# Calculating Theoretical Probability

### Probability Models

A probability model is used to determine the probability of each outcome in a sample space.
A probability model is a list of all possible outcomes of an experiment with their corresponding probabilities. The model can then be used to find probabilities of different events.
Step-By-Step Example
Creating a Probability Model
Create a probability model for the sum when rolling two dice.
Step 1
Create a sample space for rolling two dice. There are 36 different outcomes, each with an equally likely chance of being rolled.
Step 2
Determine the possible sums when two dice are rolled, and record their frequencies. The sum may be a whole number from 2 to 12.
 Sum $2$ $3$ $4$ $5$ $6$ $7$ Frequency $1$ $2$ $3$ $4$ $5$ $6$
 Sum $8$ $9$ $10$ $11$ $12$ Frequency $5$ $4$ $3$ $2$ $1$
Solution
Write the ratio for each frequency to the number of equally likely outcomes, 36.
 Sum $2$ $3$ $4$ $5$ $6$ $7$ Frequency $1$ $2$ $3$ $4$ $5$ $6$ $P(\text{sum)}$ $\frac{1}{36}$ $\frac{2}{36}=\frac{1}{18}$ $\frac{3}{36}=\frac{1}{12}$ $\frac{4}{36}=\frac{1}{9}$ $\frac{5}{36}$ $\frac{6}{36}=\frac{1}{6}$
 Sum $8$ $9$ $10$ $11$ $12$ Frequency $5$ $4$ $3$ $2$ $1$ $P(\text{sum)}$ $\frac{5}{36}$ $\frac{4}{36}=\frac{1}{9}$ $\frac{3}{36}=\frac{1}{12}$ $\frac{2}{36}=\frac{1}{18}$ $\frac{1}{36}$

### Independent and Dependent Events

Determining whether an event is independent or dependent relies on whether the probability of one event affects the probability of the other event.

Two events are independent events if the result of one event does not affect the probabilities of the other event. Rolling a die and flipping a coin are independent events, since the result of one does not affect the result of the other. When a coin is flipped twice, the two events are independent since the probabilities for the second flip are not affected by the result of the first flip.

Two events are dependent events if the result of one event affects the probabilities of another event. The events of choosing a marble from a bag and choosing a second marble without replacing the first are dependent because the first marble chosen changes the number and kind of marbles remaining in the bag. The probabilities for the second marble chosen are affected by choosing the first marble.

Note that if two marbles are chosen from a bag and the first marble is replaced after being chosen, then the events of choosing two marbles from a bag are independent since the probabilities for the second marble are not affected by the result of choosing the first marble. Look for phrases like "with replacement" or "without replacement" when determining whether events are independent or dependent.

If events $A$ and $B$ are independent, then the probability of both events occurring is given by the formula:
$P(A\text{ and }B)=P(A) \cdot P(B)$
Step-By-Step Example
Calculating the Probability of Independent Events
When selecting two numbers between one and 20 using a random number generator, what is the probability that the first number is a prime number and the second number is a multiple of three?
Step 1

Determine if the events are independent or dependent.

A random number generator will select each number at random, and selecting the first does not affect selecting the second. So, these are independent events and this formula can be used:
$P(A\text{ and }B)=P(A) \cdot P(B)$
Step 2

Identify the number of ways each event can occur and the number of equally likely outcomes.

There are 20 numbers that may be chosen in each event.

Let $A$ be the event of selecting a prime number. There are eight prime numbers in the sample space of numbers between one and 20: 2, 3, 5, 7, 11, 13, 17, and 19.
$P(A)= \frac{8}{20}=\frac{2}{5}$
Let $B$ be the event of selecting a multiple of three. There are six multiples of three in the sample space: 3, 6, 9, 12, 15, and 18.
$P(B)= \frac{6}{20}=\frac{3}{10}$
Solution
Calculate the probability.
\begin{aligned}P(A\text{ and }B)&= P(A) \cdot P(B)\\&= \frac{2}{5} \cdot \frac{3}{10}\\&= \frac {6}{50}\\&= \frac {3}{25}\end{aligned}
The probability of randomly selecting two numbers between one and 20, where the first number is prime and the second number is a multiple of three, is $\frac{3}{25}$.
If events $A$ and $B$ are dependent, then the probability of both events occurring is given by the formula:
$P(A\text{ and }B)=P(A) \cdot P(B \text{ after } A)$
Step-By-Step Example
Determining the Probability of Dependent Events
There are two final contestants in a design contest. In a bag, there are three tokens labeled spring collection and three tokens labeled fall collection. A random drawing will determine whether each contestant will design clothes for the spring collection or the fall collection. During the random drawing, one designer will choose a token from the bag, and without replacing it, the second designer will choose a token. What is the probability that both designers will choose a token labeled spring collection?
Step 1

Determine if the events are independent or dependent.

When the first designer chooses a token, there are six tokens in the bag. The token that the first designer chooses will not be replaced, so when the second designer chooses a token, there will only be five tokens in the bag.

The first event affects the probabilities of the second event. So, both events are dependent events. The formula for calculating the probability of dependent events can be used:
$P(A\text{ and }B)=P(A) \cdot P(B \text{ after } A)$
Step 2

Calculate the number of ways each event can occur and the number of equally likely outcomes for each event.

Let $A$ be the event of the first designer choosing a token labeled spring collection. When the first designer chooses a style, there are three tokens labeled with spring collection in a bag with six tokens.
$P(A)=\frac{3}{6}=\frac{1}{2}$
Let $B$ be the event of the second designer choosing a token labeled spring collection. The first designer does not replace their token. So, when the second designer chooses a style, there are two tokens labeled with spring collection in a bag with five tokens.
$P(B \text{ after } A)=\frac{2}{5}$
Solution
Calculate the probability.
\begin{aligned}P(A\text{ and }B)&=P(A) \cdot P(B \text{ after } A)\\&=\frac{1}{2} \cdot \frac{2}{5}\\&=\frac{2}{10}\\&=\frac{1}{5}\end{aligned}
The probability that both designers will choose a token labeled spring collection is $\frac{1}{5}$.

### Mutually Exclusive Events

When two events are mutually exclusive, the probability that either event will occur is the sum of the probabilities of each event.

Two events in an experiment that cannot occur at the same time are mutually exclusive events. For example, when flipping a coin, the outcome is either heads or tails—it cannot be both. When flipping a coin four times, the events of flipping a head exactly three times and flipping a tail exactly four times are mutually exclusive since they cannot happen at the same time.

If two events $A$ and $B$ are mutually exclusive, then:
$P(A \text{ or } B) = P(A) + P(B)$
Step-By-Step Example
Determining the Probability of Mutually Exclusive Events
An experiment consists of rolling two fair dice and adding the numbers that are rolled. Calculate the probability that the sum will be a prime number or a multiple of four.
Step 1
The event of the sum being a prime number and the event of the sum being a multiple of four cannot occur at the same time. So, the events are mutually exclusive and the formula for mutually exclusive events can be used:
$P(A \text{ or } B) = P(A) + P(B)$
Step 2
Use a table to display the possible sums of the numbers showing when two dice are rolled. Count the number of ways the sum can be a prime number or a multiple of four. Then, find the number of outcomes in the sample space.
Let $A$ be the event of the sum being a prime number. The numbers 2, 3, 5, 7, and 11 are prime. There are 15 ways out of 36 outcomes to roll a prime number.
$P(A) = \frac{15}{36}$
Let $B$ be the event of the sum being a multiple of four. The numbers 4, 8, and 12 are multiples of four. There are nine ways out of 36 outcomes to roll a multiple of four.
$P(B) = \frac{9}{36}$
Solution
The events are mutually exclusive. So add the probabilities.
\begin{aligned}P(A \text{ or } B) &= P(A) + P(B)\\&= \frac{15}{36} +\frac{9}{36}\\&=\frac{24}{36}\\&=\frac{2}{3}\end{aligned}
When rolling two fair dice, the probability of rolling a prime number or a multiple of four is $\frac{2}{3}$.

### Mutually Inclusive Events

When two events are mutually inclusive, the probability that either event will occur is the sum of the probabilities of each event minus the probability that both occur together.

Two events in an experiment that may occur at the same time are mutually inclusive events. For example, when rolling a die, the event of rolling a multiple of two and the event of rolling a multiple of three can occur at the same time when the outcome of the roll is six.

Consider the sample spaces for each event in rolling a multiple of two or a multiple of three. The sample space for the event of rolling a multiple of two is $\{2, 4, 6\}$. The sample space for the event of rolling a multiple of three is $\{3, 6\}$. If each outcome in the two sample spaces is counted, the outcome of rolling a 6 would be counted twice. To calculate the probability, the outcomes that occur in both lists must be removed so that they are not counted more than once.

If two events $A$ and $B$ are mutually inclusive, then:
$P(A\text{ or }B)=P(A) + P(B) - P(A \text{ and } B)$
Step-By-Step Example
Calculating the Probability of Mutually Inclusive Events
In an online survey for a pet food company, 100 customers indicated whether they have a cat or dog. Each person who answered the survey will receive a coupon from a manufacturer that makes cat food and dog food. The survey results show that 30 people have a cat, 40 people have a dog, and 10 people have both a cat and dog. What is the probability that a person chosen at random has a cat or dog?
Step 1
The event of having a cat and the event of having a dog can occur at the same time, so the events are mutually inclusive and this formula can be used:
$P(A\text{ or }B)=P(A) + P(B) - P(A \text{ and } B)$
Step 2

Calculate the number of ways each event can occur and the number of equally likely outcomes for each event. Calculate the number of outcomes that occur for both events.

Let $A$ be the event of having a cat. There are 30 people out of 100 who have a cat.
$P(A)=\frac{30}{100}=\frac{3}{10}$
Let $B$ be the event of having a dog. There are 40 people out of 100 who have a dog.
$P(B)=\frac{40}{100}=\frac{4}{10}$
There are 10 people out of 100 who have both a cat and dog. They should not be counted twice.
$P(A \text{ and } B)=\frac{10}{100}=\frac{1}{10}$
Solution
The events are mutually inclusive. So, add the probabilities and subtract the outcomes that occur for both events.
\begin{aligned}P(A\text{ or }B)&=P(A) + P(B) - P(A \text{ and } B)\\&=\frac{3}{10}+\frac{4}{10}-\frac{1}{10}\\&=\frac{6}{10}\\&=\frac{3}{5}\end{aligned}
The probability that a person chosen at random has a cat or dog is $\frac{3}{5}$.

### Types of Probability Situations

A variety of methods can be used to calculate probabilities of events.

When calculating the probability of two events, first identify how the events are related. Ask questions like:

• Does the experiment or situation involve performing two actions?
• If the results of the first action do not affect what can happen in the second action, then the events are independent: $P(A\text{ and }B)=P(A) \cdot P(B)$.
• If the results of the first action affect what can happen in the second action, then the events are dependent: $P(A\text{ and }B)=P(A) \cdot P(B \text{ after } A)$.
• Does the experiment or situation involve performing one action for which the outcome may have more than one characteristic?
• If the characteristics cannot occur at the same time, then the events are mutually exclusive: $P(A \text{ or } B) = P(A) + P(B)$.
• If the characteristics may occur at the same time, then the events are mutually inclusive: $P(A\text{ or }B)=P(A) + P(B) - P(A \text{ and } B)$.

For example, in a game that uses lettered tiles:

• The probability of drawing two tiles with vowels if the player draws a tile, puts it back, and then draws a second tile involves independent events.
• The probability of drawing two tiles with vowels if the player draws a tile, keeps it, and then draws a second tile involves dependent events.
• The probability of drawing a tile that is a blank tile or a vowel involves mutually exclusive events.
• The probability of drawing a tile that is in the word "vowel" or in the word "consonant" involves mutually inclusive events.