Permutations (Without Repitition)

Permutations (Without Repitition)

In the prior lesson, we stated that a permutation is an arrangement of items where the specific sequence is important. In these types of permutations, we assume no repetition. This means once an item is placed in a position, it cannot be selected again. The formula is:

Where:

• P: Represents Permutations (the count of possible arrangements).
• n: The total number of distinct items available to choose from.
• r: The number of items being selected and arranged. 

Note that P(n,r) is also often written as nPr

It is important to 

To understand this formula, we will first look at it's underlying concepts. 

I. Fundamental Counting Principle

The fundamental principle of counting (aka, Rule of Products) is a basic concept used to determine the total number of possible outcomes in a situation where there are multiple independent events. It allows us to count a large number of possibilities without needing to list each one individually.

The fundamental Counting Principle states that if there are a ways (choices) of doing something and b ways (choices) of doing another thing, then there are a x b ways (choices) of performing both actions.

Example: The Ice Cream Shop

You can choose a cone: Waffle or Sugar (2 options).

You can choose a flavor: Vanilla, Chocolate, or Strawberry (3 options).

How many distinct ice cream cones can you make?

2 x 3 = 6 possibilities

II. From Counting to Factorials (n!)

Now let's apply that same counting principle to arranging objects.

Example: Arranging Books
Imagine you have 3 different books (A, B, and C) and you want to put them in order on a shelf. How many different ways can you arrange them?

1. Slot 1: How many choices do you have for the first spot? 3 (A, B, or C).
2. Slot 2: You put one book down. How many are left for the second spot? 2.
3. Slot 3: You put the second book down. How many are left? 1.

According to the Fundamental Counting Principle, we multiply these choices:

3 x 2 x 1 = 6 arrangements

In mathematics, we arrange items so often that we created a shorthand symbol for "multiplying a number by every integer smaller than it down to 1." This is the Factorial (!).

5! = 5 x 4 x 3 x 2 x 1

3! = 3 x 2 x 1

1! = 1

Note the special case of the factorial of 0 = 1

0! = 1 

n! = n x (n-1) x (n-2) x ... x 1

Key Takeaway: Factorials (n!) calculate the number of ways to arrange ALL items in a set.

III. Deriving the Permutations Formula

This is where many students get lost. What if we don't want to arrange all the books? What if we have a big group, but we only want to arrange a few of them?

This is called a Permutation.

• n = Total items available.

• r = Number of items we are actually choosing and arranging.

The Logic (Without the Formula)

Let's use the Fundamental Counting Principle again.

Scenario: Eight runners are in a race (n=8). We need to award Gold, Silver, and Bronze medals (r=3). We are not arranging all 8 runners, just the top 3.

1. Gold Medal: Any of the 8 runners can win.

2. Silver Medal: The winner can't win twice, so 7 runners are left.

3. Bronze Medal: The top two are occupied, so 6 runners are left.

Using the Counting Principle:

8 x 7 x 6 - 366 ways

Finding the Formula 
How do we represent that "partial factorial" mathematically?

We calculated: 8 x 7 x 6.

We know that the full factorial is: 8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1

We want to mathematically "remove" the bold part we didn't use. In multiplication, we "remove" things by dividing.

The part we want to remove (5 x 4 x 3 x 2 x 1) is actually just 5!.

So, we can write our calculation as:

Where did the 5 come from? The 5 is the number of people who lost (did not get a medal).

Losers = Total Runners - Winners 

5 = 8 - 3

So the formula is actually

This leads us directly to the formula for permutations (nPr)

Note:

In practice, when solving these problems you will likely just use a shorthand method for what the formula formally represents. That is, you will:

1. Look at r to find how many numbers will be in your multiplication chain.

2. Then start with n as the first number in your chain.

3. Multiply by decreasing integers (n - 1, n - 2, etc)

4. Stop when you have reached r numbers.

I'll call this the countdown method

V. Solving Permutations

Example:

Solve for P(6, 3)

Step1: We look at r (which here is 3). This means you need to multiply 3 numbers.

Step 2: Start with n (for this problem is 6)

Step 3: Multiply downward until you have 3 numbers in your chain:

6 x 5 x 4 = 120

Now I'll solve using the formula:

P(6, 3) = 6 x 5 x 4 x 3 x 2 x 1/ 3 x 2 x 1

P(6, 3) = 720/6 = 120

Gemini AI

CK-12: Permutations and Combinations Compared

https://www.geeksforgeeks.org/maths/fundamental-principle-of-counting/

https://www.mathsisfun.com/data/basic-counting-principle.html

https://en.wikipedia.org/wiki/Rule_of_product

Permutations and Combinations Compared

In probability and statistics, combinations  and permutations are the two primary concepts to count how many ways a set of items can be selected or arranged. The fundamental difference between them comes down to one thing: order. If the order doesn't matter, it is a combination. If the order does matter, it is a permutation.

I. Intro to Combinations and Permutations
A. Combinations (Order Does Not Matter)
A combination is a selection of items where the sequence is irrelevant. You are only concerned with which items are picked, not how they are arranged.

A simple example would be a fruit salad. A mix of apples, grapes, and bananas is the same salad as bananas, grapes, and apples. The group is the same regardless of what went into the bowl first.

There are basically two types of combinations:

1. Combinations without Repetition

2. Combinations with Repetition

B. Permutations (Order Matters)
A permutation is an arrangement of items where the specific sequence is important. If you change the order of the items, you have a new permutation. 

A simple example would be a door lock code. If the code is 1-2-3, entering 3-2-1 will not open the door. Even though the numbers are the same, the order makes them a distinct "arrangement."

There are basically two types of permutations:

1.Permutations without Repetition

2. Permutations with Repetition

CK-12: Permutations and Combinations Compared

Gemini AI

https://www.mathsisfun.com/combinatorics/combinations-permutations.html

https://virtuallearningacademy.net/LessonDisplay/Lesson6243/MATHALGIIBU33Probability_Combinations.pdf

https://www.geeksforgeeks.org/maths/permutations-and-combinations/

Tree Diagrams

Tree diagrams are another way to show the outcomes of simple probability events. In a tree diagram, each outcome is represented as a branch on a tree.

Using a Tree Diagram to Find Probability

Example 1

If you toss a coin 2 times, what is the probability of getting 2 heads? Use a tree diagram to find your answer.

This is an example of independent events, because the outcome of one event does not affect the outcome of the second event. What does this mean? Well, when you flip the coin once, you have an equal chance of getting a head (H) or a tail (T). On the second flip, you also have an equal chance of getting a a head or a tail. In other words, whether the first flip was heads or tails, the second flip could just as likely be heads as tails. You can represent the outcomes of these events on a tree diagram.

From the tree diagram, you can see that the probability of getting a head on the first flip is 1/2. Starting with heads, the probability of getting a second head will again be  1/2. But how do we calculate the probability of getting 2 heads? These are independent events, since the outcome of tossing the first coin in no way affects the outcome of tossing the second coin. Therefore, we can calculate the probability as follows:

P(A and B) = 1/2 x 1/2 = 1/4

Therefore, we can conclude that the probability of getting 2 heads when tossing a coin twice is , or 25%.

Example 2

Irvin is getting dressed for school. He knows his sock drawer contains 4 red socks, 6 white socks, and 8 brown socks, but they are all loose and unsorted. Because the room is dim, he cannot distinguish the colors as he reaches in.

Irvin pulls out one sock at random, sees that it is red, and decides it doesn't match his blue shorts. He replaces the red sock in the drawer and mixes them up. He then reaches in a second time and pulls out a white sock.

What is the probability of this specific sequence: pulling a red sock, replacing it, and then pulling a white sock?

There are 18 socks in the drawer. The probability of getting a red sock in the first draw is :

P(red) = 4/18 

Irvin puts the sock (replacement) and draws a second sock. The probability of getting a white sock on the second draw is:

P(white) = 6/18

Therefore, the probability of getting a red sock and then a white sock when the first sock is replaced is:

P(red and white) = 4/18 x 6/18 = 24/324 = 2/27

Note that since the first sock was replaced, each draw of a sock is independent of each other

Example 3 

In the previous example, what happens if the first sock is not replaced? Since the sock is not replaced, we say the second draw is dependent on the first draw. 

The probability of the first stock being red is unchanged:

P(red) = 4/18

But since the sock from the first draw was not replaced, there are only 17 socks left in the drawer. So the probability of picking a white sock on the second pick is now:

P(white|red) = 6/17

Now the probability of selecting a red sock and then a white sock, without replacement, is:

P(red and white) = 4/18 x 6/17 = 24/306 = 4/51

CK-12: Tree Diagrams

LibreTexts: Probability Using Tree Diagrams and Combinations

https://mathleaks.com/study/kb/concept/tree_diagram

The Terminology of Probability

I. Experiment, Sample Space, Events

Probability is a measure that is associated with how certain we are of outcomes of a particular experiment or activity. 

An experiment is a planned operation or procedure that can be infinitely repeated and has a well defined set of possible outcomes. If the result is not predetermined, then the experiment is said to be a random experiment.  An outcome is the specific result of a single execution of the experiment model. 

The sample space of an experiment  is the set of all possible outcomes. Four ways to represent a sample space are: to list the possible outcomes, to create a tree diagram, a sample space diagram or to create a Venn diagram. The uppercase letter S is used to denote the sample space. For example, if you flip one fair coin, S = {H, T} where H = heads and T = tails are the outcomes.

An event is any combination of outcomes. Upper case letters like A and B represent events. For example, if the experiment is to flip one fair coin, event A might be getting at most one head. The probability of an event A is written P(A).

The probability of any outcome is the long-term relative frequency of that outcome. Probabilities are between zero and one, inclusive (that is, zero and one and all numbers between these values). P(A) = 0 means the event A can never happen. P(A) = 1 means the event A always happens. P(A) = 0.5 means the event A is equally likely to occur or not to occur. For example, if you flip one fair coin repeatedly (from 20 to 2,000 to 20,000 times) the relative frequency of heads approaches 0.5 (the probability of heads).

Equally likely means that each outcome of an experiment occurs with equal probability. For example, if you toss a fair, six-sided die, each face (1, 2, 3, 4, 5, or 6) is as likely to occur as any other face. If you toss a fair coin, a Head (H) and a Tail (T) are equally likely to occur. If you randomly guess the answer to a true/false question on an exam, you are equally likely to select a correct answer or an incorrect answer.

Calculating Probability with Equally Likely Outcomes
To calculate the probability of an event A when all outcomes in the sample space are equally likely, count the number of outcomes for event A and divide by the total number of outcomes in the sample space. 

For example, if you toss a fair dime and a fair nickel, the sample space is {HH, TH, HT, TT} where T = tails and H = heads. The sample space has four outcomes. Event A = getting one head. There are two outcomes that meet this condition {HT, TH}, so P(A) = 2/4 = 0.5.

Suppose you roll one fair six-sided die, with the numbers {1, 2, 3, 4, 5, 6} on its faces. Let event E = rolling a number that is at least five. There are two outcomes {5, 6}. P(E) = 2/6 as the number of repetitions grows larger and larger.

This important characteristic of probability experiments is known as the law of large numbers which states that as the number of repetitions of an experiment is increased, the relative frequency obtained in the experiment tends to become closer and closer to the theoretical probability. Even though the outcomes do not happen according to any set pattern or order, overall, the long-term observed relative frequency will approach the theoretical probability. (The word empirical is often used instead of the word observed.)

“OR” Event/Union
An outcome is in the event A OR B if the outcome is in A or is in B or is in both A and B. For example, let A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7, 8}. A OR B = {1, 2, 3, 4, 5, 6, 7, 8}. Notice that 4 and 5 are NOT listed twice. The set notation ∪ for union can also be used. So “A or B” and “A ∪ B” mean the same thing.

"AND" Event/Intersection

An outcome is in the event A AND B if the outcome is in both A and B at the same time. For example, let A and B be {1, 2, 3, 4, 5} and {4, 5, 6, 7, 8}, respectively. Then A AND B = {4, 5}. The set notation ∩ for intersection can also be used. So "A and B" and "A∩B" mean the same thing. 

Complement
The complement of event A is denoted A' (read "A prime"). A' consists of all outcomes that are NOT in A. Notice that P(A) + P(A') = 1.

For example, let S = {1, 2, 3, 4, 5, 6} and let A = {1, 2, 3, 4}. Then, A' = {5, 6}. P(A) =4/6 and P(A') = 2/6, and P(A) + P(A') = 4/6 + 2/6 = 1.

Conditional Probability
The conditional probability of A given B is written P(A|B).

P(A|B) is the probability that event A will occur given that the event B has already occurred. A conditional reduces the sample space. We calculate the probability of A from the reduced sample space B.

The formula to calculate P(A|B) is:

P(A|B) = P(A and B) / P(B)

where P(B) is greater than zero.

Notes:

Random experiments are often conducted repeatedly, so that the collective results may be subjected to statistical analysis. A fixed number of repetitions of the same experiment can be thought of as a composed experiment, in which case the individual repetitions are called trials.

For example, if one were to toss the same coin one hundred times and record each result, each toss would be considered a trial within the experiment composed of all hundred tosses. The outcome is the result of a single flip. 

Lumen: The Terminology of Probability

Basic Concepts of Probability

Gemini Ai

Experiment (probability theory) - Wikipedia

Defining Probability & Statistics

Probability is the branch of mathematics dealing with numerical descriptions of how likely an event is to occur. It is the science of predicting future outcomes based on a known model or set of rules.

Statistics is the discipline that concerns the collection, organization, analysis, interpretation, and presentation of data. It is the science of inferring the rules of the world based on observed data.

Statistics is generally divided into two main categories:

1. Descriptive Statistics: Summarizes data using indexes such as mean (average) or standard deviation. (e.g., "The average height in this room is 5'9".")
2. Inferential Statistics: Uses data from a sample to make conclusions about a larger population. (e.g., "Based on this survey of 1,000 people, we estimate 60% of the country supports this law.")

Mutually Exclusive & Mutually Inclusive Events

Mutually Exclusive

Events A and B are mutually exclusive events if they cannot occur at the same time. This means that A and B do not share any outcomes and P(A and B) = 0 aka P(A∩B) = 0.

Notice in the diagram below that there is no intersection between the possible outcomes of event A and the possible outcomes of event B. For example, if you were asked to pick a number between 1 and 10, you cannot pick a number that is both even and odd. These events are mutually exclusive.

Mutually Inclusive

"Mutually inclusive" (often just called inclusive events) is a term used to describe two events that can happen at the same time. If Event A and Event B are mutually inclusive, their "intersection" is not zero (P(A∩B) ≠ 0.

Notice in the diagram below that there is some overlap with events A and B indicating they can occur at the same time. For example, if a random number was chosen from 1 to 10 and event A is picking a number less than 4 and event B is picking an even number then the two events are inclusive since 2 is a number which both events share. 

Example 1

Suppose the sample space S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Let A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8}, and C = {7, 9}. 

A and B = {4, 5}. P(A and B) = 2/10 and is not equal to zero. Therefore, A and B are not mutually exclusive. 

A and C do not have any numbers in common so P(A and C) = 0. Therefore, A and C are mutually exclusive.

Calculating Probability

Probability of Mutually Exclusive Events

Example 1

Imagine rolling a standard six-sided die. What is the probability of rolling a 5 or a 6?

S = {1, 2, 3, 4, 5, 6}

A = {5}

B = {6}

P(A) = 1/6

P(B) = 1/6

P(A or B) = P(A∪B) = 1/6 + 1/6 = 2/6 = 1/3

Here, the favorable outcomes are a 5 or a 6. Since we are trying to find the probability of either of them happening, we simply add their individual probabilities. We can do this since they are mutually exclusive events. That is, neither of them can happen at the same time. 

Probability of Mutually Inclusive Events

Example 2

What is the probability of choosing a number from 1 to 10 that is less than 5 or odd?

S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = {1, 2, 3, 4}
B = {1, 3, 5, 7, 9}

P(A) = 4/10 = 2/5

P(B) = 5/10 = 1/2

P(A and B ) = 2/10 = 1/5

P(A or B) = P(A) + P(B) - P(A and B)

P(A or B) = 2/5 + 1/2 - 1/5

P(A or B) = 4/10 + 5/10 - 2/10

P(A or B) = 7/10

When we calculated the probability of the mutually exclusive events in example 1, you see that we simply added their probabilities of event A and event B together. With example 2, we have to do a little more work and also subtract out the probability of A and B. Why? Because with mutually inclusive events, events A and B share some overlap (See A and B Venn diagram above). Namely, 1 and 3 appear in both sets. If we simply added the probabilities of A and B together (4/10 + 5/10), we would be counting the numbers 1 and 3 twice—once because they are less than 5, and again because they are odd. This is called double counting.

To fix this, we must subtract the probability of the events that they share (the intersection) exactly once. This removes the extra count and gives us the correct probability.

Addition Rule

P(A or B) = P(A) + P(B) - P(A and B)

stated using set notation

P(A∪B) = P(A) + P(B) - P(A∩B)

Example 3

What is the probability of choosing a card from a deck of cards that is a club or a ten? Let A be the event of selecting a club and B be the event of selecting a ten.

P(A) = 13/52

P(B) = 4/52

P(A and B) = 1/52

P(A or B) = P(A) + P(B) - P(A and B)

P(A or B) = 13/52 + 4/52 - 1/52

P(A or B ) = 16/52 = 4/13

Example 4

Two dice are rolled, and the events F and T are as follows:
F = {The sum of the dice is four} and T = {At least one die shows a three}

Find P(F∪T).

The sample space for this problem is 6 x 6 = 36

P(F) = 3/36

P(T) = 11/36

F∩T is {(1,3), (3,1)} so the P(F∩T is 2/36

P(F∪T) = P(F) + P(T) - P(F∩T)

P(F∪T) = 3/36 + 11/36 - 2/36 = 12/36 = 1/3

8.2: Mutually Exclusive Events and the Addition Rule - Mathematics LibreTexts

K-12: Mutually Inclusive Events

K-12: Mutually Exclusive Events

https://courses.lumenlearning.com/introstats1/chapter/independent-and-mutually-exclusive-events/

Dependent Events and Sample Space

I. Intro 

Dependent events are affected by the outcome of preceding events. In other words, two events A and B, are dependent if the occurrence of A affects the probability of B. 

I. Dependent Events

A. Using the Multiplication Rule for Dependent Events

Example 1

What's the probability of drawing two sevens from a standard deck of cards if once 1 card is chosen it is not replaced?

This, of course, is a dependent event since the occurrence of A affects the probability of B. 

Let A = 1st seven chosen

Let B = 2nd seven chosen

P(A and B) = P(A∩B) = P(A) x P(B|A)

The notation with the vertical bar is referred to as a conditional probability. The vertical bar in P(B|A) means "given," so this is read as "the probability that B occurs given that A has occurred."

P(A) = 4/52 (4 sevens in deck of 52 cards)

P(B) = 3/51 (3 remaining sevens in deck with now 51 cards)

P(A∩B) = 4/52 x 3/51 =12/2652 = 1/221

The probability of drawing two sevens consecutively from a standard deck without replacement is 1/221, or approximately 0.45%.

 

Example 2

A bag contains 3 green marbles and 2 red marbles. A marble is drawn, not replaced, and then a second marble is drawn. What is the probability of drawing a green marble followed by drawing a red marble?

Let A = Drawing a green marble on first draw

Let B = Drawing a red marble on second draw

P(A) = 3/5 (3 green marbles; 5 total marbles)

P(B) = 2/4 (2 red marbles; (2 red marbles; 4 total marbles)

P(A) x P(B|A) = 3/5 x 2/4 = 6/20 = 3/10

The probability of drawing a green marble followed by a red marble from a bag containing 3 green and 2 red marbles, without replacement, is 3/10 (or 30%).

B. Conceptual Understanding

Sometimes it can be difficult to figure the correct probability for each event. Using a probability tree diagram can help us to better conceptually understand what is happening. 

Here is the diagram for Example 1

The first red branch represents P(A) of 4/52. Now notice the second red branch which is P(B|A) or stated another way, the probability of B given that A had happened. Since we are stating that A (a seven was drawn) has happened, there are now only 3 sevens left in the deck and only 51 cards remain in total. 

Next is the tree from Example 2

The first red branch represents P(A) which is 3/5. The second branch is P(B|A) stated as the probability of B given that A has occurred. We assume that event A has happened, so there are now only 4 total marbles left to draw but still 2 red marbles. So the P

CK-12: Dependent Events and Sample Spaces

Khan: The general multiplication rule

Mathbits: Independent and Dependent Events

LibreTexts: Probability Using Tree Diagrams and Combinations

https://mathspace.co/textbooks/syllabuses/Syllabus-1272/topics/Topic-23586/subtopics/Subtopic-294799/?coreTextbookSubtopicActiveTab=solidifyLesson&activeLessonTab=content

Independent Events and the Multiplication Rule

I. Intro to Independent & Dependent Events

Independent Events
In statistics, the term independent event means to have one event not dependent on the other. In other words, two events A and B, are independent if the occurrence of one does not affect the probability of the other. 

Dependent Events

Events A and B are dependent if the occurrence of Event A changes the probability of Event B.

Example

Imagine you have a standard deck of 52 cards. You want to draw a King (Event A), and then draw another King (Event B)

1. Independent (With Replacement)

• Action: You pick a card, look at it, and put it back in the deck before picking the next one.
• Event A: You draw a King. Probability is 4 in 52.
• The Reset: You put the King back and shuffle. The deck is back to 100% normal.
• Event B: You draw again. The probability of getting a King is still 4 in 52.
• Result: The first event didn't change the deck, so the probability stayed the same. Independent.

2. Dependent (Without Replacement)

• Action: You pick a card, look at it, and keep it in your pocket.
• Event A: You draw a King. Probability is 4 in 52.
• The Change: You keep the card. Now there are only 51 cards left in the deck, and only 3 Kings left.
• Event B: You draw again. The probability is now 3 in 51.
• Result: Because you didn't put the first card back, the odds for the second draw changed. Dependent.

II. Independent Events

A. Using the Multiplication Rule for Independent Events

In ABC High School, 30 percent of the students have a part-time job, and 25 percent of the students from the high school are on the honor roll. Event A represents randomly choosing a student holding a part-time job. Event B represents randomly choosing a student on the honor roll. What is the probability of both events occurring?

Event A is randomly choosing a student holding a part-time job, and event B  is randomly choosing a student on the honor roll. These 2 events are independent of each other. In other words, whether you hold a part-time job is not dependent on your being on the honor roll, or vice versa. The outcome of one event is not dependent on the outcome of the second event. To calculate the probability, you would look at the overlapping part of the Venn diagram. The region representing A and B is the probability of both events occurring. Let’s look at the probability calculation, which is done with the Multiplication Rule:

In other words, under independence, the probability of randomly selecting a student who both has a part-time job and is on the honor roll is 7.5%.

Multiplication Rule For Independent Events

If A and B are independent events, the probability of both events occurring is the product of the probabilities of the individual events.

P(A∩B) = P(A and B) = P(A) • P(B)

Example 1

2 coins are tossed one after the other. Event A consists of the outcomes when tossing heads on the first toss. Event B consists of the outcomes when tossing heads on the second toss. What is the probability of both events occurring?

Event A consists of the outcomes when getting heads on the first toss, and event B consists of the outcomes when getting heads on the second toss. What would be the probability of tossing the coins and getting a head on both the first coin and the second coin? We know that the probability of getting a head on a coin toss is 1/2, or 50%. In other words, we have a 50% chance of getting a head on a toss of a fair coin and a 50% chance of getting a tail.

Therefore, there is a 25% chance of getting 2 heads when tossing 2 fair coins.

Example 2

The following table represents data collected from a grade 12 class in DEF High School.

Suppose 1 student was chosen at random from the grade 12 class.

(a) What is the probability that the student is female?

(b) What is the probability that the student is going to university?

Now suppose 2 people both randomly chose 1 student from the grade 12 class. Assume that it's possible for them to choose the same student.

(c) What is the probability that the first person chooses a student who is female and the second person chooses a student who is going to university?

Answers

(a)  P(female) = 80/164 = 0.4878 or 48.78% chance the student is female

(b) P(going to university) = 71/164 = 0.4329 or 43.29% chance the student is going to university

(c) P(female) x P(going to university)

     48.78% chance female x 43.29% chance going to university = 21.11%
    

Therefore there is a 21.11% probability that the first person chooses a student who is female and the second person chooses a student who is going to university.

Example 3

2 cards are chosen from a deck of cards. The first card is replaced before choosing the second card. What is the probability that they both will be sevens?

Let A = 1st seven chosen

Let B = 2nd seven chosen

A deck of cards consists of 52 cards. Each deck has 4 suits (with each suit having13 cards). Each suit has one seven, therefore there are 4 sevens in each deck of 52 cards.

Since the card is replaced after the first selection, these events are independent.

P(A) = 4/52

P(B) = 4/52

P(A and B) = 4/52 x 4/52 or P(A∩B) = 4/52 x 4/52

P(A∩B) = 16/2704 = 1/169

B. Conceptual Understanding

One way to help conceptually understand the multiplication rule is with the use of a tree diagram.

Let's go back to example 1. What is the probability of flipping a fair coin and getting "heads" twice in a row? That is, what is the probability of getting heads on the first flip AND heads on the second flip?

Imagine we had100 people simulate this and flip a coin twice. On average, 50 people would get heads on the first flip, and then 25 of them would get heads again. So 25 out of the original 100 people — or 1/4 of them — would get heads twice in a row.

The number of people we start with doesn't really matter. Theoretically, 1/2 of the original group will get heads, and 1/2 of that group will get heads again. To find a fraction of a fraction, we multiply.

We can represent this concept with a tree diagram like the one shown below.

We multiply the probabilities along the branches to find the overall probability of one event AND the next even occurring.

Khan: The general multiplication rule

Mathbits: Independent and Dependent Events

Venn Diagrams

I. Intro

In probability, a Venn diagram is a graphic organizer that shows a visual representation for all possible outcomes of an experiment and the events of the experiment in ovals. Normally, in probability, the Venn diagram will be a box with overlapping ovals inside. Look at the diagram below:

Sample Space (S)

The Sample Space (𝑆) represents all of the possible outcomes of a random experiment. 

Events 

The ovals (in this example 𝐴 and 𝐵) are referred to as events. An event is a subset of the sample space, representing one or more possible outcomes of a random experiment. 

Outcome

A single result of an experiment.

Example

Two coins are tossed one after the other. Event A consists of the outcomes when tossing heads on the first toss. Event B consists of the outcomes when tossing heads on the second toss. Draw a Venn diagram to represent this example.

We know that:

Notice that event A  and event B  share the Heads + Heads outcome and that the sample space contains Tails + Tails, which is neither in event nor event.


II. Intersection & Union

Event A represents randomly choosing a student from High School who holds a part-time job. Event B represents randomly choosing a student from High School who is on the honor roll. Draw a Venn diagram to represent this example.

We know that:

Notice that the overlapping oval for A  and B  represents the students who have a part-time job and are on the honor roll. The sample space, S , outside the ovals represents students neither holding a part-time job nor on the honor roll.

In a Venn diagram, when events A and B occur, the symbol used is ∩. Therefore, A∩B  is the intersection of events A and B and can be used to find the probability of both events occurring. In this example the intersection of A & B are the students both holding a part-time job and that are on the honor roll.

If, in a Venn diagram, either A or B  occurs, the symbol is ∪. This symbol would represent the union of events A and B , where the outcome would be in either A or B. In this example, the union of A & B would be the students that are holding a part-time job or that are on the honor roll. 

Intersection

The intersection of events A and B, denoted A ∩ B, includes only the outcomes common to both A and B, representing "A and B" occurring together.

Union
The union of two events A and B, denoted A ∪ B, consists of all outcomes that belong to A, or B, or both.

Example

You are asked to roll a die. Event A  is the event of rolling a 1, 2, or a 3. Event B is the event of rolling a 3, 4, or a 5. Draw a Venn diagram to represent this example. What is A∩B? What is A∪B?

https://flexbooks.ck12.org/cbook/ck-12-basic-probability-and-statistics-concepts/section/1.1/primary/lesson/venn-diagrams-bsc-pst/

https://www.siyavula.com/read/za/mathematics/grade-11/probability/10-probability-01

Definition of Basic Geometry & Measurement

Basic Geometry serves as the introduction to understanding physical space and visual forms. Taught throughout elementary and middle school, it focuses on observation and measurement rather than formal mathematical proof.

The subject first establishes a core vocabulary, teaching students to identify, name, and classify elements such as points, lines, angles, 2D polygons, and 3D solids. It then connects these visual concepts to arithmetic through the study of Measurement. This begins with spatial formulas for perimeter, area, and volume, but extends to other practical systems of quantification—specifically time and currency. By grouping these topics, the curriculum demonstrates that the arithmetic skills used to measure physical space apply equally to measuring value and duration.

Basic Definition of Pre-Algebra

Pre-algebra serves as the transitional math course between basic arithmetic and full algebra. It is typically taught as part of the middle school curriculum in the United States. 

Pre-algebra first expands knowledge of how numbers are expressed—through tools like exponents, roots, scientific notation—and how they relate to one another via ratios, rates, proportions, and percents. It then introduces core algebraic concepts, such as variables as placeholders for unknowns, basic expressions, and one-step equations, shifting from concrete arithmetic to abstract problem-solving. This two-phase structure solidifies arithmetic fluency while building the symbolic thinking essential for Algebra 1, ensuring students grasp both numerical representation and relational balance before tackling multi-step equations.

Basic Definition of Arithmetic

Arithmetic is an elementary branch of mathematics that deals with fundamental numerical operations of addition, subtraction, multiplication, and division. In a wider sense, it also includes exponentiation, extraction of roots, and taking logarithms (This basic definition is the one these notes will adhere to).

Wikipedia: Arithmetic