4.2: Theoretical Probability
- Page ID
- 58263
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- Define theoretical probability as based on known, equally likely possible outcomes.
- Understand that probabilities range from 0 (impossible) to 1 (certain).
- Apply key rules such as the complement rule: the probability of an event not occurring equals 1 minus the probability that it does occur.
Theoretical probability is the branch of probability that calculates the likelihood of an event occurring based on mathematical reasoning and logical analysis, rather than on actual experiments or observations. It assumes that all outcomes in a sample space are equally likely.
What is a Sample Space?
In probability theory, a sample space (S) is the set of all possible outcomes of a random experiment. It represents every outcome that could occur when an experiment is performed.
Examples of Sample Spaces:
- Flipping a Coin 🪙
- Possible outcomes: {Heads, Tails}
- Sample space: S = {H, T}
- Flipping Two Coins 🪙🪙
- Possible outcomes: {Heads and Heads, Heads and Tails, Tails and Heads, Tails and Tails}
- Sample space: S ={HH, HT, TH, TT}
- Rolling a Six-Sided Die:🎲
- Possible outcomes: 1, 2, 3, 4, 5, 6
- Sample space: S = {1, 2, 3, 4, 5, 6}
- Rolling Two Six-Sided Dice 🎲🎲
- Possible Outcomes—It is best to write them out as ordered pairs. The first value is an outcome for die#1, and the second value is an outcome for die#2. There are 36 outcomes in total. See the table below.
| 1 | 2 | 3 | 4 | 5 | 6 | |
| 1 | (1,1) | (1,2) | (1,3) | (1,4) | (1,5) | (1,6) |
| 2 | (2,1) | (2,2) | (2,3) | (2,4) | (2,5) | (2,6) |
| 3 | (3,1) | (3,2) | (3,3) | (3,4) | (3,5) | (3,6) |
| 4 | (4,1) | (4,2) | (4,3) | (4,4) | (4,5) | (4,6) |
| 5 | (5,1) | (5,2) | (5,3) | (5,4) | (5,5) | (5,6) |
| 6 | (6,1) | (6,2) | (6,3) | (6,4) | (6,5) | (6,6) |
- Drawing a Card from a Standard Deck
- Sample space: All 52 cards in a deck. The table below lists all 52 outcomes. Jokers are excluded, and the deck consists of four suits, each containing 13 cards.
| Hearts (♥) | Clubs (♣) | Diamonds (♦) | Spades (♠) |
|---|---|---|---|
| A♥ | A♣ | A♦ | A♠ |
| 2♥ | 2♣ | 2♦ | 2♠ |
| 3♥ | 3♣ | 3♦ | 3♠ |
| 4♥ | 4♣ | 4♦ | 4♠ |
| 5♥ | 5♣ | 5♦ | 5♠ |
| 6♥ | 6♣ | 6♦ | 6♠ |
| 7♥ | 7♣ | 7♦ | 7♠ |
| 8♥ | 8♣ | 8♦ | 8♠ |
| 9♥ | 9♣ | 9♦ | 9♠ |
| 10♥ | 10♣ | 10♦ | 10♠ |
| J♥ | J♣ | J♦ | J♠ |
| Q♥ | Q♣ | Q♦ | Q♠ |
| K♥ | K♣ | K♦ | K♠ |
Table \(\PageIndex{1}\): Sample Space for a Deck of Cards
Probability Properties
- The probability of any event in the sample space is between 0 and 1, inclusive as shown here \(0 \leq P(\text { event }) \leq 1\)
- If the P(event)=1, then it will happen and is called a certain event.
- If the P(event)=0, then it cannot happen and is called an impossible event.
- The sum of all the probabilities in the sample space is one as shown here \(\sum P(\text { outcome })=1\)
Suppose you conduct an experiment where you pull a card from a standard deck. Write the answers as a fraction or a decimal rounded to three place values.
- What is the probability of getting a Spade?
- What is the probability of getting a Jack?
- What is the probability of getting a Spade and an Ace?
- What is the probability of getting a Spade or an Ace?
- What is the probability of getting a Jack and an Ace?
- What is the probability of getting a Jack or an Ace?
Solutions
- Let A = getting a spade = {2♠, 3♠, 4♠, 5♠, 6♠, 7♠, 8♠, 9♠, 10♠, J♠, Q♠, K♠, A♠} therefore,
\(P(A)=\dfrac{13}{52} = 0.25\)
- Let B = getting a Jack = {J♠, J♣, J♥, J♦} therefore,
\(P(B)=\dfrac{4}{52} = 0.077\)
- Let E = getting a Spade and an Ace = {A♠} therefore,
\(P(E)=\dfrac{1}{52} = 0.019\)
- Let F = getting a Spade or an Ace ={2♠, 3♠, 4♠, 5♠, 6♠, 7♠, 8♠, 9♠, 10♠, J♠, Q♠, K♠, A♠, A♣, A♦, A♥} therefore,
\(P(F)=\dfrac{16}{52} = 0.308\)
- Let G = getting a Jack and an Ace = { } since you can’t do that with one card. therefore,
\(P(G)=0\)
- Let H = getting a Jack or an Ace = {J♠, J♣, J♦, J♥, A♠, A♣, A♦, A♥} therefore,
\(P(H)=\dfrac{8}{52} = 0.154\)
Assume that two dice are tossed and we desire to compute the probability of an outcome. Typically the outcomes are computed by adding the two numbers (represented by the pips/dots on the sides of the dice) together. Assume that one pair of dice is tossed one time. Compute the following probabilities. Write the answers as a fraction or a decimal rounded to three place values.
- The sum is 5.
- The sum is less than or equal to 5.
- The sum is 4 or 9.
- The sum is 13.
- The probability of getting doubles.
Solutions
- Let A = getting a sum of 5 = {(1,4), (2,3), (3,2), (4,1)} therefore,
\(P(A)=\dfrac{4}{36} = 0.111 \)
- Let B = sum less than 5 = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)} therefore,
\(P(B)=\dfrac{10}{36} = 0.278 \)
- Let C = sum is 4 or 9. = {(1,3), (2,2), (3,1), (3,6), (4,5), (5,4), (6,3)} therefore,
\(P(C)=\dfrac{7}{36} = 0.194 \)
- Let D = sum is 13 = { } since there is no sum of 13 for two dice, therefore,
\(P(D)=0\)
- Let E = getting doubles = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}
\(P(E)=\dfrac{6}{36} = 0.167 \)
Complementary Events
In probability, complementary events are pairs of outcomes that cover all possibilities but cannot happen at the same time. If one event occurs, the other cannot. For example, flipping a coin results in either heads or tails—these are complementary events. The key idea is that the probability of an event and its complement always adds up to 1, which makes complements useful when it's easier to calculate the chance of something not happening.
If two events are complementary events, then to find the probability of one, subtract the probability of the other from one. The notation used to complement A is not A or \(\bar{A}\).
\(P(A)+P\left(\bar{A}\right)=1, \text { or } P(\bar{A})=1-P(A )\) where \( P(\bar{A}) = P( \text{not } A) \)
- Suppose you know that the probability of it raining today is 0.45. What is the probability of it not raining?
- Suppose you know the probability of not getting the flu is 0.24. What is the probability of getting the flu?
- In an experiment of picking a card from a deck, what is the probability of not getting a card that is a Queen?
Solutions
a. Since not raining is the complement of raining, then
\(P(\text { not raining })=1-P(\text { raining })=1-0.45=0.55\)
b. Since getting the flu is the complement of not getting the flu, then
\(P(\text { getting the flu })=1-P(\text { not getting the flu })=1-0.24=0.76\)
c. You could solve this problem by listing all the ways to not get a queen, but that set is fairly large. One advantage of the complement is that it reduces the workload. You use the complement in many situations to make the work shorter and easier. In this case it is easier to list all the ways to get a Queen, find the probability of the Queen, and then subtract from one. Queen = {Q♠, Q♣, Q♦, Q♥} so
\(P(\text { Queen })=\dfrac{4}{52}\) and
\(P(\text { not Queen })=1-P(\text { Queen })=1-\dfrac{4}{52}=\dfrac{48}{52}\)
- In an experiment of rolling a fair die once, find the probability of rolling a 4 on the die. Write the answer as a fraction or a decimal number rounded to three places.
- In an experiment of pulling a card from a fair deck, find the probability of not getting a jack. Write the answer as a fraction or a decimal number rounded to three places.
Solutions
a. The sample space for this experiment is {1, 2, 3, 4, 5, 6}. You want the event of getting at least a 4, which is the same as thinking of getting 4 or more The event space is {4, 5, 6}. The probability is
\(P(\text { at least } 4)=\dfrac{3}{6} = 0.5\)
b. The sample space for this experiment is the entire deck of cards.
Getting a card that is not a jack would involve listing all of the cards that are not jacks, it would be much easier to list the outcomes that make up the complement. The complement is all the jacks. The event space for the complement would be {J♠, J♣, J♦, J♥}. The probability of the complement would be \(\dfrac{4}{52}\). The probability of not getting a jack is
\(P(\text { not a jack } )=1-\dfrac{4}{52}=\dfrac{48}{52} = 0.923\)
Authors
"4.2: Theoretical Probability" by Toros Berberyan, Tracy Nguyen, and Alfie Swan is licensed under CC BY-SA 4.0
Attributions
"4.2: Theoretical Probability" by Kathryn Kozak is licensed under CC BY-SA 4.0
Exercises
- Suppose you conduct an experiment where you pull a card from a standard deck. Round the answers to 3 decimal places.
- What is the probability of getting a club?
- What is the probability of getting an Ace?
- What is the probability of not getting an Ace?
- What is the probability of getting a Jack or a Spade?
- What is the probability of getting a Jack and a Spade?
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- A coin is flipped two times. Find the following probabilities. Round the answers to 3 decimal places.
- Getting two tails.
- Getting no tails.
- At least one head.
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- Assume that two dice are tossed and we desire to compute the probability of an outcome. Typically, the outcomes are determined by adding the two numbers (represented by the pips/dots on the sides of the dice) together. Assume that one pair of dice is tossed one time. Compute the following probabilities. Write the answers as a fraction or a decimal rounded to three place values.
- The sum is 7.
- The sum is greater than or equal to 10.
- The sum is 5 or 12.
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- Two dice are tossed. Find the probability of rolling a pair of dice and getting doubles (the same number on the first and second die). Write the answer as a decimal. Round the answer to three decimal places.
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- Answers
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If you are an instructor and want the solutions to all the exercise questions for each section, please email Toros Berberyan.