a heart based only on five trials. If Stanley wants to estimate the chance of selecting a heart using empirical probability, he should perform the experiment a lot more than five times.