Today during class we did our "Pre-test" for our current unit of probability. The pre-test prepares us for our upcomming unit test which takes place tomorrow March 19. The format of our pre-test was 2 multiple choice questions, 2 short answer questions, and 1 3-step problem. We were given about 35-40 minutes to write our pre-test and after that Mr. K split us up into groups to compare answers. After about ten minutes of comparing answers, Mr. K went over the questions and showed us step by step how it came out to the answer.


In my group there were a couple a questions that stumpted my group members and I, but now I understand why and how we got the answer.

Here is one question that we didn't quite get right..

Using the word FOOD, find the probability that an arrangement of this word will begin with the two O's if all the letters are used. Correct to the nearest hundreth.

So to start off, you need to make sure that the two O's are stuck together at the front of the four letter combination. A way that Mr. K would refer it to would be "take the two O's and put them in a bag."


And this is what the 4 letter word combination can look like..




The first 1 stands for the 2 O's which are "stuck together," they are first because the question requires for them to be at the front of the four letter combination. The 2 stands for the choice of either F or D. And the last 1 stands for the left over of the previous two. (Say I choose F for the second space, then the last space would be a D and vice versa.)

So now you are required to multiply 1 by 2 by 1 in order to find out the number of ways for the combination. The answer is 2.

Now you take the the amount of ways that the four letter word can be rearranged(2 differnt ways) and you divide it by 4 factorial(representing the total number of letters) divided by 2 factorial(representing the 2 O's)

You would do it like so..



The result would be 2/12 then would reduce into 1/6, and because the question says around to the nearest hundreth, you must express the faction into a decimal which is 0.17.



* JUST A REMINDER * our unit test is tomorrow!
For further review Mr. K posted up extra questions.

I choose Kayla for the next scribe!

Today's Slides: March 18

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Applied 40S March 18, 2009

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Hi everybody, just a heads up there is a pre-test tomorrow so be prepared.

Any who today in class we learned two new topics. These topics are called Mutually exclusive and Non Mutually exclusive events. The way something can be mutually exclusive is if it is impossible for them to occur together. Basically like saying you can't be the age 31 and at the same time be 16, there is just no way. Or like saying can you roll a die and get both a positive and negative number at the same time.

Examples

Drawing one card either, a black ace or a red two

When randomly selecting two animals from a barn either a, cat or dog

Randomly selecting a person either a, girl or a boy

Now non mutually exclusive events are the exact opposites when things can happen together, like either drawing from a deck of cards a king or a spade. You can actually draw a king of spades.

Examples

Selecting a person for your basketball team who is either, fast or tall

Sleeping with either a, long pillow or comfy pillow

Rolling a two dice and either getting a sum of an odd number or a double

Although you may just think of this as the exact opposite to mutually exclusive events there is more thinking involved. Mutually exclusive event formula is (A) n (B) = zero set, done. For non mutually exclusive events there is a big formula you have to follow.

Using Mr.K's example I will explain

You wish to draw either a king or a spade from a single deck
A represents kings
B represents spades
Now remember one card is both keep this in mind

So the formula looks like this P ( A U B ) = P(A) + P(B) - P(AnB)
Plug in the numbers and you it looks like
P ( A U B ) = 4/52 + 13/52 - 1/52
= 1/16

WHOA, why did we subtract the 1/52 people may ask, reason being because the card gets counted so we want to subtracted it so that it doesn't get counted twice.

So that is what we learned in class what I recommend doing is try to make up your own mutually and non mutually exclusive events and practise further with the formula. Now the next scribe I choose is IRIS......

Today's Slides: March 17

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Applied 40S March 17, 2009

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Today's Slides: March 16

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Applied 40S March 16, 2009

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Today in class Mr. K started off by putting us into groups. We were then given problems to solve. You can check the slide for today.



Independent and Dependent event

My understanding about Dependent Variable it's called dependent variable because it's distribution of values depends upon the distribution of the other values or in other words It is something that depends on other factors while Independent Variable is a variable whose value determines the value of other variables or it is a variable that stands alone and isn't changed by other variable.


Here are some examples

The Probability of heads landing up when you flip a coin is 1/2

What is the probability of getting tails if you flip it again? It is still 1/2
The two events do not affect each other because it is an Independent event.

click to open the box

Two balls are drawn successively without replacement from a box which contains 4 blue balls and 3 red balls.
Find the probability that:

(a) the first ball drawn is blue and the second is red
(b) both balls are red.

The second event is dependent on the first

P(blue) = 4/7

There are 6 balls left and out of those 6, three of them are red. So the probability that the second one is red is given by:

P(red) = 3/6 = 1/2 reminder always simplify

dependent event so 4/7 × 1/2 = 2/7

Also dependent event. Using same method, but realizing there will be 2 red balls on the second draw, we have:



The next scriber I choose is : HENSON

Today's Slides: March 13

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Applied 40S March 13, 2009

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Today in class Mr.K started off by putting us into groups. We went over the homework that was assigned to us the night before. The answers to the questions are posted on todays slides, March 12th.


The new lesson we learned today was COMBINATIONS ( the "choose" formula). The formula is as follows:





"n" is the number of objects availabl to be arranged

"r" is the number of objects tht are being arranged




EXAMPLES:















You would read this as 10 "choose" 5 equals 10 divided by 5 factorial multiplied by 5 factorial.


By choosing 5 people to make a team which equals 5!, that means there is remainder of 5 other people from the 10.



The answer to this question is : 252

Another example:

This you would read as 15 "choose" 3 equals 15 divided by 3 factorial multiplied by 12 factorial.

This shows that you would choose 3 people(3!) out of the 15 people (15!) given which leaves you with 12 people (12!) left.

The answer is: 455

Tomorrow is Pi day. So don't forget to bring your pies. Also remember to find a cool math joke to share with the class.

The scriber I choose is:

MAC

Today's Slides: March 12

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Applied 40S March 12, 2009

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Today in class we started off being broken down into groups. We then quickly checked our homework. We were then given a problem to solve, it's on the slides today March 11, slide 9.

The question was:
Suppose that, when you go home from school, you like to take as great a variety of routes as possible, and that you are equally likely to take any possible route. You will walk only east or south.

We used the Pascal's Triangle to solve how many ways we can get from School to home. The answer was 180 ways to get from School to Home.

This is how it was done:














Then we went into more depth of the question, Daniel went up and explained that there was really only a certain way to go home AND pass the post office. This is what he means. The red lines shows how many ways you can pass the post office, from school to home. The blue just shows the extra routes going from school to home, WITHOUT passing through the post office.
















So now getting back to the question which was, what is the probability of passing the post office on your way home.

The answer would be:

Number of ways from School to Home passing the post office: 72
Over the total number of ways from School to Home: 180


SO: 72 / 180 would be the probability of passing through the post office on your way home!

Reminder, always reduce fractions if possible!
Reduced fraction = 2 / 5

And the next scribe would be..

DON !

Today's Slides: March 11

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Applied 40S March 11, 2009

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Today's Slides: March 10

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Applied 40S March 10, 2009

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Hello There !

Today Mr. K started off with putting us in groups and went over what we did yesterday. He talked and went over the permutations (the 'Pick' formula) examples and the homework that he assigned yesterday. The homework from yesterday took most of the time in class, he also added questions to our homework from yesterday.

The question is: What is the probability that one of these numbers is even if the digits are randomly chosen?, that question goes with question number two part c. We can solve the question by finding the solution of question 2a and 2b and dividing them together to find the probability of one of those numbers to be even when the digits are randomly chosen.

After he went over the homework we had to work on new questions that deals with.....

PERMUTATIONS of NON DISTINGUISHABLE OBJECT. It's the number of ways to arrange n objects that contain K1,K2,K3 set of non distinguishable objects.

Examples:

Book

Since there are 4 letters and 2 letters that are the same the equation would be 4!/2! = 12

Mississippi

Since there are 11 letters and 4S's, 4I's, 2P's, the equation would be 11!/(4!*4!*2!) = 34650.

Non distinguishable object is an easier way of solving problems. We didnt really get to look into it a lot so Mr. K said we would do some more tomorrow !

The next scriber would be....
KYLE ! =)

Today we started off with talking about PI DAY. We get to party this Friday & eat PIE! Not only are we going to eat pie but we get to share math jokes. So everyone don't forget to bring pie along with a stupid math joke.

Our lesson for today was Permutations (the 'Pick' Formula). A permutation is an ordered arrangement of objects.
n - # of objects available to be arranged
r - # of objects that are being arranged

Factorial notation is really an easier way of multiplying all the natural numbers from a particular number down to 1.

We went over our homework as well which are posted on the blog.

Kathrine already knows she's the next scriber.

Today's Slides: March 9

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Applied 40S March 9, 2009

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Counting

Hello everyone! For today's class we started off with Mr. K teaching us how to count in bases. 

First off, we started off with learning how to count with base ten. 

We had the numbers : 0 1 2 3 4 5 6 7 8 9  to work with.  

What he showed us was that from a 10, you start off from the right going to the left. To make this more clear I'll show you all. 

From here we started off with the zero. What we know is that ten to the power of zero is equaled to 1 from there we take the zero and multiply it to ten to the power of zero. 

That there we find out that we would have one-zero. 

Then we would move on to the left, to the one and you would do the exact same thing. 

So now instead of doing ten to the power of zero, we would have ten to the power of one. 

Ten to the power of one would equal to 10. So you have one-ten.  

Here is another example to try to make this more clearer. 

Say I put down 1 2 3 5. 

Then you would have 5 "ones"

3 "tens"

2 "hundreds"

1 "thousand"

Each of these steps is a multiplied by ten. Showing you that it is a base ten. But the one that we will use most would be base two. 

x2^4 = 16     x2^3=8     x2^2=4     x2^1=2     x2^0=1     

That's what I believe we ended with for counting bases. 

Then we moved on to the fundamental principle of counting. 

We first went over our homework that was assigned to us. 

We were taught about factorial. 

Ex. 

8*7*6*5*4*3*2*1 = 40 320

Instead of going doing all that work, he also showed us how to do it on the calculator. 

What you have to do is punch in 8. 

Once you see the number 8 on your screen, you press the button MATH. 

In there you press your arrow key to the left and that would take you to probability. 

When you are there, you would see a exclamation mark. "!" That's what you want to press. 

Hit enter and it would lead you to the main screen. If you hit enter again, it would give you the answer. 

8! = 40 320

What we also found out is that when you do 0!

Then it would equal to one. 

0! = 1 Has been assigned as a DEFINITION. 

When given a question like that, do not write it like so:

10! = 10*9*8*7*6*5*4*3*2*1 = 3 628 800

This is not the proper way to show it. 

After all of the homework review, we also did a group work. 

First question was, How many 4 digit numbers are there in which all the digits are different. 

_  _  _  _ 

Well what I did was that I figured that there are ten numbers, but since you can't use 0 as the first number of the digits, then you would have 9 numbers that you are able to use.  So there are 9 possible numbers that you can use for the first digit. 

9  _  _  _ 

For the second digit slot, you can now use the 0. So you have another 9 options that you are able to put in the second slot because you already used a number in the first slot and you are not able to use it again. So it would look like this:

9  9  _  _

Then for the next one it would be an 8 because you have used another number that is not able to be used again, and so on. 

Then it would be :

9  9  8  7

What you would do with these numbers is you would multiply them together. 

9 * 9 * 8 * 7

That would give us 4536 different ways to put four different digit numbers together. 

The next question is: How many of these digits are odd?

_  _  _  _

What you have to do is figure what odd numbers are there. It would be 1, 3, 5, 7, and 9. There are five odd numbers, so what you would have to do is place a five at the end of the digits like so: 

_  _  _  5

Now that you know how many odd numbers, then you can figure out how many numbers can go into the first digit. Since you used 1 odd number and you can not use a 0 in the first digit, then you only have 8 possibilities of any of the other numbers being in the first digit.

8  _  _  5

For the second digit, you can now use the 0 so the number of possibilities would be 8 again. 

After that the numbers cannot be used again, so the third digit has 7 possibilities. Then it would be : 

8  8  7  5 

Now you multiply them together and you get 2240 possibilities. 

That is what I basically remembered from today's class. 

It's the weekend so have fun! 

But don't forget to also DO YOUR HOMEWORK! 

Next scribe will be ... CAMILLA! =)

Today's Slides: March 6

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Applied 40S March 6, 2009

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The Fundamental Principle of Counting

Overview of Today

We started off today's class taking a quick look at the blog and then discussed the Bento picture (The first Slide for today). We discussed different ways to arrange the three vegetables in the yellow dish (came to the conclusion it was six). After that we reviewed the homework for last night, and moved on to The Fundamental Principle of Counting.

The Fundamental Principle of Counting

The Fundamental Principle of Counting states: If there are M ways to do a first thing and N ways to do a second thing then there are M x N ways to do both things (found on slide 7).
For Example:
You need to run your weekly errands and need to make a stop at a mall, a grocery store and a hardware store. In your area there are two malls, three grocery stores and two hardware stores approximately the same distance from your house. How many different possible routes are there that include stopping at one of each store.

First pick out the given information:
2 Malls
3 Grocery Stores
2 Hardware stores

Then multiply:
2 Malls x 3 Grocery Stores x 2 Hardware Stores
=12 Different Routes

For more practice:
http://regentsprep.org/regents/Math/counting/praccnt.htm
This site contains simple questions and explanations.

Important Things to Know:

Today there was not a great number of people who completed their homework and Mr. K was not thrilled. He has implemented Homework Quizzes. Mr. K will be checking our homework from a past date so it is important that we date our work as well as label the questions properly. The quiz will just be to hand in the question when it is asked.

There are a couple homework questions at the end of the slides.

Tomorrows scribe will be Alvina :)

Today's Slides: March 5

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Applied 40S March 5, 2009

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Scribe Post for March 2nd

Hello everyone, today was a good day. It was a good day because Mr. K was back in class. I'm glad Mr. K is back and you probably are too.


Today's class was a slack class we did not really learn anything new, all we did was listen to Mr. K clarify what Dr. Eviatar was teaching us in the past week he was away.
He clarified the randBin feature on our graphing calculators. He also clarified The patterns in pascal's triangle.

Sorry the post is so short, there is not really much I can say other than that the class being as slack as it was it felt like a day off of math class.





The next scribe post will be... Jason