4.2: Theoretical Probability

 

Learning Objectives

• 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:

1. Flipping a Coin 🪙

• Possible outcomes: {Heads, Tails}
• Sample space: S = {H, T}

2. Flipping Two Coins 🪙🪙

• Possible outcomes: {Heads and Heads, Heads and Tails, Tails and Heads, Tails and Tails}
• Sample space: S ={HH, HT, TH, TT}

3. Rolling a Six-Sided Die:🎲

• Possible outcomes: 1, 2, 3, 4, 5, 6
• Sample space: S = {1, 2, 3, 4, 5, 6}

4. 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.

Figure \(\PageIndex{1}\):This table shows the sample space for rolling two six-sided dice. The numbers across the top represent the outcomes of the first die, and the numbers along the left side represent the outcomes of the second die. Each cell contains an ordered pair \((x, y)\), where \(x\) is the result of the first die and \(y\) is the result of the second die. Since each die has 6 possible outcomes, there are \(6 \times 6 = 36\) equally likely outcomes in total. This table can be used to determine probabilities of different events. For example, to find the probability of rolling a sum of 7, count the outcomes that add to 7: \((1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\). There are 6 such outcomes out of 36, so the probability is \( \frac{6}{36} = \frac{1}{6} \). This visual helps identify outcomes and calculate probabilities for sums, doubles, or other events involving two dice.

5. 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.

Sample Space for a Deck of 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 \(\PageIndex{1}\): The deck contains four suits—hearts, diamonds, clubs, and spades. Hearts and diamonds are red, while clubs and spades are black. Each suit includes thirteen ranks: ace, numbers two through ten, and the face cards jack, queen, and king. The cards are rectangular with white backgrounds, suit symbols (pips), and corner indices showing the rank and suit. Face cards display illustrated portraits, while numbered cards show repeated suit symbols arranged in patterns. The deck may appear stacked, fanned, or spread to show multiple cards.

Probability Properties

1. The probability of any event in the sample space is between 0 and 1, inclusive, as shown here \(0 \leq P(\text { event }) \leq 1\)
2. If the P(event) = 1, then it will happen and is called a certain event.
3. If the P(event) = 0, then it cannot happen and is called an impossible event.
4. The sum of all the probabilities in the sample space is one, as shown here \(\sum P(\text { outcome })=1\)

Example \(\PageIndex{1}\) calculating theoretical probabilities

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.

1. What is the probability of getting a Spade?
2. What is the probability of getting a Jack?
3. What is the probability of getting a Spade and an Ace?
4. What is the probability of getting a Spade or an Ace?
5. What is the probability of getting a Jack and an Ace?
6. What is the probability of getting a Jack or an Ace?

Solutions

1. Let A = getting a spade = {2♠, 3♠, 4♠, 5♠, 6♠, 7♠, 8♠, 9♠, 10♠, J♠, Q♠, K♠, A♠} therefore,

\(P(A)=\dfrac{13}{52} = 0.25\)

2. Let B = getting a Jack = {J♠, J♣, J♥, J♦} therefore,

\(P(B)=\dfrac{4}{52} = 0.077\)

3. Let E = getting a Spade and an Ace = {A♠} therefore,

\(P(E)=\dfrac{1}{52} = 0.019\)

4. 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,

\(P(F)=\dfrac{16}{52} = 0.308\)

5. Let G = getting a Jack and an Ace = { } since you can’t do that with one card. therefore,

\(P(G)=0\)

6. Let H = getting a Jack or an Ace = {J♠, J♣, J♦, J♥, A♠, A♣, A♦, A♥} therefore,

\(P(H)=\dfrac{8}{52} = 0.154\)

Example \(\PageIndex{2}\) calculating theoretical probabilities

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.

1. The sum is 5.
2. The sum is less than or equal to 5.
3. The sum is 4 or 9.
4. The sum is 13.
5. The probability of getting doubles.

Solutions

1. Let A = getting a sum of 5 = {(1,4), (2,3), (3,2), (4,1)} therefore,

\(P(A)=\dfrac{4}{36} = 0.111 \)

2. 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,

\(P(B)=\dfrac{10}{36} = 0.278 \)

3. Let C = sum is 4 or 9. = {(1,3), (2,2), (3,1), (3,6), (4,5), (5,4), (6,3)} therefore,

\(P(C)=\dfrac{7}{36} = 0.194 \)

4. Let D = sum is 13 = { } since there is no sum of 13 for two dice, therefore,

\(P(D)=0\)

5. Let E = getting doubles = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}

\(P(E)=\dfrac{6}{36} = 0.167 \)

Example \(\PageIndex{3}\)

A liberal arts club is organizing a cultural showcase. Students randomly select one event ticket from a box. The box contains the following tickets:

• 5 Music Performance tickets
• 4 Poetry Reading tickets
• 3 Art Exhibition tickets

1. What is the probability that a student selects a poetry reading ticket?
2. What is the probability that a student selects a music performance ticket?

Solution

1. Probability of selecting a Poetry Reading ticket

P(Poetry)= \(\dfrac{4}{12} = \dfrac{1}{3}\)

2. Probability of selecting a Music Performance ticket

P(Music)=\(\dfrac{5}{12}\)

 

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.

Definition \(\PageIndex{2}\): complementary events

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 \( P(\bar{A}) = P( \text{not } A) \)

Example \(\PageIndex{4}\) complementary events

1. Suppose you know that the probability of it raining today is 0.45. What is the probability of it not raining?
2. Suppose you know the probability of not getting the flu is 0.24. What is the probability of getting the flu?
3. 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

\(P(\text { not Queen })=1-P(\text { Queen })=1-\dfrac{4}{52}=\dfrac{48}{52}\)

Example \(\PageIndex{5}\) using the complement to find probabilities

1. 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.
2. 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

\(P(\text { not a jack } )=1-\dfrac{4}{52}=\dfrac{48}{52} = 0.923\)

Exercises

Exercise \(\PageIndex{1}\)

Two fair six-sided dice are rolled at the same time. Each die has six possible outcomes: 1, 2, 3, 4, 5, and 6.

1. What is the probability that the sum of the two dice is 8?
2. What is the probability that the sum of the two dice is 10?
3. What is the probability that the sum of the two dice is greater than 9?

Answer

1. The outcomes that sum to 8 are \((2,6), (3,5), (4,4), (5,3), (6,2)\), so there are 5 outcomes. \( P(\text{sum } = 8) = \dfrac{5}{36} \)
2. The outcomes that sum to 10 are \((4,6), (5,5), (6,4)\), so there are 3 outcomes. \( P(\text{sum } = 10) = \dfrac{3}{36} = \dfrac{1}{12}\)
3. Sums greater than 9 are 10, 11, and 12. The outcomes are: 10 → 3 outcomes, 11 → 2 outcomes, 12 → 1 outcome (total = 6). \( P(\text{sum } > 9) = \dfrac{6}{36} = \dfrac{1}{6} \)

Exercise \(\PageIndex{2}\)

A standard deck of playing cards contains 52 cards. The deck has four suits: hearts, diamonds, clubs, and spades. Each suit contains 13 cards (Ace through King).
One card is randomly selected from the deck.

1. What is the probability of selecting a heart?
2. What is the probability of selecting a king?
3. What is the probability of selecting a black card (clubs or spades)?

Answer

1. There are 13 hearts out of 52 cards. \( P(\text{heart}) = \dfrac{13}{52} = \dfrac{1}{4} \)
2. There are 4 kings out of 52 cards. \( P(\text{king}) = \dfrac{4}{52} = \dfrac{1}{13} \)
3. There are 26 black cards (clubs and spades) out of 52 cards. \( P(\text{black card}) = \dfrac{26}{52} = \dfrac{1}{2} \)

Exercise \(\PageIndex{3}\)

An economics professor prepares a box of topic cards for a class activity. The cards represent different areas of economics:

• Macroeconomics: 6 cards
• Microeconomics: 5 cards
• International Economics: 3 cards
• Behavioral Economics: 2 cards

A student randomly selects one card from the box.

1.  What is the probability that the student selects a Macroeconomics card?
2. What is the probability that the student selects a Microeconomics card?
3. What is the probability that the student selects either an International Economics card or a Behavioral Economics card?

Answer

1. Total cards = \(6 + 5 + 3 + 2 = 16\). \( P(\text{Macroeconomics}) = \dfrac{6}{16} = \dfrac{3}{8} \)
2. \( P(\text{Microeconomics}) = \dfrac{5}{16} \)
3. Favorable cards = \(3 + 2 = 5\). \( P(\text{International or Behavioral}) = \dfrac{5}{16} \)

Exercise \(\PageIndex{4}\)

A survey is conducted on a college campus. One student is selected at random. The results show that 60% of students participate in a campus club, 50% work part-time, and 30% do both.

1. What is the probability that the selected student participates in a club and works part-time?
2. What is the probability that the selected student participates in a club or works part-time?
3. What is the probability that the selected student does neither activity?

Answer

1. \( P(\text{club and work}) = 0.30 \)
2. \( P(\text{club or work}) = 0.60 + 0.50 - 0.30 = 0.80 \)
3. \( P(\text{neither}) = 1 - 0.80 = 0.20 \)

Exercise \(\PageIndex{5}\)

A standard deck of playing cards contains 52 cards. The deck has four suits: hearts, diamonds, clubs, and spades. There are 13 cards in each suit.
One card is randomly selected from the deck.

1. Find the probability of not selecting a heart.
2. Find the probability of not selecting a king.
3. Find the probability of not selecting a red card.

Answer

1. \( P(\text{not a heart}) = 1 - P(\text{heart}) = 1 - \dfrac{13}{52} = \dfrac{39}{52} = \dfrac{3}{4} \)
2. \( P(\text{not a king}) = 1 - P(\text{king}) = 1 - \dfrac{4}{52} = \dfrac{48}{52} = \dfrac{12}{13} \)
3. \( P(\text{not a red card}) = 1 - P(\text{red}) = 1 - \dfrac{26}{52} = \dfrac{26}{52} = \dfrac{1}{2} \)

 

 

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