Wilred & Alvin's Math Project '08

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APPLIED MATH 40S

DEVELOPING EXPERT VOICES

2008

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PROBABILITY&

STATISTICS

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PROBABILITY& STATISTICS

What is Statistics?What is Probability?a measure of how likely it is that some event will occur; a number expressing the ratio of favorable cases to the whole number of cases possible.

a mathematical science relevant to the collection, analysis, interpretation or explanation, and presentation of data.

Good evening CounterTerrorist. I have commanded my terrorist team to plant a bomb somewhere in your training facility. The only way to defuse the bomb is to answer these 5 questions that will help you figure out the 5 digit codes.

Good evening CounterTerrorist. I have commanded my terrorist team to plant a bomb somewhere in your training facility. The only way to defuse the bomb is to answer these 5 questions that will help you figure out the 5 digit codes.Remember... if the bomb has been successfully defused, then we will surrender our weapons to you. However, if you fail to defuse it, the victory will be ours. You will surrender the rest of your weapons and you will be in our control.

Click on the link on top and set your timerto 15 minutes only. When time runs outthen you're out of luck and the last laugh will be mine.

http://www.onlinestopwatch.com/bombcountdown/fullscreen/

http://www.online-stopwatch.com/bomb-countdown/full-screen/

CHALLENGE #1

1. In the game of Counter Strike there are two teams battling against each other. The Counter Terrorist and the Terrorist. In this case the Terrorist need to plant the bomb at the BOMBSITE. In order to do this they need to take short routes to reach their destination and plant the bomb. So now the terrorists need to find the shortest route.

a.) How many ways can you go to T SPAWN to the Bombsite?

b.)How many ways can you get to the bombsite without going through the enemies territory?

c.) What is the probability that you will not walk through the enemy territory?

a.) How many ways can you from T SPAWN to the Bombsite?

BOMBSITE

T SPAWN

a.) How many ways can you go from T SPAWN to the Bombsite?

1

1 1 1

1

1

1

1

1 1 1

2 3

3

4

5

6

10

15

21 56

35

4 5

70

126

6 7

21

64

6

T SPAWN

BOMBSITE

28

36

106 170

232 402

10 15

20 35

The only way to solve this problem is by using Pascal's Triangle.


1

1 1 1

1

1

1

1

1 1 1

2 3

3

4

5

6

10

15

21 56

35

4 5

70

126

6 7

21

64

6

T SPAWN

BOMBSITE

28

36

106 170

232 402

10 15

20 35

The only way to solve this problem is by using Pascal's Triangle.Pascal's Triangle works by adding the top square from the square that you want to figure out by the left side's number.


1

1 1 1

1

1

1

1

1 1 1

2 3

3

4

5

6

10

15

21 56

35

4 5

70

126

6 7

21

64

6

T SPAWN

BOMBSITE

28

36

106 170

232 402

10 15

20 35

The only way to solve this problem is by using Pascal's Triangle.Pascal's Triangle works by adding the top square from the square that you want to figure out by the left side's number.

*You have 402 ways to get from T SPAWN to the BOMBSITE

b.)How many ways can you get to the bombsite without going in the enemy's territory?

T SPAWN

BOMBSITE

CT SPAWN


1 1 1 1 11T SPAWN

1

1

1

1

1

2 3

3

4

5

6

10

15

21 51

25

4 5

40

86

6 7

11

44

6

BOMBSITE

18

26

66 110

152 262

1510

5


1 1 1 1 11T SPAWN

1

1

1

1

1

2 3

3

4

5

6

10

15

21 51

25

4 5

40

86

6 7

11

44

6

BOMBSITE

18

26

66 110

152 262

1510

5

The enemy's territory is in blue. To do this you need to use the Pascal's Triangle. This time we got 10 in red font by just copying the 10 below the 6


1 1 1 1 11T SPAWN

1

1

1

1

1

2 3

3

4

5

6

10

15

21 51

25

4 5

40

86

6 7

11

44

6

BOMBSITE

18

26

66 110

152 262

1510

5


Now we do the same thing for the number 5 since we're considering this whole BLUE box as one SQUARE so we just add them up and use the Pascal's triangle.


1 1 1 1 11T SPAWN

1

1

1

1

1

2 3

3

4

5

6

10

15

21 51

25

4 5

40

86

6 7

11

44

6

BOMBSITE

18

26

66 110

152 262

1510

5



* Therefore, after you use the Pascal's triangle you should get an answer of 262 ways of going to the BOMBSITE without going in the enemies territory.


1 1 1 1 11T SPAWN

CT SPAWN

1

1

1

1

1

2 3

3

4

5

6

10

15

21 51

25

4 5

40

86

6 7

11

44

6

BOMBSITE

18

26

66 110

152 262

1510

5



* Therefore, after you use the Pascal's triangle you should get an answer of 262 ways of going to the BOMBSITE without going in the enemies territory.

CT SPAWN

1

1 1 1

1

1

1

1

1 1 1

2 3

3

4

5

6

10

15

21 51

25

4 5

40

86

6 7

11

44

6

T SPAWN

BOMBSITE

18

26

66 110

152 262

1510

5

c.) What is the probability that you will not walk in the enemy's territory?

1 1 1 1 1 1

1

1

1

2 3

3

4

5

6

10

15

21 56

35

4 5

70

126

6 7

21

64

6

BOMBSITE

28

36

106 170

232 402

10 15

20 35

262

402

131= 201

262

402

131= 201

* For this problem you just need to simplify 262 and 402 to get the probability

CHALLENGE #2

2. There are 20 CounterTerrorists all around the map. 5 of themcarry a grenade, which is 25 percent, and the rest of them do not have enough money to buy some.

a.) Calculate the 95 percent confidence interval for the number of people carrying a grenade.

b.) Calculate the 95 percent confidence interval for the percent of people carrying a grenade.

c.) Calculate the percent margin of error.

2. There are 20 CounterTerrorists all around the map. 5 of them carry a grenade, which is 25 percent, and the rest of them do not have enough money to buy some.


±1.95 = .9488±1.96 = .9500±1.97 = .9505n = 20

p= .25q= .75m= np = 20 = 5nq= 15

Dist. is approx. normal

20(.25)(.75)

= 1.9364

N = Number of trialsP= probability of successQ= probability of failurem= np= means number of trials x the probability that they carry a grenade.

Sigma sign (a.k.a) Standard Deviation

±1.96 = .9500 The Magic Number for finding the 95% confidence interval



n = 20p= .25q= .75m= np = 20 = 5nq= 15


20(.25)(.75)

= 1.9364Remember: The magic number for finding the 95% confidence interval is 1.96

5 1.96(1.94) = 1.25 + 1.96 (1.94= 8.8

±1.95 = .9488±1.96 = .9500±1.97 = .9505



n = 20p= .25q= .75m= np = 20 = 5nq= 15


20(.25)(.75)

= 1.9364

Remember: The magic number for a 95% confidence interval is 1.96

5 1.96(1.94) = 1.25 + 1.96 (1.94= 8.8

Therefore, confidence interval for the number of people carrying a grenade is1.2 to 8.8

±1.95 = .9488±1.96 = .9500±1.97 = .9505


b.) Calculate the 95 percent confidence interval for the percent of people carrying a grenade.

1.2 to 8.8 confidence interval in percent1.220

= .06x100 = 6%

8.820

= .44x100 = 44%

(6 , 44) is the confidence interval for the percent of people carrying a grenade


c.) Calculate the number margin of error.

±1.96(1.94) = 3.80

* 3.80 is the number of margin error

CHALLENGE #3

In the counterterrorist's side. Leet, Chet, Montrose, Cortez, Ahmad and Player. There are 6 different kinds of weapons on the floor in a row.


a.) How many ways can they arrange themselves if they don't move from the place where they picked up there guns.

Gun 1 Gun 2 Gun 3 Gun 4 Gun 5 Gun 6




First off, we lay down 6 underlines symbolizing the guns and how they are lined up.




First off, we lay down 6 underlines symbolizing the guns and how they are lined up.Now, looking at "Gun 1". Since there are 6 players that are currently without a weapon. It means that anyone of them can have the gun.




First off, we lay down 6 underlines symbolizing the guns and how they are lined up.Now, looking at "Gun 1". Since there are 6 players that are currently without a weapon. It means that anyone of them can have the gun.

6

Then the number over it will be a 6, symbolizing that 6 players have a chance to aquire "Gun 1".




6




6Now for the "Gun 2", since there are only 5 players who do not have a gun yet, the number on it shall be 5.





5 symbolizes how many players have a chance on aquiring the second gun.

5






5

The same thing will also be done in "Gun 3", since 2 players now have a gun, the remaining 4 players now have the chance to pick the third gun.






5

The same thing will also be done in "Gun 3", since 2 players now have a gun, the remaining 4 players now have the chance to pick the third gun.

4

Therefore, the number 4 symbolizes how many players are left to have the chance to aquire the third gun.




6 5 4 As the players choose there weapons, fewer and fewer selections are left for the remaining players.





...lets go ahead and fill in the rest.






3

3 on Gun 4






3

3 on Gun 4

2

2 on Gun 5






3

3 on Gun 4

2

2 on Gun 5

...and 1 on Gun 6

1




6 5 4 3 2 1 Now to find out how many ways the players can be positioned after picking up there guns, we will have to multiply them.

x x x x x





x x x x x

...multiplying them will give us...





x x x x x

...multiplying them will give us...

Therefore there are 720 ways that they can be arranged!





x x x x x

...multiplying them will give us...= there are 720 ways that they can be arranged!

Another way of solving this problem is by using Factorial Notation.





x x x x x



Factorial Notation is used when we want to multiply all natural numbers from a particular number down to 1. Like so...





x x x x x




6! = 6 x 5 x 4 x 3 x 2 x 1 = 720





x x x x x




6! = 6 x 5 x 4 x 3 x 2 x 1 = 720The "!" is the Factorial Symbol. You can find this on your TI83 calculator by following this button sequence.





x x x x x




6! = 6 x 5 x 4 x 3 x 2 x 1 = 720The "!" is the Factorial Symbol. You can find this on your TI83 calculator by following this button sequence.

[MATH] > Press the left arrow to move to the far right (Prb) > [4]

CHALLENGE #4

Jason, Rob, Allan and Kyle were in a marksman competition. General Rad made a bet that Jason and Rob will be in the top 2. General McArthur, on the other hand, made a bet that Allan will be in first place and Kyle on second place. They all have the same level of skills.


a.) What is the probability that General Rad will win the bet?


b.) What is the probability that General McArthur will win the bet?

To find the answers for this question, we will have to construct a Tree Diagram.


A tree diagram is used to show all of the possible outcomes or combinations of an event.

For example, this tree represents 3 coins.

H

T

H = HeadsT = Tails




H

T

H

H

T

T

H = HeadsT = Tails




H

H

T

H

H

T

T

T

T

T

T

H

H

H

You can tell that there are 3 coins because there are 3 columns of branches spread out after another.

H = HeadsT = Tails




H

T

H

H

T

T

T

T

T

T

H

H

H

H

You can tell that there are 3 coins because there are 3 columns of branches spread out after another.The RED branches represent the first coin.

H = HeadsT = Tails

1st coin




You can tell that there are 3 coins because there are 3 columns of branches spread out after another.

The RED branches represent the first coin.

BLUE the second.

H = HeadsT = Tails

H

T

H

H

T

T

T

T

T

T

H

H

H

H

1st coin

2nd coin




T

You can tell that there are 3 coins because there are 3 columns of branches spread out after another.The RED branches represent the first coin. BLUE the second. GREEN the third.

H = HeadsT = Tails

H

T

H

H

T

T

T

T

T

H

H

H

H

1st coin

2nd coin

3rd coin



H

T

H

H

T

T

T

T

T

T

H

H

H

H

HHH

HHT

HTH

HTT

THH

THT

TTH

TTT

H = HeadsT = Tails

Now, to find out the all the different outcomes of flipping 3 coins...



H

T

H

H

T

T

T

T

T

T

H

H

H

H

HHH

HHT

HTH

HTT

THH

THT

TTH

TTT

H = HeadsT = Tails


...all we have to do is trace the branches like so...



H

T

H

H

T

T

T

T

T

T

H

H

H

H

HHH

HHT

HTH

HTT

THH

THT

TTH

TTT

H = HeadsT = Tails


...all we have to do is trace the branches like so...

...and so on.



H

T

H

H

T

T

T

T

T

T

H

H

H

H

HHH

HHT

HTH

HTT

THH

THT

TTH

TTT

H = HeadsT = Tails

If we want to find out what the probability is that 2 out of 3 coins will be heads, all we have to do is check how many of the outcomes have 2 "H"s on them.



H

T

H

H

T

T

T

T

T

T

H

H

H

H

HHH

HHT

HTH

HTT

THH

THT

TTH

TTT

H = HeadsT = Tails


3/8

Once we have the number of them with 2 "H"s, we then divide it by the number of all the outcomes.



H

T

H

H

T

T

T

T

T

T

H

H

H

H

HHH

HHT

HTH

HTT

THH

THT

TTH

TTT

H = HeadsT = Tails


3/8

Once we have the number of them with 2 "H"s, we then divide it by the number of all the outcomes.

When you have the answer, move the decimal down by 2 numbers, and that should give you the percentage.

= .375=37.5%

JasonRobAllanKyle

RobAllanKyle

JasonAllanKyle

JasonRobKyle

JasonRobAllan

AllanKyle

RobAllan

RobKyle

AllanKyleJasonKyle

AllanJason

RobKyle

JasonKyle

RobJason

RobAllanJasonAllanRobJason

KyleAllanKyleRobAllanRobKyleAllanKyleJasonJasonAllanKyleRobKyleJasonJasonRobAllanRobAllanJasonJasonRob

a.) What is the probability that General Rad will win the bet?

JasonRobAllanKyle

RobAllanKyle

JasonAllanKyle

JasonRobKyle

JasonRobAllan

AllanKyle

RobAllan

RobKyle

AllanKyleJasonKyle

AllanJason

RobKyle

JasonKyle

RobJason



Now thanks to the Tree diagram, We can clearly see the number of outcomes.

All we really have to look at the branches that has the chosen peoples names on the first column or/and second.

AllanKyle


All we really have to look at the branches that has the chosen peoples names on the first column or/and second.

JasonRobAllanKyle

RobAllanKyle

JasonAllanKyle

JasonRobKyle

JasonRobAllan

RobAllan

RobKyle

AllanKyleJasonKyle

AllanJason

RobKyle

JasonKyle

RobJason



2/24= .08= 8%

a.)

JasonRobAllanKyle

RobAllanKyle

JasonAllanKyle

JasonRobKyle

JasonRobAllan

AllanKyle

RobAllan

RobKyle

AllanKyleJasonKyle

AllanJason

RobKyle

JasonKyle

RobJason




All we really have to look at the branches that has the chosen peoples names on the first column or/and second. 2/24

= .08= 8%

a.)

b.)

1/24= .04= 4%

CHALLENGE #5

A pistol factory produces 42 pistols per month. Two percent of all the pistols are defective. What is the probability of getting four defective pistols in one week?

The above problem is an example of a binomial probability experiment. Some properties of a binomial experiment are:

1. There are set number of trials in this experiment. In this case, there are 42 trials



2. Each trial has exactly two possible outcomes: pistol fires pistol fails



3. Events are independent. This means that the probability of success is the same for each trial. i.e., each pistol has a 2 percent probability of being defective.



4. We will be looking for the probability of successes. i.e. P(four defective alarms) = ?



5. The total of all probabilities is 1.This means that the pistol works or fails, and that no other possibility exists.



6. The data in a binomial problem are always discrete. In this case, it means that the number of pistols is an integral value (1,2,3,4, et cetera) which can be counted, and is not a continuous (i.e., measured) value, as is the case in a normal distribution.


'S' and 'F' (Success and Failure) are the possible outcomes of atrial in a binomial experiment, and 'p' and 'q' represent theprobabilities for 'S' and 'F.'

• n = the number of trials• x = the number of successes in n trials• p = probability of success• q = probability of failure• P(x) = probability of getting exactly x successes in n trials

Remember that 'Success' in this case, is the probability of selecting adefective pistol.

• P(S) = p • P(F) = q = 1 p


'S' and 'F' (Success and Failure) are the possible outcomes of atrial in a binomial experiment, and 'p' and 'q' represent theprobabilities for 'S' and 'F.'

• n = the number of trials• x = the number of successes in n trials• p = probability of success• q = probability of failure• P(x) = probability of getting exactly x successes in n trials

Remember that 'Success' in this case, is the probability of selecting adefective pistol.

• P(S) = p • P(F) = q = 1 p

So how do we answer now? I need help...


Binomial PD. Data: VariableX = 4Num.Trial=42P=0.02 WE DO THIS BY USING THE CALCULATOR FUNCTION

BINOMPDF

Press: 2nd Func (VARS) press the number (0) on your calculator then enter in the following in order

binompdf(42,0.02,4) then lastly press ENTER and you'll see the probability of getting four defective pistols in one week


Binomial PD. Data: VariableX = 4Num.Trial=42P=0.02

binompdf(42,0.02,4)=0.0083 x 100=0.83 %

binompdf(42,0.06,5) .0083110781


Binomial PD. Data: VariableX = 4Num.Trial=42P=0.02

binompdf(42,0.02,4)=0.0083 x 100=0.83 %

binompdf(42,0.06,5) .0083110781

∴ The probability of getting four defective pistols in one week is 0.83%

FINAL CHALLENGE:(BOMB DEFUSING)

To find out what the 5 digit codes are, you will need to figure out the riddle of the whole presentation. You must be able to answer the codes correctly in order for the bomb to be defused.

1. What is 8*8 divided by 2 five times?



2. What do you call the black ball in the game of billiards that is part of the first question?



3. What is Mr. K's favourite number + add .5 and .5 five and you get the third code?



4. Mr.K's cat passed away in 2004 by deleting the first 3 digits what do you get as how many cats does Mr.K have left?



5. Mr.K's favourite number divide it by 3 and add one to it.

2 8 7 4 3

IF YOU GET THESE CODES CORRECTLY YOUR MISSION IS DONE AND THERE WILL BE NO MORE TERRORISM IN THIS WORLD.

Wilred & Alvin's Math Project '08

Technology

Transcript of Wilred & Alvin's Math Project '08