MDM4U1@FMG - Recently Updated Pages

6.1 Basic Probability Concepts

rajdua — Mon, 28 Jan 2008 07:31:11 CST

Probability: The relative possibility that an event will occur, as expressed by the ratio of the number of actual occurrences to the total number of possible occurrences.

It attempts to predict answers such as:

What are the chances of you rolling 7 or 11 in a row?
You may also be able to predict how likely something is to happen, for example: if the weather report forecast a 90% chance of rain, there is still a slight possibility that sunny skies will prevail. While there are no sure answers, in this case it probably will rain.

The Probability Scale
We can look at probability on a scale from 0 - 1.
0 - event never will occur
.5- the event and "not event" are likely to occur
1- always will occur

We are able to calculate probability in two different situations:

The first being Experimental. Experimental probability is when data is collected and the probabilities are calculated. For example, Xhovana and Nina are playing Tic Tac Toe, they play 10 games and it was observed that Nina won 7 out of the 10 games, so the probability of nina winning is 7 out of 10 or 70%.

The second situation of probability is Theoretical. Theoretical probability are when results are predicted using counting principles. For example, if Xhovana goes skiing and goes off a difficult jump 10 times you would expect half of the jumps for her to fall andGre the other half for her to land, so the probability of her to fall is 5 out of 10 or 50%.

TERMS:

Statistical Fluctuation: If the number of repetitions is small, the two probabilites can be different.

Experiment: Probability problems involving repetition i.e. rolling a pair of die or flipping a coin

Trial: A step in a probability experiment in which an outcome is produced and tallied (each repetition is called a trial)

Event: Grouped outcomes with something in common

Probability of an Event A, P(A): A quantified measure od the likelihood that event A will occur. The probability of an event is always a value between 0 and 1

Sample Space (S): The set of all possible outcomes in a probability experiment
i.e. If Player A draws a 1 and Player B draws 2, you can label this outcome (1,2) in this particular game the result is the same for the outcomes (1,2) and (2,1) but with different rules it might be important who draws which number. So, it makes sense to view the two outcomes differently.

* Remember that the probability of P(A) is between 0 and 1

The probability of an event A, P(A) is:

P(A)= n(A)

n(S)
n(A)= # of ways event A can occur n(S)= total # of possible outcomes P(A')= probability of A not happening (compliment of A) P(A')= 1-P(A)

There are three basic types of probabilty, empirical, theoretical and subjective.

Empirical: The number of times that an event occurs in an experiment divided by the number of trials. The empirical probability is also known as the experimental or relative-frequency probability i.e if you had found that in the first ten trials, the product was greater than the sum four times, then the empirical probability of this event would be:
P(A)= 4

10

= 0.4

Theoretical: The probability of an event deduced from analysis of the possible outcomes. Theoretical probability is also called classical or a priori probability i.e What is the probability of flipping a coin 20 times and getting heads?
As we all know their are two sides to a coin (a head and a tail) we are expected to get heads 10 times if we flip the coin 20 times. This would be half or 50% probability. Since for each time we flip the coin once thier is 1/2 chance of getting heads.

Subjective: An estimate of the likelihood of an event based on intuition and experience (an educated guess) i.e estimate the probability that
a) the next pair of shoes you buy will be the same size as the last pair you bought
Solution: there is a small chance that the size of your feet has changed significantly or that different styles of shoes may fit you differently, so 80-90% would be a reasonable subjective probability that your next pair of shoes will be the same as your last pair.

The below PowerPoint Wiki contains examples that involve basic probability concepts

5.2 Combinations...

rajdua — Wed, 24 Oct 2007 21:05:11 CDT

5.2 Combinations

A combination is a selection of objects in which order is not important. The number of combinations of objects chosen from a set of objects is:

In general, the sum of all the combinations of n distinct things is 2n.

Example: You have 5 shirts, but you can only select 3 for your vacation. In how many different ways can you do this?

=10, the order in which you select the shirts, does not matter.

Example: You are going to draw 7 cards from a standard deck of 52 cards. How many different 7 card hands are possible? This is a combination problem, because a hand is a group of cards where order does not matter. First we need to find n and r:

n= 52, there are 52 cards in a deck of cards.
r = 7, the number of cards that need to be drawn
nCr = n! / (n-r)!r!

52C7 = 52! / (52-7)!7!

= 52! / (45! 7!)

= (52 * 51 * 50 * 49 * 48 * 47 *46 * 45! ) / (45! * 7!)

= (52 * 51 * 50 * 49 * 48 * 47 * 46) / (7 * 6 * 5 * 4 * 3 * 2 * 1)

= 133784560

Example: There are 12 boys and 14 girls in Mrs. Smith’s math class. Find the number of ways Mrs. Smith can select a team of 3 students from the class to work on a group project.
The team consists of 1 girl and 2 boys.

Boys: Girls:

References

Canton, Barbara J., Erdman, Wayne, Irvine, Jeff, Lim, Louis, Mclaren, Fran, Meisel,

Roland W., Miller, David T., Speijer, Jacob. (2002). Mathematics of Data
Management. Toronto: McGraw - Hill.

6.6 Applying Matrices to Probability Problems

Bhupz — Wed, 13 Jun 2007 14:47:47 CDT

Pascal's method

neniz — Tue, 12 Jun 2007 14:51:50 CDT

Pascal's Method can be applied to counting paths or routes between two points in a variety of arrays and grids.

Pascal's Triangle:

Where do we use Pascal's triangle:
Pascal's Triangle is more than just a triangle filled with numbers. Many times, one does not know what to do with these numbers, or even how to use them. However, there are two major areas where these numbers in the Pascal Triangle can be used, and these areas are Algebra and Probability/Combinations.

PROBABILITY/COMBINATIONS:

The Pascal triange can be used to find combinations.

Example:
Jon has five hats on a rack, and he wants to know how many different ways he can pick two of them and wear them. It doesn't matter which hat is on top, it just matters which two hats he pick. This problem leads to the question "how many different ways can you pick two objects from a set of five objects?"
The answer is the number in the second place in the fifth row, i.e. 10. 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1. Due to choosing this property, the binomial coefficient [6:3] is usually read "six choose three." If you want to find out the probability of choosing one particular combination of two hats, then that probability is 1/10.

For further explanation on how Pascal's triangle works:

Each row added is equivalent to the sum of numbers in that row.
That sum can be multiplied by 2 to get the sum of the next row.
Example: 1x2=2... 2x2=4... 4x2=8... 8x2=16... etc.

A Video to further explain Pascal's Triangle:

7.3 Geometric Distribution

BrendanS — Sun, 10 Jun 2007 21:44:30 CDT

Geometric Distribution

Geometric distribution is a probability model that help us determine how many failures occur before a single success.

Random Variable

X = number of failures before success

Formula

q = probability of failure for a single trial ( = 1-p )
p = probability of success for a single trial
x = number of failures ( = 0, 1, 2, ...)
p+q = 1

Expectation of a Geometric Distribution

q = probability of failure for a single trial
p = probability of success for a single trial

Examples

1. Jason is rolling a die. Calculate the probability of getting a 1 on the 5th roll.

First. find "p" and "q"
p = probability of success for a single trial
p = getting a 1
p = 1/6

q = probability of failure for a single trial ( = 1-p )
q = getting anything but 1
q = 1 - p
q = 1 - 1/6
q = 5/6

Next, x is the number of failures before a success
In the question it said getting a 1 on the 5th roll, meaning the 5th roll is the success. Than there are 4 failures before it.
x = 4

The probability of getting a 1 on the 5th roll is 625 / 7776.

2. What is the expected number of rolls before a 1 come out?

p = probability of success for a single trial
p = getting a 1
p = 1/6

q = probability of failure for a single trial
q = getting anything but 1
q = 1 - 1/6
q = 5/6

The expected number of rolls before a 1 come out is 5

3. A top NHL hockey player scores on 93% of his shots in a shooting competition.

a) What is the probability that the player will not miss the goal until his 20th try?

First, find q.
- Since the questiong is asking, the player does not miss the goal UNTIL his 20th
try, this means he will miss on the 20th try. Therefore, the question is asking the
probability of the NHL player missing on his 20th shot. So, the player's
percentage of scoring is actually probability of failure.

Therefore,
q = 0.93

Second, find p
- Since we are given the probability of failure we must determine the probability of success by subtracting 1 - q.

p = probabilty of success in each single trial
p = 1-q
p = 1-0.93
p = 0.07

Therefore,
p = 0.07

Third, determine X
- Because the player misses the 20th shot, it means he scored 19 of the previous shots. Since the scoring of the player is measured as unsuccessful, the number of unsuccessful trials is 19.

Therefore,
x = 19

Forth, solve.

The probability of the NHL hockey player not missing until his 20th try is 0.0716%.

b) What is the expected number of shots before he misses?

First, find p & q
p = probability of success in each single
p = 0.07

q = probability of failure
q = 0.93

Next, Solve

Therefore, the NHL hockey player will shoot 13.3 shots before he misses.

By: Jason K., Shahnaz S., Andrew K.,
Improved By: Amrit L., Brendan S.

References:
Mr. D'Onofrio's notes, 7.3 handout.
Ms. Richardson's notes, 7.3 handout
Gr. 12 Data Textbook, pg. 395

4.1 Organized Counting

tofu-la — Sun, 10 Jun 2007 17:43:41 CDT

¤ The total number of possibilities can be determined without actually counting each one individually.

Combinatorics: the method of counting the possibilities
For example: a tree diagram

Tree Diagram

Example 1:
A radio station announces that they are giving away a prize package to the first caller. The package involves 1 movie, 1 gift certificate, and 1 car. The movie choices are: Sin City and Saw III. The possible gift certificates are to Hollister and Buffalo. The possible choices of cars are a Lexus and Ford. What are all the possible packages a winner can choose from?

Task: figuring out the number of possible combinations of prize packages

The results of the tree diagram can also be determined using the Fundamental or Multiplicative Counting Principle because drawing a tree diagram can be time consuming with large amounts of data.

Therefore, there are 8 possible prize packs.

Example 2:
For lunch David has to have to choose a sandwich, drink and dessert. The cafeteria sells BLT, Tuna, and Club sandwiches. For drinks they offer, orange juice, iced tea, apple juice and coke, and as for dessert, they sell brownies and cookies. How many different combinations of sandwich, drink and dessert can one person make?

As the diagram illustrates, there are many different combinations David can have for lunch, a total of 24 different combinations. The tree diagram allows us to see all the different combinations. Using the tree diagram approach can be a hassle sometimes once the amount of choices increase drastically. So a simpler way to calculate the amount of possible choice was found.

3 X 4 X 2 = 24

Total # of sandwiches X Total # of drinks X Total # of desserts = Total # of possibilities

Fundamental or Multiplicative Counting Principle

A SIMPLER METHOD

¤ If a task is made up of stages with separate choices in each stage, the total number of ways of doing the task is m x n x p..., where 'm' is the number of choices for the 1st stage, 'n' is the number of choices for the 2nd stage and so on.

The Fundamental Counting Principle is applied in the following example.

Example 3:
Telephone numbers in Canada used to be 7 digits and they did not include an area code.

(a) How many 7 digit phone numbers are possible?

Task: writing down 7 numbers
Number of stages: 7
Write 7 blanks

___ ___ ___ ___ ___ ___ ___

*Remember: Numbers can be repeated; however the phone number can NOT begin with 0

9 10 10 10 10 10 10 = 9 x 10 x 10 x 10 x 10 x 10 x 10

= 9 000 000
Therefore, there are 9 000 000 possible phone numbers.

(b) How many possible phone numbers are there that do not have two of the same numbers next
to each other?

*Remember: the phone number can NOT begin with 0; however the other numbers in the phone number can include 0.

9 9 9 9 9 9 9 = 478 2969
Therefore, there are 478 2969 phone numbers that do not have two of the same numbers next to each other.

(c) How many phone numbers do not repeat the same number at all?

9 9 8 7 6 5 4 = 544 320

Therefore, there are 544 320 phone numbers that do not repeat the same number at all.

Additive Counting Principle

If one task can be performed m ways, but can also be performed n ways, and if the two tasks and mutually exclusive, meaning they cannot be performed at the same time, then there are m + n ways of doing the task.

(The Additive Counting Principle is applied in the following example.)

Example 4:
Jim is playing Monopoly with his friends. They are playing with one die. Jim, during his turn, either wants to roll a 1, or a number higher than 4. How many ways are there for Jim to roll either a 1, or a number higher than 4?

Task: rolling the die
Number of Stages: 1

How many ways can Jim roll a 1? 1 way (just the number 1)

In how many ways can Jim roll a number higher than 4? 2 ways (5 and 6)

Can both tasks be done at the same time? No (Jim can only roll once during his turn) Therefore, the tasks are mutually exclusive.

By adding the two possible ways, we can figure out the total number of ways for Jim to roll a 1 or a number higher than 4.

Jim can roll the number he wants in 3 ways.

First tree diagram provided by <http://www.gliffy.com/>
Second tree diagram by tofu-la

4.2 Factorials and Permutations

ewakabat14 — Sat, 09 Jun 2007 11:57:06 CDT

Factorial: Determines the number of different ORDERS in which one can arrange or place set of items

n!= n x (n-1) x (n-2) x (n-3)...3 x 2 x 1

A factorial indicates the multiplication of consecutive natural numbers.

Example 1:
The senior choir has rehearsed five songs for an upcoming assembly.
In how many different orders can the choir perform the songs?

Solution:

There are five ways to choose the first song, four ways to choose the second, three ways to choose the third, two ways to choose the fourth, and only one way to choose the final song. Using the fundamental counting principle, the total number of different ways is

5x4x3x2x1 = 5!
=120
The choir can sing the five songs in 120 different orders.

Example 2: Evaluate

=84

Example 3:

Example 4:

Homework Page 239 # 3a.

Place in nPr form:

6x5x4

=6P3

A Permutation of 'n' distinct items is an arrangement of all the items in a definate order.

n!= nPn

A permutation of 'n' distinct items taken 'r' at a time is an arrangement of 'r' of the 'n' items in a definate order.

nPr = P (n,r) = n!
(n-r)!

A rearrangement of the elements of a set.

Example 1: The word bookkeeper is unusual in that it has three consecutive double letters. How many permutations are there of the letters in bookkepper?

Solution:
If each letter were different, there would be 10! permutations, but there are two os, two ks, and there es. You must divide by 2! twice to allow for the duplication of the os and ks and then divide by 3! to allow for the three es:

=151 200

There are 151 200 permutations of the letters in bookkeeper.

The following are all the permutations of DOLE:

DOLE,DOEL,DLOE,DLEO,DEOL,DELO
ODLE,ODEL,OLDE,OLED,OEDL,OELD
LODE,LOED,LDOE,LDEO,LEOD,LEDO
EOLD,EODL,ELOD,ELDO,EDOL,EDLO

There are 24 permutations of the letters in DOLE. This number matches what you would calculate using 4P4 = 4!

7.2 Binomial Distributions

Ayman13 — Fri, 08 Jun 2007 23:24:49 CDT

Binomial Distributions occur when an experiment is repeated and a particular outcome (Success and failure) is counted. Experiments that are repeated are called the Bernoulli Trials.

However, there are certain conditions for a Bernoulli Trial to occur:

Only two outcomes are possible (Success and failure)
The outcome of each trial does not depend on the previous trial
The probability for success and failure is the same for each trial
Trials are repeated a specified number of times

An example of when binomial distributions can be used when the number of defects is counted during quality control.

The formula for binomial distributions:

x = Number of single successes
n = Number of Bernoulli Trials
p = Probability of successes outcome
q = Probability of single failure
Also, the sum of the probability of successes and failures always equal one; therefore, p + q = 1.

In addition,
The formula for the expectations for a binomial distribution is:

Example I:
When flipping a coin 6 times, what is the probability of getting 4 heads?

You could make a tree diagram to figure this out but the tree would be difficult to construct. Though if you were to make it, you would see that there are 64 possible outcomes of flipping a coin 6 times and 15 of them give you the event of flipping 4 heads. From which you can find that the probability of flipping 4 heads in 6 flips of a coin is: 15/64

The more efficient way to solve this problem is to use the formula:

x=4 heads (number of sucesses)
n=6 flips ( number of bernoulli trials)
p=1/2 (probubility of sucesses)
q=1/2 (probibility of failure)

P(4 heads)=6C4*(1/2)^4*(1/2)^2
P(4 heads)=15*(1/16)*(1/4)
P(4 heads)=15/64
You receive the same answer of 15/64 that you would get if you manually made a tree diagram and counted the branches. This example shows why the formula is more efficient.

Example II (indirect method):

What is the probability of tossing at lease two sixes in 6 rolls of a die?

Because the question includes the phrase "at least", it would be highly advisable to use the indirect method.

n = 6 p = 1/6 q = 5/6 x = 0,

P(x = 0, 1) = 1 - {[6C0 x (1/6)^0 x (5/6)^6 ] + [6C1 x (1/6)^1 x (5/6)^5]}

P(x = 0, 1) = 1 - 0.736775549

P(x = 0, 1) = 0.263224451

Therefore, one has a 26.3% of tossing at least two sixes in 6 rolls of a die

Example III:

For example II, what would the expectations for the binomial distribution?

n = 6 p = 1/6 E(X) = 6(1/6) = 1 If one rolls a die six times, one should expect to get a six only once.

By: Edith, Chrystalin, and Daniel
Edited by: Ayman and Gabriela

References:
Mr. D'Onofrio's lesson sheet

Ms. Richardson's lesson sheet
Data Management Text

Encoding and decoding using matrices

DMeghan — Fri, 08 Jun 2007 19:58:45 CDT

Matrix multiplication can be used to “encode” and “decode” messages. For that a coding matrix is required which is known to both the sender and the recepient. The following flowchart illustartes this in a vivid way:-

Encoding a Message & Decoding a Message

Example:

Return to Problem Solving With Matrices Page

1.7 Problem solving with Matrices

DMeghan — Fri, 08 Jun 2007 19:56:06 CDT

Multiplying Matrices

Identity matrix have entries of 1 along the main diagonal and zeroes for all other entries.

Inverse Matrix

NOTE:

Property of Inverse Matrix

1.5 Graph Theory

amerjad — Fri, 08 Jun 2007 15:19:15 CDT

What is Graph Theory?

Graph theory is a branch of mathematics concerned about how networks can be encoded and solved.

Applications for Graph Theory:

1- Making and organizing timetables.
2- Studying molecules in physics and chemistry.
3- Finding optimal paths for transportation purposes.
4- Colouring maps.

Important Terms and Definitions:

Term	Defintion
Graph	Also called a network is a collection of lines joined together.
Edge	Lines in a graph are called edges.
Vertex	Dots or nodes are referred to as vertices.
Path	Connected series of edges.
Circuit	A path that begins and ends at the same vertex.
Connected Network	Is a network that has at least one path joining each pair of vertices.
Complete Network	Is a network in which every pair of vertices is joined by an edge (line).

* A Graph is considered traceable if it has all the vertices connected to at least one other node (vertex) and ALL edges can be traveled exactly once in a continuous path.

* If all the verticies of a graph are of even degree, then the graph is tracable. Also, if a graph has two verticies of odd degree and the rest are even, then the graph is tracable.

Example 1

Make networks of the following maps:

Map 1

Solution

- Start by lettering the areas in the map as follows:

- Then make a network by drawing a point for each letter and then connect the dots of the letters which areas are touching or are adjacent to each other, as follows:

Map 2

Try this one out...

Solution

Example 2

Refer to the network in the figure.
a) How many degrees does each vertex have?
b) Is the network traceable?

Solution

a) Number of degrees of each vertex:

T: 3
U: 2
V: 4
W: 4
X: 2
Y: 3
Z: 4

b) Yes, the network is traceable because only two verticies have odd degrees
(T and Y), and the rest of the verticies have even degrees.

Example 3

The unknown town of Networki was situated on three islands and the banks of Circuitry River. Networki had 10 bridges as shown in the map. The people wanted to find a way to traverse the town by only crossing each bridge once. Is this possible?

Solution

Make a network of the map:

Count the degrees of each vertex:

A: 4
B: 4
C: 4
D: 4
E: 4

Therefore, it is possible to traverse all the bridges without double crossing because each vertex has an even number of degrees.

6.4 Independent and Dependent events

antzie — Fri, 08 Jun 2007 01:07:35 CDT

If A and B are independent events, then probability of both occurring is:
P(A) x (B)

Example:
What is the probability of rolling a 3 with a dice, and drawing a 3 from a deck of cards?

Solution:

P(A n B) = 1/6 x 4/52 = 1/78

If event B is dependent upon even A then:
P(A and B) = P(A) x P(B | A)

Conditional Probability:
P(A | B) = P(A n B) ÷ P(B) =

The second event is conditional upon the first event's result:
(B given A has already occurred)

Example:
A family has three children. What is the probability that the family has 2 boys and a girl given the middle child is a boy?

Solution:
A. 2 boys and a girl
B. Middle is a boy

Given the formula:
P(A | B) = P(A n B) ÷ P(B)
= 2/8 ÷ 4/8 = 1/2
= 2/8 ÷ 4/8
= 1/2

Still Having trouble finding the probability for Independent Events?

Use the Product Rule.

When trying to find the probability of independent events use the product rule:

P(A&B) = P(A) x P(B)

A and B are independent of each other, which means that they do not affect each other in any way.

Independent Events

Example 1: Find the probability of rolling a six on two different dice.

1. First ask yourself, " Are these events Independent or Dependent? "
Since the outcome of one die does not affect the outcome of the other, then we
establish that these are independent events.

2. Now find the probability for each die using P(A)= n(A) ÷ n(S). You will have to
refer back to 6.1 Basic Probability Concepts.

A(1st die) = getting a 6
B(2nd die) = getting a 6
S(1st die) = rolling anything
S(2nd die) = rolling anything

Therefore,
n(A) = 1 n(B) = 1
n(S) = 6 n(S) = 6
So,
P(A) = 1/6 P(B) = 1/6

Once the probabilities of each event is found, multiply them together to achieve the probability of A and B (rolling a 6 on two different dice).

Since, P(A n B) = P(A) x P(B) then

P(A n B)
= 1/6 x 1/6
= 1/36

Therefore the probability of rolling two dice and getting a 6 on each die is 1/36.

Conditional Probability

The conditional probability of B, P(B/A), is the probability that B occurs, given that A has already occurred.

Example 2: Out Take is an email program.
Probability of out take hanging and then operating system crashing afterwards: 0.01
If out take hanging, what is the probability that operating system will soon crash?

A = the operating system crashing
B= Out Take hanging
P(A n B) / P(B)
= P(A) / P(B)
= 0.01/0.025
= 2/5

Question #17.
Laurie, an avid golfer, gives herself a 70% chance of breaking par (scoring less than 72 on a round of 18 holes) if the weather is calm, but only a 15% chance of breaking par on windy days. The weather forecast gives a 40% probability of high winds tomorrow. What is the likelihood that Laurie will break par tomorrow, assuming that she plays one round of golf?

Solution:
P (par in wind or no wind) = P (par in wind) + P (par in no wind)
= (0.15)(0.4) + (0.7)(0.6)
= 0.06 + 0.42
= 0.48

Therefore, the likelihood that Laurie will break par tomorrow, assuming that she plays one round of golf is 0.48 or 48%.

Canton, Erdman, et al. Mathematics of Data Management. Toronto: McGraw-Hill Ryerson, 2002.
Page 335, Question 17.

Mrs. Richardson's notes: 6.4 Independent and Dependent Events.

1.6 Matrices Algebra

Dianne121 — Thu, 07 Jun 2007 23:01:19 CDT

What is a Matrix?

A matrix is a rectangular array of numbers used to manage and organize data. It is made up of horizontal rows and vertical columns.

Dimensions of the Matrix (Am,n)
A matrix with m rows and n columns has dimensions of m ×n. For example, the matrix A2,3 has 2 rows and 3 columns.

Example 1

Types of Matrices

A matrix with only one row is called a row matrix. A matrix with only one column is called a column matrix. A matrix with equal dimensions (m=n) is called square matrix.

Identity Matrix

It is a matrix with 1's on the main diagonal (the diagonal which runs from the upper-left to the lower-right) and 0's everywhere else.

Transpose of the Matrix (At)

You can find the transpose of the matrix by interchanging the corresponding rows and columns.

Addition and Subtraction

Two or more matrices can be added or subtracted only if they have the same dimensions. To add or subtract matrices, add or subtract the corresponding elements of each matrix. In the example seen below, if you wanted to add Matrix A and Matrix B, you would add [a 1,1] with [b 1,1]. Since [a 1,1] is 20 and [b 1,1] is 11, you would simply add 20 to 11, giving you an answer of 31, which would be the [c 1,1] element of Matrix C.

Multiplication and Division

1. Multiplying Matrices

You can ONLY multiply the matrices if the inner dimensions are the same. For instance, Am,n and Bn,p. Since Matrix A is 2 by 2, and Matrix B is 2 by 2, you can multiply these two matrices because the column number for Matrix A = 2 and the row number for Matrix B = 2.

Example 1

Find Matrix A * Matrix B.

Solution

To find A * B, the matrices A, B must have the same inner-dimensions. When multiplying two matrices, the resulting matrix will have the dimensions of the row number of Matrix A and the column number of Matrix B. So a 2 by 2 matrix * a 2 by 2 matrix will result in a 2 by 2 matrix. Similarly, a 2 by 3 matrix * a 3 by 4 matrix will result in a 2 by 3 matrix.

c 1 by 1= first row in A * first column in B
=6*2+-5*7= -23
c 1 by 2 = first row in A * second column in B
= 6*9+-5*4= 34
c 2 by 1=3*2+0*7=6
c2*2=3*9+0*4=27

Example 2

Multiplying Matrices Video

2. Finding the inverse of a 2 by 2 matrix.

Example 1

Finding the inverse of a 2 by 2 matrix Video
(The video uses the formula shown above.)

Example 2

To find if any 2 by 2 matrix has an inverse or not.

It has an inverse when ad≠bc, and the flowchart shows that .

3. Multiplying matrices by a scalar

To multiply a matrix by a scalar, multiply each element by the scalar quantity. For the example below, you would multiply 5 (the scalar quantity) by each element in the matrix.

4. Dividing Matrices

Dividing Matrices isn't like multiplying. It's different in some ways. If you are to divide by a scalar, each element is divided by the scalar quantity. However, a matrix divided by another matrix is ALWAYS undefined.

Example 1

Simplify the following expression and find A:

Solution

To isolate for A we divide each element by 2. You should get this:

Example 2

Simplify the following expression:

Solution

Since you are trying to divide a matrix by another matrix, the answer is: UNDEFINED

Matrices with real world problems

As defined before, matrices are used to organize and store collections of data. So far we've only discussed how to solve matrix problems with textbook questions. However, this does not mean that matrices are not useful in the real world.

1. Solving real world matrix problems by hand

Example 1

Matrix A represents the proportion of students at a high school who have part-time jobs on Saturdays and the length of their shifts. Matrix B represents the number of students in each grade.

Solution

Since Matrix A and Matrix B have the same dimensions, you are able to multiply. The result will be a 3 by 2 matrix.

Around 73 males and 77 males work up to 4 hours, 140 males and 148 females work 4 to 6 hours, and 65 and 71 females work more than 6 hours on Saturdays.

2. Multiplying matrices in spreadsheets

Using a spreadsheet program, such as Microsoft Excel, it is possible to do matrix functions in them such as multiply.

Example 1

The following table shows the number and sex of full-time students enrolled each university in Ontario one year.

Solution

First, we fill in the spreadsheet in Microsoft Excel with the following information like this (values in the spreadsheet go up to T3 but the picture cuts it off):

There are two functions that we are going to use. We are going to use the MMULT function and the INDEX function. The problem with the MMULT function is that it only displays the first entry of the product matrix. We then use the INDEX function to look at a specific element in the product matrix. To use the MMULT function, click on the cell that you want the result to be displayed in. At the top, in the entry bar, type in the following: =MMULT(array1,array2)

array1 = the range of cells defining the first matrix
array2 = the range of cells defining the second matrix

To use the INDEX function, in the entry bar at the top, type in the following: =INDEX(MMULT(array1,array2),row,column)

row = the row you want the product matrix to display
column = the column you want the product matrix to display

The result should look like this:

3. Finding the inverses of matrices in Excel

Example 2

Using Excel, we will calculate the inverse of:

Solution

First, type the matrix into Excel. It should look like this:

Click on the cell D1. This is where the first element of the inverse will be. While that cell is clicked on, at the entry bar on the top, type in the following: =MINVERSE(A1:B2)

A1:B2 defines where the matrix is located on the spreadsheet. However, alike the MMULT function, the MINVERSE function only displays the first element of the matrix. In order to display the whole matrix at one time, click on the cell to the right of the one displaying the first element and type in the following: =INDEX(MINVERSE(A1:B2),1,2)

The ",1,2)" sets which element to be displayed. Since it says 1,2, the element of the first row and second column will be displayed. Continue this procedure until the whole matrix is shown. It should look like this:

Links
Definitons regarding matrices

References
Mathematics of Data Management , Grade 12, (MDM4U). McGraw-Hill Ryerson

6.2 Odds

martaa — Thu, 07 Jun 2007 21:57:42 CDT

Probability is used to express a specific and predictive value (in the form of a percentage or fraction) while odds compare the degree of confidence about a future outcome (in the form of a ratio). It is the chances in favour of an event occurring versus the chances against the same event from occurring. The odds against an event is the reverse ratio. Keep in mind that odds are NOT the same as probabilities.

_______________________________________________________________

ODDS IN FAVOUR of an event occurring:

The probability of selecting the purple box is ¼ or 1 box out of a total of 4 boxes.
The odds in favour of the purple box is 1:3 or 1 purple box against 3 white boxes

or

Note:

Odds in favour are written as P(A) : P(A')

Example 1: Determining The Odds.
David came across a box of cell phones. The box contained four blue phones, seven red phones, and two pink phones. What are the odds in favour of David picking a pink cell phone out of the box?

Event A will be picking a pink phone out of the box. The probability of event A is:

The probability of not picking a pink phone from the box is:

Recall that the odds in favour or A can be written as P(A) : P(A'), therefore the odds in favour of event A are:

Therefore, the odds in favour of David picking a pink cell phone out of the box are 2:11.

_________________________________________________________________

ODDS NOT IN FAVOUR of an event occurring:

Odds not in favour are written as P( A') : P(A)

As you notice, this formula is the reciprocal (complementary) of the formula used for finding odds in favour of an event occurring.

Example 2: Odds Against An Event.

The probability of a certain student being late in one week for Mr. D'Onofrio's fourth period Data Management class is 0.6. This data is based on the number of detentions the student has had so far. What are the odds that the student will arrive on time detention free this week?

The probability of the student arriving on time for Mr. D'Onofrio's class:

Therefore it is a 2:3 chance that the student will arrive on time detention free this week

__________________________________________________________________

Example 3: Odds in favour of A are .

The odds of Jack getting accepted to the University of Toronto is 3:8. What is the probability of Rico not getting accepted to the University of Toronto?

Therefore, the probability of Jack not getting accepted to the University of Toronto is .

_________________________________________________________________

Remember that odds give the degree of confidence that an event will or will not occur. So that means that the higher the ratio, then the higher the degree of confidence that event will or will not occur.

Example 4: Ratios.

If someone was to tell you that the odds of the Toronto Raptors making the playoffs is 5:1 then you are more confident than if they were to tell you the odds of the Toronto Raptors making the playoffs is 4:1.
__________________________________________________________________
Example 5: P( A') =

What are the odds in favour of A happening?

Therefore, the odds in favour of A happening is 1:3.
__________________________________________________________________

Try This: Homework Help

Mathematics of Data Management Textbook page 319 #12

George estimates that there is a 30% chance of rain the next day if he waters the lawn, a 40% chance if he washes the car, and a 50% chance if he plans a trip to the beach.
Assuming George's estimates are accurate, what are the odds

a. in favour of rain tomorrow if he waters the lawn?

b. in favour of rain tomorrow if he washes the car?

c. against rain tomorrow if he plans a trip to the beach?

References

Canton, Erdman, et al. Mathematics of Data Management. Toronto: McGraw-Hill Ryerson, 2002.

6.5 Mutually Exclusive Events

Morriechell — Thu, 07 Jun 2007 20:39:07 CDT

Mutually Exclusive Events: Events that cannot occur at the same time.

This equation is used when one sees the word "or" and ACP (Additive Counting Principle) is used because we are adding probabilities.
Ex. We would use this equation if a question asked: What is the probability that in a group of teachers we randomly select either a male or a female. [Basically a teacher cannot be a male and a female at the same time(as far as we know), so it is said that these two events are mutually exclusive.]

Non-Mutually Exclusive Events: Events that can occur at the same time.

This equation is used when one sees the word "or" and events from A and B can happen at the same time. Again ACP is used because we are adding probabilities.
Ex. We would use this equation if a question asked: What is the probability that in a group of teachers we randomly select either an Italian or a male. [Basically a teacher can be an Italian and a male at the same time(ex.Mr.D'Onofrio), so it is said that these two events are non-mutually exclusive. We subtract the intersection of A and B to get rid of any duplicates.]

Example 1 From Homework:
Handout - Exercise 5.3:
Question # 5:

The following table illustrates the distribution of elementary and secondary school enrolment in public and private institutions in Canada in 1977.

Province/Territory
Public
Private

CANADA

5,285,274

185,035

Newfoundland
157,803

293

PEI
27,919

-

Nova Scotia
201,759

1,410

New Brunswick
163,317

393

Quebec
1,319,511

86,110

Ontario
1,974,266

58,226

Manitoba
225,854

7,642

Saskatchewan
219,327

1,573

Alberta
441,255

6,070

British Columbia
536,481

23,318

Yukon
4,866

-

NWT
12,916

-

If a Canadian student was selected at random, what would be the probability that the student was:
(a) from Ontario?

(b) from either Ontario or Quebec?

(c) from the public system?

(d) from either Ontario or the Private system?

(e) from either Manitoba or the Yukon?
Refer to (b)
Ans = 0.044

Example 2 from Class Notes:

A Mutually Exclusive Event..
If a committee of five is to be chosen randomly from six males and eight females, what is the probability that the committee is either all male or all female?

Example 3 from Class Notes:

A Non-Mutually Exclusive Event..
Rolling 2 dice. What is the probability of rolling a sum greater than 7 or doubles?

Example Chart: Classify each pair of events as mutually exclusive or non-mutually exclusive.
Lesson 6.5- Homework Question 1- pg 340

Event A Event B Event Type

-randomly drawing a grey sock from a drawer -randomly drawing a wool sock from a drawer non-mutualy exclusive

-randomly selecting a student with brown eyes -randomly selecting a student on the honour roll non-mutually exclusive

-having an even number of students in your class -having an odd number of students in your class mutually exclusive

-rolling a six with a die -rolling a prime number with a die mutually exclusive

References: Mr. D'Onofrio's notes, Exercise 5.3 work sheet; Mrs. Richardson's notes, 6.5 Question 1 page 340.

5.4 The Binomial Theorem

Doris88 — Wed, 06 Jun 2007 08:53:49 CDT

When we expand a power of a binomial expression we get a polynomial which can be considered as a series. It is not an arithmetic or geometric one but there is definitely a pattern.

Clearly, doing this by direct multiplication gets quite tedious and can be rather difficult for larger powers or more complicated expressions.

Now that we are familiar with combinations, we know that each term in Pascal's triangle corresponds to a value of nCr.

In comparing the two triangles above, we will be able to observe that
Recall Pascal's method for creating his triangle:

With this comparison, the method must also apply to combinations, giving Pascal's formula:

Proof:

This proof allows us to see that the values of nCr do follow the pattern that creates Pascal's triangle.

Example 1: Applying Pascal's Formula to Combinations
Rewrite the following as the sum of 2 other combinations

The Binomial Theorem

When we expand a power of a binomial expression, the result is a polynomial, which can be considered as a series. It is not an arithmetic or geometric series, but there is a visible pattern occurring. This pattern leads us to the Binomial Theorem.

The Binomial Theorem is the formula used to obtain the expansion of powers of a binomial and is as follows:

Before we move on to a few more examples, let us look at some of the properties of the binomial expansion (a + b)n.

There are n + 1 terms
Progressing from the first term to the last, the exponent of a decreases by 1 from term to term while the exponent of b increases by 1. In addition, the sum of the exponents of a and b in each term is n.
If the coefficient of each term is multiplied by the exponent of a in that term, and the product is divided by the number of that term, we obtain the coefficient of the next term.

Example 2: Applying the Binomial Theorem
Use combinations to expand the following

Example 3: Homework Question - # 20a on page 29

When finding the general term we apply the formula:

References

Canton, Barbara J., Erdman, Wayne, Irvine, Jeff, Lim, Louis, Mclaren, Fran, Meisel,

Roland W., Miller, David T., Speijer, Jacob. (2002). Mathematics of Data
Management. Toronto: McGraw - Hill.

7.4 Hypergeometric Distribution

Jimin — Tue, 05 Jun 2007 20:59:23 CDT

1 Hypergeometric Distribution

Is a discrete probability distribution, involving a series of dependant trials, with more than one type of success or failure. Involves a number of n draws from a limited population, without replacement. Phrases such as "not replaced" and "not put back" hint at hypergeometric distribution to be required. The sum of xP(x) is expected outcome.

Hypergeometric Trials vs. Bernoulli Trials

Bernoulli trials are independent of one another and all have the same probability, whereas hypergeometric trials are dependant and have varying probabilities. Bernoulli trials have only one type of success and one type of failure, whereas hypergeometric trials have more than one type of success and more than one1 type of failure.

Formula

The formula of a hypergeometric distribution is given by:

where:
x1= # of successful trials of type a
x2= # of successful trials of type b
x3= # of successful trials of type c
a= total possible # of elements in type a trial
b= total possible # of elements in type b trial
c= total possible # of elements in type c trial
n= total # of trials = x1+x2+x3

Example 1

What is the probability of a Formula 1 race finishing with; 2 Ferrari, 2 Renault, and 1 Honda in the top 5 if each team has 5 cars in the race and the race consists of only those teams?

let x1 = # of successful trials of type a
= # of times Ferrari finishes in the top 5
= 2 Ferrari's

let x2 = # of successful trials of type b
= # of times Renault finishes in the top 5
= 2 Renault's

let x3 = # of successful trials of type c

= # of times Honda finishes in the top 5
= 1 Honda

let a = total possible # of elements in type a trial
= # of Ferrari's in the race
= 5

let b = total possible # of elements in type b trial
= # of Renault's in the race
= 5

let c = total possible # of elements in type c trial
= # of Honda's in the race
= 5

n= total # of trials = x1+x2+x3 = 5

= 0.1665

Therefore, probability of this happening is 0.1665.

Example 2

There are five bananas and seven oranges in the refrigerator. Four fruits are chosen at random to serve guest. What is the probability that exactly two of the fruits will be oranges?

a=# of Oranges
=7

b=# of Bananas
=5

n=# of trials(# of fruits to be chosen)
=4

x=# of successful trials(# of oranges require)
=2

Expected Values of Hypergeometric Distribution

The expected values of a hypergeometric distribution can be given by:

where:
X= number of successes
s= total # of items in a population which would be seen as a success
f= total # of items in a population which would be seen as a failure
n= # of trials

Example 2

A jar of jellybeans contains 20 yellow jellybeans and 25 red jellybeans. If 5 jellybeans were drawn from the jar randomly, what is the expected number of red jellybeans drawn?

let n = 5
let s = 25
let f = 20

Therefore, if you were to withdraw 5 jellybeans you would expect 3 to be red.

Example 2

There are five bananas and seven oranges in the refrigerator. Four fruits are chosen at random to serve guest. What is the expected number of oranges chosen?

a=# of Oranges
=7

b=# of Bananas
=5

n=# of trials(# of fruits to be chosen)
=4

Therefore, you can expect 2.3 oranges.

References
Wgman, Diane. Mathematics of Data Management. Toronto: McGraw-Hill Ryerson, 2002.

5.3 Problem Solving With Combinations (Part 1)

Tyndale11 — Mon, 04 Jun 2007 22:29:49 CDT

Overview

This chapter will look at situations where you want to know the total number of possible combinations of any size that you could choose from a given number of items, where some may be identical.

Diction

Total set of items is referred to as ‘n’.

Items being chosen from a set are referred to as ‘r’.

A null set of items is a set that has no elements.(ie. 6C0=1)

-Retain the null set when the question implies that no object at all can also be selected or states: up to, maximum of…
- Remove the null set when the question implies that some objects (at least 1) have to be selected.

· When using combinations you can:

1. Choose a specific number of objects ‘r’ from the set ‘n’.
2. Choose a different ‘r’ in each case.
-Use cases in questions with terminology like:
-Either/or
- At least 1,
-1 or more
-2 or more
- Minimum of
- Maximum of
- Up to

The terminology has a great importance in getting an answer correct or incorrect. Pay attention to it.

Formula

Combination notation nCr--> = n! (see example 2 and flowchart)

(n-r)!r!
2. Combinations in which some items are identical
§ (p+1) (q+1) (s+1) …-1 [to remove the null set] (see example 3)
§ (p+1) (q+1) (s+1) … [to include the null set] (see example 3)
3. Not choosing exactly ‘r’ objects, all distinct objects and including the null set: (2^n)
4. Not choosing exactly ‘r’ objects, all distinct objects and excluding the null set: (2^n) -1
5. Not choosing exactly ‘r’ objects, all distinct objects and at least 2 objects are selected: (2^n)- nCr = (2^n) – (nCr0+nCr1+…+ nCri)

NOTE: 2^n = 2 to the power of n

Combinations Lesson (Flowchart)

Examples

Example 1. Using exactly r objects, with identical items.

Problem - If you are dealing from a standard deck of 52 cards, how many 3 card hands could have at least one spade? Solution - recall from the flowchart, we use Cases.
Solution-

Step 1- Write out the cases.

Case 1: Only one spade from 13 spades, two other cards from remaining 39 cards.

13C1 x 39C2 = 13 x 741 = 9 633

Case 2: Two spades from 13 spades, one other card from remaining 39 cards.

13C2 x 39C1 = 78 x 39 = 3042

Case 3: Three spades from 13 spades, zero other cards from remaining 39 cards.

13C3 x 39C0 = 286 x 1 = 286

Step 2 - Use Additive counting principle (ACP)

9 633 + 3042 + 286 = 12 961

Therefore there are 12 961 combinations possible of dealing a 3 card hand from 52 cards and deal out at least 1 spade.

**this can be done with a 5 card hand. Just use the 3 card hand as a simpler model.**

Example 2. Using exactly ‘r’ objects, with no identical objects.

Problem - In how many ways can a 4 colors be drawn from a pack of 24 pencil crayons?
Solution - Recall from the flow chart, we use the formula nCr = n!

(n-r)!r!

n=24; r=4

Way1-performed using a calculator
Way 2-calculated by hand

nCr = n!
(n-r)!r!

= 24C4 = 24!
(24-4)! 4!

= 10 626 ways = 24!
20!4!

= 10 626 ways

Either way there are 10 626 ways of drawing 4 pencils from a pack of 24 pencil crayons.

Example 3. Not using exactly ‘r’ objects, but with some identical objects.

Problem - How many sums of money can i make with 1 penny, 2 nickles and 1 dime .
Solution - Recall from flow chart we use the formula

(p+1)(q+1)(s+1) … or
(p+1)(q+1)(s+1) …-1

Step 1 -

Let p represent the number of pennies
Let q represent the number of nickles
Let s represent the number of dimes

Step 2 -
Lets substitute values into (p+1)*(q+1)*(s+1)

Step 3 -
= (1+1)*(2+1)*(1+1)
= (2)*(3)*(2)
=12

Since a combination of money is required, the null set cannot be used. [ (p+1)(q+1)(s+1) …-1 ]

Step 4 -
=12-1
=11

Therefore 11 sums of money can be made.

Example 4. Not using exactly ‘r’ objects and with no identical objects.

Problem - The school band consists of 11 members. In how many ways can a committee of at least one member be formed from the school band?
Solution - Recall from the flow chart, the formula (2^n)-1

n=11
2^11-1
= (2x2x2x2x2x2x2x2x2x2x2) -1
=2048 - 1
= 2047

Therefore 2047 committees can be formed from the group of 11 band members with at least one member on a committee.

References

Canton, Barbara J., Erdman, Wayne, Irvine, Jeff, Lim, Louis, Mclaren, Fran, Meisel,

Roland W., Miller, David T., Speijer, Jacob. (2002). Mathematics of Data
Management. Toronto: McGraw - Hill.

(2006, July 7). Combinations Flowchart. Retrieved 2006, Nov 18 from

http://www.gliffy.com/gliffy/#d=1107901&t=combinations

1.1 Iterative Process

Pacek89 — Fri, 01 Jun 2007 19:23:40 CDT

What is an Iterative Process?

An iterative process is a process where the same procedure is repeated over and over again to obtain a desired result.
Methods of showing an iteration or the iterative
process are as follows:

6.3 Probabilities Using Counting Techniques

Humza_89 — Fri, 18 May 2007 20:49:26 CDT

In a number of different situations, it is not easy to determine the outcomes of an event by counting them individually. Alternatively, counting techniques that involve permutations and combinations are helpful when calculating theoretical probabilities.

This section will examine methods for determining theoretical probabilities of successive or multiple events.

Permutation? or Combination?

The following flow chart will help determine which formula is suitable for any given question. By simply following a series of "yes" or "no" questions, the appropriate formula can be determined.

Ex. 1 - Using Permutations:

The specific outcome of Mike starting in lane 1 and the other two starting in lane 2 and lane
3 can only happen one way, so n(A) = 1. Therefore,

The probability that Mike will start in the first lane next to his other brothers in lane 2 and 3 is
approximately 0.00101.

Ex. 1(a) - Using Permutations:

Exactly Three People form a line at a grocery store. What is the probability that they will line up in descending order of age? (I.e. oldest, middle and youngest)

→Solution using the blank like method:

n(A): # of ways they will line up in descending order of age, thus:

n(S): # of ways all three can line up in any order, thus:

To calculate the probability that the three people will line up in descending order is:

Ex. 2 - Using Combinations:

However, the group may consist of any 3 members from the group of 13. Thus, n(S) is the total
number of possible outcomes.

The probability of selecting a group consisting of runners only is:

Therefore, the probability of selecting a committee with 3 runners is approximately 0.03497.

b) Either the committee is comprised of runners only, or it isn't. Therefore, the probability of the complement A, P(A'), gives the desired result.

So the probability of selecting a committee not comprised of runners is approximately 0.96503.

Ex. 2(b) – Using Combinations:

A group of 12 people is going out town on Saturday night. The group will take three cars with four people in each car. If they distribute themselves among the cars at random, what is the probability that Chantal and Rafael will be in the same car? (*note: f#: friend)

→Solution using the blank line method:

n(A):# of ways Chantal and Rafael will be in the same car, thus:

n(S):# of ways Chantal and Rafael will be in any one of the three cars, thus:

Ex. 3 - Using the Fundamental Counting Principle:

For the third employee, since there are 363 ways this person can have a different birthday from
the other two, the probability that all three employees have different birthdays is:

Now continue the process with the 12 people in the office:

The probability that at least two people in the office have the same birthday is approximately
0.167.

Key Concepts:

In probability experiments with many possible outcomes, you can apply the fundamental

counting principle and techniques using permutations and combinations.

Permutations are useful when order is important in the outcomes; combinations are

useful when order is not important.

References:

“Gliffy: BETA.” Gliffy. San Francisco, California, 2005. 18 Dec 2006.

<http://Gliffy.com/>.

Wgman, Diane. Mathematics of Data Management. Toronto: McGraw-Hill Ryerson, 2002.

Province/Territory	Public	Private
CANADA	5,285,274	185,035
Newfoundland	157,803	293
PEI	27,919	-
Nova Scotia	201,759	1,410
New Brunswick	163,317	393
Quebec	1,319,511	86,110
Ontario	1,974,266	58,226
Manitoba	225,854	7,642
Saskatchewan	219,327	1,573
Alberta	441,255	6,070
British Columbia	536,481	23,318
Yukon	4,866	-
NWT	12,916	-

Event A	Event B	Event Type
-randomly drawing a grey sock from a drawer	-randomly drawing a wool sock from a drawer	non-mutualy exclusive
-randomly selecting a student with brown eyes	-randomly selecting a student on the honour roll	non-mutually exclusive
-having an even number of students in your class	-having an odd number of students in your class	mutually exclusive
-rolling a six with a die	-rolling a prime number with a die	mutually exclusive

Way1-performed using a calculator	Way 2-calculated by hand
nCr	= n! (n-r)!r!
= 24C4	= 24! (24-4)! 4!
= 10 626 ways	= 24! 20!4!
	= 10 626 ways

MDM4U1@FMG - Recently Updated Pages

6.1 Basic Probability Concepts

Probability: The relative possibility that an event will occur, as expressed by the ratio of the number of actual occurrences to the total number of possible occurrences.

TERMS:

5.2 Combinations...

References

6.6 Applying Matrices to Probability Problems

Pascal's method

7.3 Geometric Distribution

Geometric Distribution

Random Variable

Formula

Expectation of a Geometric Distribution

Examples

4.1 Organized Counting

4.2 Factorials and Permutations

7.2 Binomial Distributions

Encoding and decoding using matrices

1.7 Problem solving with Matrices

Multiplying Matrices

Inverse Matrix

Property of Inverse Matrix

1.5 Graph Theory

6.4 Independent and Dependent events

1.6 Matrices Algebra

What is a Matrix?

Types of Matrices

Identity Matrix

Transpose of the Matrix (At)

Addition and Subtraction

Multiplication and Division

Matrices with real world problems

6.2 Odds

6.5 Mutually Exclusive Events

5.4 The Binomial Theorem

The Binomial Theorem

References

7.4 Hypergeometric Distribution

1

Hypergeometric Distribution

Hypergeometric Trials vs. Bernoulli Trials

Formula

Expected Values of Hypergeometric Distribution

5.3 Problem Solving With Combinations (Part 1)

Overview

Diction

Formula

Combinations Lesson (Flowchart)

Examples

References

1.1 Iterative Process

What is an Iterative Process?

6.3 Probabilities Using Counting Techniques