January 29, 2007

Podcapsule Last Thursday morning the students in this class wrote their final exam. That afternoon we wrapped up our class and recorded this podcast. It's called a podcapsule because, like a time capsule, it'll stay here as a permanent record of the students reflections and recommendations to themselves on how to guarantee their success in their next math class.

Although the class is over and the podcast is done you can keep it going as long as you like ...

First listen to the podcast (6.3 Mb, 13 minutes 11 seconds), then, if you like, you can keep it alive by adding an audio comment or text comment on this post.

January 23, 2007

Exam Rehearsal for Identities and Counting

Hey people! It's JessiccaI_198 as scribe today. Today we had two exam rehearsals; one for identities in the morning and the other one was for counting in the afternoon. We talked a bit about the arc sine and about functions, but I didn't really get many notes on that. But here are the questions and solutions for the Exam Rehearsal Identities:

1) The expression (sinx + cosx)^2 - 1 is equivalent to:

a) sin2x

b) cos2x

c) tan^2x

d) 0

Solution:

sin^2x + 2sinxcosx + cos^2x - 1

2sinxcosx

sin2x

2) If sinA = 3/5 and sinB = 2/3, and A and B are acute angles, what is the value of cos(A - B)?

a) -2/3

b) (4√5 - 6)/15

c) (4√5 + 2)/5

d) (4√5 + 6)/15

Solution: cos(A - B) = cosAcosB + sinAsinB

= (4/5)(√5/3) + (4/5)(√5/3)

= (4√5 +6)/15

3) Find all values of x in the interval 0 < x < cos2x =" -3sinx">

The answer is x = 7π/6, 11π/6

Solution:

cos2x = -3sinx - 1

1 - 2sin^2x = -3sinx - 1

0 = 2sin^2x - 3sinx - 1

= (2sinx + 1) (sinx - 2)

sinx = -1/2 sinx = 2

2 is rejected because in the unit circle, no value can be bigger than 1.

x = 7π/6, 11π/6

4) If sinΘ = √5/3, what is the value of cos2Θ?

Solution:

cos2Θ = 1 - 2sin^2Θ

= 1 - 2(5/9)

= 1 - 10/9

= 9/9 - 10/9

= -1/9

5) Prove the identity: sinΘ/(1-cosΘ) + sinΘ/(1+cosΘ) = 2cscΘ Here are the problems from this afternoons class on counting:

1) The number of different arrangements of 3 boys and 4 girls in a row, if the girls must stand together, is represented by:

a) 4!x4!

b) 3!x4!

c) 4!x4!x2!

d) 3!x4!x2!

2) The students in a music department have practised 6 contemporary and 5 traditional choruses. For their concert, they will choose a program in which they present 4 of the contemporary and 3 of the traditional choruses. How many different programs can be presented, if the order of the choruses does not matter?

a) 25

b) 35

c) 150

d) 330

Solution:

6C4 . 5C3

= 150

3) All telephone numbers are preceded by a 3-digit area code. In the original Bell Telephone System of assigning area codes, the first digit could be any number from 2 to 9, the second digit was either 0 or 1, and the third digit could be any number except 0. In this system, the number of different are codes possible was:

The answer is 144 different ways.

Solution:

8C1 . 2C1 . 9C1 = 144 different ways

8C1 - there are eight numbers in the first digit - 2, 3, 4, 5, 6, 7, 8, and 9.

2C1 - two number in the second digit - 0 and 1

9C1 - all the numbers, excluding 0 4) A paperboy who delivers papers on his bike can travel only on the trails represented in the diagram. The number of different trails that the paperboy can take to get from house A to house B without backtracking is:

Solution: There are 60 different trails

5. a) How many groups of 3 chairs can be chosen from 7 chairs if the chairs are all different colours?

b) How many different ways can 7 chairs be arranges in a row if 2 of the chair are blue, 3 are yellow, 1 is red and 1 is green? (Assume that all of the chairs are identical except for colour.)

c) How many different ways can the chairs in (b) above be arranged in a circle?

a) 35
b) 420
c) 60

Solution:
5a) 7C3 = 35
- since order doesn't matter. 7 is for the total number of chairs, and 3 is for the number of chairs to be chosen.

b) 7! / 3!2! = 420
- 7! represents the total number of arrangements, and 3! 2! represent the non-distinguishable objects.

c) 6! / 3!2! = 60
- one of the chairs is a reference point, and the rest are seated in a circle. 3!2! are the non-distinguishable objects.

January 22, 2007

scribe post: Exam Rehearsal Probability

Hello guys!!!!!
Our topics in mathematics just ended. For us it is a relief but we must always put in mind that learning in a never ending process so there are still maths stuffs out there waiting for us. Anyway, this morning we did a rehearsal examination in probability. Mr. K gave us 30 minutes to work on it then after that we had a group discussion where we talked about our answers and did some brainstorming. Afterwards, Mr. K discussed with to us the right answers with their respective solutions. The questions given are:

(1) A certain soccer player has scored on 82% of his penalty kicks throughout his career. Given this information, the probabality that he will score on exactly 4 of his next 5 penalty kicks, correct to the nearest hundredth, is

(a) 0.80
(b) 0.66
(c) 0.41
(d) 0.08

The answer is (c) 0.41 because the solution for this problem is (2) If P(A) = 3/4 and P(A and B)= 1/2, where A and B are dependant events, then P(BA) equals:

(a) 1/4
(b) 3/8
(c) 2/3
(d) 5/4

The answer is (c) 2/3. The problem indicated that A and B occured in dependant events, which means that both have to happen.

Solution:

P(A) * P(B) = 1/2

3/4 * P(B) = 1/2

P(B) = 1/2 * 4/3

P(B) = 2/3

(3) Peter places 5 equal-sized tiles in a cloth bag. Each tile has a letter on it. The letters are P, E, T, E and R. The probability that Peter selects the 5 tiles, one at a time, in order such that they spell PETER, correct to the nearest hundredth, is

>  (4) A child 2 quarters, 2 dimes, and 3 nickels in his pocket, but he does not understand the value of any of the coins. He puts 35cents worth of candy on the counter at a store and randomly selects two coins from his pocket. The probability that the two coins he selects will have a total at least as high as the value of the candy is: (5) It is known that 53% of graduating are boys. Three are chosen at random. Given that at least two of the three grads are boys, determine the probability that all three of the grads are boys. ( Answer accurate to at least 4 decimal places. ) Well that's all folks.....
The next scribe is JESSICA!!!!!!!!
I already told her that she is next scribe...
Anyway guys, good luck to all...
May the good news be ours after the provincial exam....
God bless us all!!!!!!!!!

Sequences & Series Notes

Here are the notes to add to your math dictionaries for this last unit on Sequences and Series. The notes must be copied into your dictionaries by hand.

Here they are: Page 1 of 3, Page 2 of 3, Page 3 of 3.

January 18, 2007

Jefferson's Scribe post

the next scribe is JHO-ANN.

a) what is the sum of the integers from 1 to 5000?

T1 = 1
D = 1
N = 5000

‘the initial term is 1
‘the to from getting from 1 to 5000 is 1. on account that each term increases by one. And that it doesn’t say what the difference really is so its 1
‘n is the rank of the rank of the term we want to find. The rank is the sum in this case and we want to find what the sum is when the sequence reaches 5000.

Sn = n/2[2(t1) + (n – 1)d]
S5000 = (5000/2)[2(1) + (5000 – 1)1]
S5000 = 2500[2 + 4999]
S5000 = 2500 = 
S5000 = 12 502 500

b) what is the sum of all multiple of 7 between 1 & 5000?

T1 = 7
D = 7
N = 714

Since it says multiples of 7,the initial term is 7 because you start with 7 when it comes to “multiples of 7”. The difference is 7 because it increases by 7 each time.
Now we want to find the sum of sequence when it reaches to 5000. but we don’t really want to find 5000. since we’re looking for the sum of the multiples 7 between 7 & 5000 then there really aren’t 5000 terms if we’re increasing by 7 each time. To find what the rank is, we need to 5000/7. and you get 714. whatever. Don’t need to worry about the decimal answers if I were you.

Sn = n/2[2(t1) + (n – 1)d]
S714 = (714/2)[2(7) + (714 – 1)7]
S714 = 357[(14 + 7130)]
S714 = 1 784 785

c) What is the sum of all integers from 1 to 5000 inclusive that are not multiple of 7?

12 502 500 – 1 786 785
= 10 715 715

This is a no brainer but I’ll explain it to you anyways. To get the answer you take the sum of all integers from 1 to 5000 and subtracted by the sum of the multiples of 7.

A super ball is dropped from a height of 200 cm. It rebounds to ¾ of the distance it fell each time it hits the ground. What is the total vertical distance traveled by the ball when it hits the ground for the fourth time?

We know that each time it hits the ground,the ball comes back ¾ of the last distance it traveled. So it traveled down, up, down, up, down, up, and down. So in order to calculate the total distance it traveled. We need to brake this into 2 solutions first. We need to solve the “down” distance and then the “up” distance and then add the together.

Down:

So what are the sequences?

Well since the first term is 200 and it decreases every time by a ¼( bounces back to ¾ of the last term) and that it hits the ground for times…..

200 + 150 + 112.5 + 84.375

(but we really didn’t need to know that only the first term)

*DID YOU KNOW: that all the things we need to know to solve the question was in the question itself? Go check it out ; - )

T1 = 200
R = ¾
N = 4

S4 = 200(1 – (3/4)4)/(1 – ¾)
S4 = 546.875

(thanks to the help of our righteous hero- the CALCULATOR!!)

The 3 term for the “down” distance is:
‘think everything opposite of what I said from the other part

150 + 112.5 + 84.375

T1 = 150
R = ¾
N = 3

S3 = 150( 1 – (3/4)3)/ (1 - ¾)
S3 = 346.875

546.875 + 346.875 = 893.75 (anybody else noticed that the is a difference of 200 for the 2 answers. Me thinks there is a shortcut for this problem….)

I've voice recorded mr k's lecture today. It covered about half the class. Now i have listened to it a bit and you have to crank up the volume super high to listen on account that the mic was far from mr k. I just wanted to test out if it worked or not. Feel free to add some comments about it. Apologies to the loud scratching sounds on account i was moving the mp3 play on the desk, i assumed that it picked up the sound.

test voice lecture

January 17, 2007

Sum of Arithmetic / Geometric Sequence

This is RICHARD and I'm your scribe for today.

Before we started our lesson in Geometric sequences, Mr. K discussed the solutions on the on-line quiz on probability.

(1) A card is drawn randomly from a deck of 52 playing cards. What is the probability that it will be a spade or a face card? (3) A certain partial deck of cards contains 6 red cards and 4 black cards. Two cards are chosen at random. What is the probability that both cards are black?

R = 6

B = 4

4C2
____ = 6/45

10C2

(6) The diagram shows a partial map of a certain city park, with walking paths located on the grid lines. A tourist starts at point A and randomly selects a path to point C, walking only to the south and east. What is the probability the tourist passes through point B? The number of ways to get to point C is RRRDD which is 5!/3!2!. The number of ways to get to point B is RRD which is 3!/2!. The number of ways to get to point B over the number of ways to get to point C is 3/10

(7) It is known that about 5% of coffee cups used in a particular restaurant will be chipped. What is the probability that out of 10 cups selected at random, exactly 2 will be chipped?

CNCNNNNNNN

C - represents the chipped
N- represents not chipped

10!/8!2! (o.o5)² (0.95)^8
= 0.0746

(8) Three coins are tossed. What is the probability that at least two heads turn up?

These are the possibilities that you can get at least two heads

HHT = 1/8

HTH = 1/8

THH = 1/8

HHH = 1/8

The sum of these events is 1/2

(10) A jar contains 5 red and 7 blue marbles. What is the probability of pulling out 2 blue marbles in a row, without replacement? (2) A certain kind of die has four sides, labelled 1 to 4. When the die is rolled, each side has a probability of 1/4 that it will appear on the bottom. If the die is rolled 5 times, what is the probability that the side labelled “3" appears on the bottom in exactly two of the five rolls?

3N3NN

= 5!/ 2!3! (1/4)² (3/4)³

(4) Over the course of his basketball career, Julius succeeded in making a free throw in 75% of his attempts. In a randomly selected game, he attempted to make a free throw 12 times. What is the probability that he succeeded in 8 of the 12 attempts? (You don't need to know anything about basketball to answer this problem.)

12C8 (0.75)^8 (0.25)^4

(5) Three different names are randomly selected from the following list of five names. Max, Kim, Codie, Lee and, Alex. Determine the probability that “Kim” is one of the three names selected.

(1· 4C2)/ 5C3

= 3/10

(9) Assuming equally likely probabilities for male and female births, what is the probability that a four-child family will have at least one boy?

P(all girls) = (1/2)^4 = 1/16

P( at least 1 boy) = 1 - 1/16 = 15/16

After we had discussed about the on-line quiz, we proceed to our topic in Geometric sequences.

A mathematician named Carl Friedrich Gauss is a brilliant student. One day his teacher got angry because the class is so noisy. She gave an activity to keep them busy. She asked them to add the numbers from 1 to 100. Gauss came out with a better solution. He instantly computed the answer 5050. t1, t1 + d, t1 + 2d, t1 + 3d, . . . . . . t1 +97d , t1 + 98d, t1 + 99d Find the sum of the arithmetic sequence 2, 7, 12 17, . . . . S 51

S51 = 51/2 [2(2) + (51-1) 5 ]

= 51/2 [ 4 + 250 ]

=51/2 (254)

= 6477 given the geometric sequence 16, 8, 4, 2, . . . .Find the sum of this sequence up to th 10th term.

r = 1/2

t1 = 16 that's it for today!! the next scribe will be Jefferson..

Ashley's BoB

I can't believe I totally forgot about this! Anyways Probability is alright. There are some questions that I find I'm able to do fast and other questions where I really have to think about it and I get stuck because I can't find the compliment. I actually think probablitity is a lot easier than combinations because probability uses combinations so I'm a lot more familiar with it. But overall I understand how to get some of the questions. But questions like the one on the pre-test I still need to work on. I'm just glad I made it on time to do the BoB. 2 minutes before math starts haha.... Do well guys (:

January 16, 2007

Richard's BOB on PROB

One thing I can say about probability is that you need to understand the situation to get the right solution...
we follow some basic rules such as if it is "and" it means that we multiply probabilities and if it says "or", we add probabilities. But we also need to know if it is mutually exlcusive or not, independent or dependent...
...but there's only one thing that we shoulld always keep in mind regarding the probabilities, it is a number between 0 and 1 and the sum of the probabilities of the events will always equal to 1...

I hope everyone will do good on the test..Good luck and God Bless!!

Jefferson's blogging on blogging

WEEEEEEELLLLLLLLLLLLL, This sort of hard to explain on my part. In probability it's easy to understand after the teacher explains it to you. But when you apply it, It's all a puzzle, and i hate puzzles! I know how to do probability questions that involve the word "given that", but I'm just a little confused with the independent and dependent and mutually exclusive and not mutually exclusive. I mean what do you do after you find that out? what sort of math is involved if these events are mutually exclusive. I have the hardest time understanding it. They rest of probability is A-OK for me! the tree diagram always helps and it gives you more than you're looking for if you know what I mean.

What was the probability that after taking the day off school to sleep so that maybe i'd get my voice back, that i would remember to do my blogging on blogging?? all i can say is Thanks to the kids that posted them before now, cause when i went to check the scribe post to see what i missed, BOBs were posted!! Anywho enough about that, how about that probability unit?? I don't know about any of you, but I thought it was a good unit... I think there were only a couple times that i didn't know what i was doing, and those times were cleared up rather quickly. The other day when we did the pretest my group did so well.... minus a slip up where we were soo close. My favourite thing about the unit was doing the "what is the probability of A given that you already B?" those ones were no sweat... but the combinatoric ones still trip me up every once in a while... especially when i get distracted :D ... so i hope that i can keep focused and do well on that test tomorrow... and of course that everyone else does well too.... ok that's all for me tonight cause i'm getting some more sleep... who knows i might even get my whole voice back tomorrow.... till then, like mr. k says "Learn hard!!"

jho's blog

This unit had it's pros and cons. In the beggining i was getting confused, but since we were exposed to this unit for quite some time i was able to understand the concept, and therefore i am confident i will do well in this test. Hmmm maybe..haahha can't be too confident..but that pre-test was good, i was able to answer most of the questions without a problem. Hopefully i will do well tomorrow. Man i have been missing a lot of work lately, because of my loss of internet connection, got to earn some marks. GOOD LUCK everybody =D

Tenny's BOB

wow this unit has it's up and down. You would have to read twice ,to be able to write down your answer. sometime the answer you think its right might be even wrong.The first couple of part of unit was easy but it gets harder ,and harder. These questions you would have to do alot to be able to be familar with it. So practice,practice. Well thats all i have to say. bye

Bloggin on Blogging

In probability well I'm not so good at it and let's just leave it at that. On plenty of questions I end up making plenty of mistakes, whether it's to using the incorrect method or leaving something out when solving the problem. While practice does help improve something, I've had close to a decade of problem solving questions and I doubt another few days is going to get me ready for those questions.

Scribe post

Hey everyone its Tennyson ,well today we just did some questions but i was unable to finish. i will just put down the one that we did only.The next scriber can carry out the other questions.

1). List the 1st four terms in each sequence

a).tn= 3(n-1)
b).tn=(n-1)2
c).tn=2(power of)n-1

a).tn= 3(n-1)

• -tn=3(1-1)
• -tn=3(0)
• -tn=0

• -tn=3(2-1)
• -tn=3(1)
• -tn=3

• -tn=3(3-1)
• -tn=3(2)
• -tn=6

• -tn=3(4-1)
• -tn=3(3)
• -tn=9

b).tn=(n-1)2

• -tn=(1-1)2
-tn=(0)2
• -tn=0

• -tn=(2-1)2
• -tn=(1)2
-tn=1

• -tn=(3-1)2
• -tn=(2)2
-tn=4

• -tn=(4-1)2
• -tn=(3)2
-tn=9

c).tn=2(power of)n-1

• -tn=2(power of)1-1
• -tn=2(power of)0
• -tn=1

• -tn=2(power of)2-1
• -tn=2(power of)1
-tn=2

• -tn=2(power of)3-1
• -tn=2(power of)2
• -tn=4

• -tn=2(power of)4-1
• -tn=2(power of)3
• -tn=8

Later on , we talked if the question was a Geometric sequences ,Arithmetic sequences, or neither.

a).This question is an arithmetic sequence because when the number is the same ,you add or subtract from each of the numbers to get the next number in the sequence.

b).This question is neither because They dont involve in multiplying ,divid,adding,or subtract , it just does have the patterns like the other sequences.

c). This question is an geometric sequences, because the common ratio is a number that you can multiply term in order to get the other next term like,Also can be divided by the previous term.

Those question are all related to a graph,tables, and many other things

next person is ricardo , but the list was confusing so yeah , hope u didnt do 4 scribs

by3by3by3

ruschev Blogging on Blogging

I don't really know where I am right now in the probability unit....T_T.. I get some of the answers sometimes correct and sometimes I get some of the answers wrong and sometimes I know I have the correct answer and got it wrong.....sometimes I have the right answer but get the part where you solve for the correct answer wrong............. I guess I just need a Little more practice!

good luck tomorrow~

Oliver B's blogging on blogging.

Well now, I sorta like probability, but the part I didn't or well, I wasn't fond of much was, the part where combinatorix was combiend with probability, only because I don't think I did that good during combinatorix, which I need to review on. I guess it is kinda neat though how they're together like that. Overall I did like this probability unit, I hope I have as much luck tomorrow during the test though! I wish everyone luck on that, okay bye bye.

daph's seventh BOB

Probability for me is one of the scariest topics in mathematics. I found it hard to solve questions about cards and its decks. Moreover, thinking for the starting solution such as the tree diagram is another burden for me because I cannot really imagine things to be fitted in the diagram. On the contrary, I found it a little challenging. Some of the questions are really tough but sometimes I managed to solve it. Another is I want to acknowledge the effort that Mr. K gave to us in order to understand the probability well because he explained everything in details and still willing to explain it more if some of us was not able to understand it.
Honestly saying I'm not doing my assignments in probability. It is the reason why I cannot push myself in to a higher level when it comes to probability.
GOOD LUCK!!!!!!!!!!!!!
God bless us all in the upcoming test!!!!!!!!

JessiccaI_198's BOB for Probability

I think that this unit was interesting, and I enjoyed doing it. I like making the trees for the problems which make everything easier and more organized. I just have to remember the methods to be used for certain problems. Using some of the Counting Principles helps a lot in some of the problems, and makes it easier to calculate too, so I'm glad we get to use some of that. Hopefully I won't forget to use some of the rules in the problems, but I'll probably just look at some of the problems we did in class so I would probably know which one I would need to use for the ones that will be on the test. The pre-test was a little hard, but I think I'm prepared for the unit test. Everyone Good Luck!!

BOB

This unit is very interesting and I get most of it. It's just that sometimes when it seems like you have the right answer, you don't..does that make any sense? What I mean by that is that a wrong answer can seem right as to the point where it looks possible so you end up getting the wrong answer. So multiple choice questions for this unit are pretty difficult. I really like the long answers where something is assumed given that an event happens, because they're a breeze. What's really hard to remember is that you always have to have the probability over the sample space and I always forget that. The pre-test was okay so I'm sure the test will be okay as well. Good luck everyone.

January 15, 2007

Probability Quiz Here is it! You have until midnight tomorrow to complete it. We'll talk about in class on Wednesday morning. The test is Wednesday afternoon.

Learn Hard!

Wouldn't it be nice if the world really worked like this ... ;-)

scribe post; geometric sequences.

today we started off with mr. k giving us questions on the board.
find the next three numbers..
1, 2, 4, 8, 16, 32
how you would get the next set of numbers was multiplying the number by two
OR 1, 2, 4, 7, 11, 16
how you would get this one was to find the pattern in the sequence. so you would have to find the difference between the numbers. for example: 2 and 1, the difference is 1. and the difference between 4 and 2 is 2. so you would get the pattern. you know that the number is increasing sequentially. so the next number must be a 7 - since the next number is a 7, the difference between 7 and 4 is three. and so on..

6, 11, 16, 21, 26, 31
for this one, the difference between the numbers is 5 and it's always constant.

3, 6, 12, 24, 48, 96
this one is basically like the first one, you multiply the numbers by two.

1, 1, 2, 3, 5, 8, 13, 21
for this one.. you find a pattern by adding the numbers.. for instance: 1 plus 1 is 2. then 2 plus 3 is 5. and 3 plus 5 is 8. so you get the point..

1, 3, 6, 10, 15, 21, 28
this one is basically like the first one.. but the second answer.. but this one doesn't start at 1. it starts at two - meaning.. the difference between 3 and 1 is 2. 6 and 3 is 3. and it's going up sequentially, and so on..

x, -x2, x3, -x4, x5, -x6, x7
for this one, you would multiply by -(-x)

3, 4, 7, 11, 18, 29, 47
this one, you add the numbers together. 3 and 4 is 7. 4 and 7 is 11. 7 and 11 is 18. etc etc..

77, 49, 36, 18, 8, 8, 8
this one was kinda tricky. you would split the numbers - so you would get 7 and 7. you would then multiply them together and you would get 49. 4 times 9 is 36. 3 times 6 is 18, and soo on..

in the afternoon class, we had a pretest on probability. and here are the questions and solutions to them.
1. twelve people, including you, are members of a choir. the choir director is going to choose three members to attend a workshop. the probability you and two other members will be chosen is:
a) 1/4 b) 3/10 c) 1/12 d) 1/10

first of all, YOU have to go. so you would get 1 x 11C2 because you're part of that twelve so there are eleven people remaining to choose from - and you're going to choose two. and the denominator would be 12C3 since there are twelve people and you have to choose three. so that would look like..
1 x 11C2 ALL OVER 12C3
and that would all work out to 1/4, which would be a.

2. rex is playing a guessing game. the probability he will guess each question correct is 0.3. what is the probability he will guess exactly 5 out of 10 questions correctly?
a) 0.15 b) 0.10 c) 0.29 d) 0.80

so the P(correct)= 0.3 and the P(wrong)= 0.7
since there can be different ways of getting the questions wrong or right, but you still have to get five correct. so you would get..
10!/5!5!
the 5! represents the correct ones and the other 5! represents the wrong ones. you would then get..(C)5(W)5
(0.3)5(0.7)5
and when you work that all out, you would get 0.10 and that's the answer b.

3. you choose four digits from the numbers 0-9 (ten digits in total), with no repeated digits. to win a prize, all four of your numbers need to math the four selected by a computer. the probability that you will win a prize is:
P(all four digits)= 4C4 ALL OVER 10C4
and when you work that all out, you would get 1/210 (and that's the answer)

4. the serial nuber of a \$10 bill contains 8 digits. if your \$10 bill contains the digit 7 at least once, you will win a prize. what is the probability that your \$10 bill will win?
first find the probability that you wouldn't get any 7's. so, the probability that you would get a 7 is 1/10 and the probability that you wouldn't get a 7 is 9/10. soo..
P(not seven)= (0.9)8
since you now have the probability of not getting any 7's, you can now get the probability of getting at least one 7.
P(7 at least once)= 1-(0.9)8

5. jeff takes a lunch to school on two days, and on the other three days he buys it. if he takes lunch - he is late for his period 4 class 15% of the time, but if he buys lunch - he is late 35% of the time.
a) what is the probability that jeff arrives on time?

b( if jeff was late, what is the probability that he took his lunch to school?
first make a tree diagram = (0.4)(0.85) + (0.6)(0.65)
= 0.34 + 0.39
=0.73
= 73%

= 0.06/0.06 + 0.21
= 0.06/0.27
= 0.2222
= 22%

oh yeah, the test is now on wednesday.
umm.. the next scriber is tennyson.
gooday&goodnight (:

Challenge It's a new year with new units, new tests, and new learning. And how will you meet those challenges?!

January 13, 2007

Scribe Post    Okay, sorry this scribe is so late but I got it done heh. Leave some comments if you think there could have been some changes to make it more funny or just funny. Click on the pictures to view. And the next scribe is on my other scribe post which is up to date.

Scribe Post

Okay, today Mr.K started and finished off, with the following questions:

1)Tickets numbered 3,6,9,12,15 and 18 are placed in Box A. Tickets numbered 6,12,18,24 and 30 are placed in Box B. A ticked is chosen at random from each box. Find the probability:

a)That both tickets have the same number?

First find the numbers that are in both Box A and Box B. You will see that only 6,12, and 18 are in both Box A and Box B. Then calculate the probability for each, as shown:

P(A6B6)=(1/6)(1/5)= (1/30)
P(A12B12)= (1/30)
P(A18B18)= (1/30)

(1/30)+(1/30)+(1/30)= (3/30)= (1/10)

Therefore the probability that both numbers are the same is (1/10).

b) That there are different numbers on the two tickets?

This question is simple to answer, because we have just figured out how many times the numbers will be the same and it was (1/10), so now all we have to do is subtract this fraction from 1, resulting in the probability of the numbers not being the same.

1-(1/10)= (9/10)

2)John uses Google 50% of the time, Lycos 30% of the time and Altavista 20%.If he is using Google, there is a 40% chance he is searching for info about cell phones. If he's using Lycos or Altavista the probability he is researching cell phones is 30%.

First we should create a tree to organise our information, and making the question more easier to calculate. SORRY BUT MY PICTURE WILL NOT LOAD AT THE MOMENT, I will quickly look into it and post it as soon as possible.

a)What is the Probability that John decides for info about cell phones?

This will be simple we just need to look at our tree (which i currently don't have up so the change to this question will be made once i clear up my problem.)

b)You see John looking at a website that is not about cell phones. What is the probability he used Google to find it?

P(G/NCP)= (P(GNCP))/(P(GNCP)+ P(LNCP)+P(ANCP)
=(30/100)/((30/100)+(21/100)+(14/100))
=(30/100)/(65/100)
=(30/100)X(100/65)
=(3000/6500)
=(3/65)
=4.6%

Therefore the probability he used Google is 4.6%

*P(G/NCP) meaning: prob. used Google but not for cell phones.
*P(GNCP) meaning: prob. used Google not for cell phones.
*P(LNCP) meaning: prob. used Lycos not for cell phones.
*P(ANCP) meaning: prob. not used Altavista.

3)The letters AADKRRV are written on separate pieces of paper and placed in a bag. If they are drawn from the bag at random, what is the probability that the order of the letters spell AARDVARK?

There where many ways to calculate this problem. I will show you how I have completed the question.

First I reflected on the unit on Counting, and the question became simple. A method we where tought involved lines and multiplying the factorials that go in them. Then I applied it to my knowledge of going about solving probability questions. I came up with this:

3!/8 chances of A 1st.
2!/7 chances of A 2nd.
1!/6 chances of R 3rd.
1!/5 chances of D 4th.
1!/4 chances of V 5th.
1!/3 chances of A 6th.
1!/2 chances of R 7th.
1!/2 chances of K 8th.

Then I simplified the answer and this is what I got:

(3!2!)/(8!)

4)Two traffic lights operate independently. The probability that the first is red is 0.4. The prob. that the second is red is 0.7. What's the probability that neither are red?

P(NR1)= 0.6
P(NR2)= 0.3

(0.6)(0.3)=.18

Unfortunately the Big Space has come back,heh, I was able to kinda fix but its still there.

THE NEXT SCRIBE IS THE JESSICA THAT HAS'NT DONE 3 SCRIBES YET.

January 08, 2007

Scribe Post

First day of class after the winter break and of course not much happened today, as we went through a few questions which are listed below. Other then that we discussed more about probability.

Suppose a test for cancer is 98% accurate. This means that the result of the test is correct 98% of the time. Suppose that 0.5% of the population has cancer. What is the probability that a person who tests positive has cancer? Suppose 1, 000, 000 randomly selected people are tested. There are 4 possibilities.

• A person with cancer tests positive
• A person with cancer tests negative
• A healthy person tests positive
• A healthy person tests negative
1) a) How many of the people tested have cancer? (1, 000, 000)(.005) = 5000
b) How many do not have cancer? 1, 000, 000 - 5000 = 995, 000

2) Assume the test is 98% accurate for people with cancer
a) How many people with cancer test positive? (5000)(.98) = 4900
b) How many people with cancer test negative? 5000 - 4900 = 100

3) Assume the test is 98% accurate for people without cancer
a) How many people without cancer test positive? 995, 000 - 975, 100 = 19, 900
b) How many people without cancer test negative? (995, 000)(.98) = 975, 100

4) a) How many people tested positive for cancer? 4900 + 19, 900 = 24, 800
b) How many of these people do have cancer? 4900
c) What is the probability that a person who tested positive for cancer actually does have cancer?

P(C|P) = 4900/(4900 + 19900)

P(P) = 0.0049 + 0.0199

= 0.0248

P(C|P) = P(CP)/(P(CP) + P(HP))

= 0.0049/0.0248

= 0.1976

P(H|N) = P(HN)/(P(CN) + (PHN))

= 0.9751/(0.975 + 0.0001)

= 0.9999

I'm not too sure about what's going to happen tomorrow since I believe that a lot of people from class will be taking the English Provincial Exams so unless the scribe will be skipped to Wednesday the next person is
Gerald.

Peter Donnelly

Oxford statistician Peter Donnelly explores the common mistakes humans make in interpreting statistics, and the devastating impact these errors can have on the outcome of criminal trials.

You learned the math he's talking about here today.

Click on the picture. (22 min. 6 sec.) January 05, 2007

Scribe post

Probability

1st period class

it was the day before the Holiday break so we didn't do much but as usually we started the day by doing question......^_^

Blood types in North America are distributed as follows:

Type O: 44% Type A: 42% Type B: 10% Type AB: 4%

a) are the blood types mutually exclusive?

b)what is the probability that a couple both have type O blood?

a) no (because the blood type can be both be "A" "B" or "AB" and "O" is the universal blood Donner meaning that it can fit into any blood type category)

b) "tree diagram"!! HO = "has blood type "O" NO = No blood type "O" the first tree branch representing one of the couple has 0.44 chance of getting blood type "O" and the second branch representing the second couple has 0.44 chance getting type "O" also. we use "Multiplying Probability" since we want to find out what is the probability that both couples have the same type "O". (HO)HO) = Santa clause? HO NO HO NO I'm just kidding the answer is (0.44)(0.44) = .1936 or 19% that both of them have the same blood type "O".

2nd period class

In a group of 30 students what is is the probability that at least two have the same birthday?

*find the probability that nobody else have the same birthday

* use Pick formula since the order matters

*(# of days in a year)P(# of students) divided by (# of days in a year)^(# of students)

365P30/365^30 = 0.26 then subtract it from 1 ( 1- 0.26 =74%)

therefore the probability of at least two students have the same birthday is 74% (3o students)

and then one of my classmate asked him if our answer for this question was really true....in a 30 people group is there really a 74% chance of picking at least two people that have the same birthday? but most of our classmates are "missing"*wink**wink* so he started getting students on the hall way to make a group of 30 students and then he asked us one by one our birthdays but in our 30 group of students no one has the same birthday so he went back on the hallway to get some more people to raise the chances............were like HO HO HO HO laughing but we forgot to make a list and everyone was like HO NO HO NO no one made a list oi that was fun..............

sorry I couldn't get all the notes for this question :)

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THE SCRIBE IS TENNYSON .....................FOR MONDAY