Wilred & Alvin's Math Project '08

DEVELOPING
EXPERT VOICES
2008
APPLIED MATH 40S

Please click on this link

http://youtube.com/watch?v=sSvE_dG7BGk

PROB
ABIL &
TIST ICS
ITY STA

Flickr by: bcsia Flickr by: mac steve

PR
OB
AB I CS
ILI I ST
TY A T
ST
&
What is Probability? What is Statistics?
a measure of how likely it is that a mathematical science relevant
some event will occur; a number to the collection, analysis,
expressing the ratio of favorable interpretation or explanation, and
cases to the whole number of presentation of data.
cases possible.

Flickr by: bcsia Flickr by: mac steve

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.

http://www.onlinestopwatch.com/bombcountdown/fullscreen/
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 timer
to 15 minutes only. When time runs out
then you're out of luck and the last laugh will be
mine.

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?
T SPAWN

BOMBSITE

a.) How many ways can you go from
T SPAWN to the Bombsite?
T SPAWN 1 1 1 1 1 1

The only way to 2 3 4 5 6 7
solve this 1

problem is by
using Pascal's
3
Triangle. 6 10 15 21 28
1

4 10 20 35 36 64
1

5 15 35 70 106 170
1

BOMBSITE
6 21 56 232 402
126
1

T SPAWN 1 1 1 1 1 1

solve this 1

problem is by
using Pascal's
Triangle. 3 10 15
6 21 28
1
Pascal's Triangle
works by
adding the top
4 10 20 35 36 64
square from the 1
square that you
want to figure
out by the left 5 15 35 70 106 170
side's number. 1

BOMBSITE
6 21 56 232 402
126
1

T SPAWN 1 1 1 1 1 1

solve this 1

problem is by
using Pascal's
Triangle. 3 10 15
6 21 28
1
Pascal's Triangle
works by
adding the top
4 10 20 35 36 64
square from the 1
square that you
want to figure
out by the left 5 15 35 70 106 170
side's number. 1

BOMBSITE
6 21 56 232 402
126
1

*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

CT SPAWN

BOMBSITE


1 1 1
T SPAWN 1 1 1

2 3 4 5 6 7
1

3
6 11 18
1 5

4 10 10 15 26 44
1

5 15 25 40 66 110
1

BOMBSITE
6 21 51 152 262
86
1


1 1 1
T SPAWN 1 1 1
The enemy's territory is
in blue. To do this you
need to use the Pascal's 2 3 4 5 6 7
1
Triangle. This time we
got 10 in red font by just
copying the 10 below 3
6 11 18
the 6 1 5

4 10 10 15 26 44
1

5 15 25 40 66 110
1

BOMBSITE
6 21 51 152 262
86
1


1 1 1
T SPAWN 1 1 1
1
6 11 18
the 6 1 5

Now we do the same 15
thing for the number 5 1 4 10 10 26 44

since we're considering
this whole BLUE box as
one SQUARE so we 5 15 25 40 66 110
just add them up and 1

use the Pascal's triangle. BOMBSITE
6 21 51 152 262
86
1


1 1 1
T SPAWN 1 1 1
1
6 11 18
the 6 1 5


this whole BLUE box
5 66 110
as one SQUARE so we 15 25 40
1
just add them up and
6 21 51 152 262
86
1

* 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
T SPAWN 1 1 1
1
6 11 18
the 6 1 5


this whole BLUE box
5 66 110
as one SQUARE so we 15 25 40
1
just add them up and
6 21 51 152 262
86
1

* 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.

c.) What is the probability that you will not walk in
the enemy's territory? 1 1 1 1 1 1

2 3 4 5 6 7
1

262   131 1
3
6 10 15 21 28

=
402   201 4 10 20 35 36 64

5 15 35 70 106 170
1

BOMBSITE
6 21 56 232 402
126

T SPAWN 1 1 1 1 1 1

262   131 1
2 3 4 5 6 7

= CT SPAWN
3
6 11 18

402   201
1 5

4 10 10 15 26 44
1

* For this problem you just need to 1
5 15 25 40 66 110

simplify 262 and 402 to get the probability
BOMBSITE
6 21 51 152 262
86
1

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.
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
a.) Calculate the 95 percent confidence interval
for the number of people carrying a grenade.
±1.96 = .9500
n = 20 ±1.97 = .9505
p= .25
q= .75 N = Number of trials
m= np = 20 = 5 P= probability of success
nq= 15 Q= probability of failure
Dist. is approx. normal m= np= means number of trials x the probability
that they carry a grenade.
20(.25)(.75)

= 1.9364
Sigma sign (a.k.a)
Standard Deviation

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

±1.95 = .9488
for the number of people carrying a grenade. ±1.96 = .9500
n = 20 ±1.97 = .9505
p= .25
q= .75 5 1.96(1.94) = 1.2
m= np = 20 = 5 5 + 1.96 (1.94= 8.8
nq= 15
Dist. is approx. normal

20(.25)(.75) Remember: The magic number for
= 1.9364 finding the 95% confidence
interval is 1.96

±1.95 = .9488
for the number of people carrying a grenade.
±1.96 = .9500
n = 20 ±1.97 = .9505
p= .25
q= .75 5 1.96(1.94) = 1.2
m= np = 20 = 5 5 + 1.96 (1.94= 8.8
nq= 15
Dist. is approx. normal

20(.25)(.75) Remember: The magic number for
a 95% confidence interval is 1.96
= 1.9364

Therefore, confidence interval for the
number of people carrying a grenade is
1.2 to 8.8

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.

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

1.2 to 8.8 confidence interval in percent
1.2 = .06x100
20 = 6% (6 , 44) is the confidence
interval for the percent of
8.8 = .44x100 people carrying a grenade
20 = 44%

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.

c.) Calculate the number margin of error.

±1.96(1.94) = 3.80

* 3.80 is the number of margin error

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.

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

in a row.

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


in a row.

guns and how they are lined up.
Now, looking at quot;Gun 1quot;. Since there are 6 players
Gun 1 Gun 2 Gun 3 Gun 4 Gun 5 Gun 6 that are currently without a weapon. It means that
anyone of them can have the gun.

in a row.

6 guns and how they are lined up.
Now, looking at quot;Gun 1quot;. Since there are 6 players
Gun 1 Gun 2 Gun 3 Gun 4 Gun 5 Gun 6 that are currently without a weapon. It means that
anyone of them can have the gun.

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

in a row.

6

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

in a row.
6 5 5 symbolizes how many players have a chance on
Gun 1 Gun 2 Gun 3 Gun 4 Gun 5 Gun 6 aquiring the second gun.

in a row.
6 5 5 symbolizes how many players have a chance on

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

in a row.
6 5 4 5 symbolizes how many players have a chance on

The same thing will also be done in quot;Gun 3quot;, since 2
players now have a gun, the remaining 4 players now have
the chance to pick the third gun.
Therefore, the number 4 symbolizes how many players
are left to have the chance to aquire the third gun.

in a row.

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

in a row.

6 5 4 As the players choose there weapons, fewer and
...lets go ahead and fill in the rest.

in a row.

6 5 4 3 As the players choose there weapons, fewer and

3 on Gun 4

in a row.

6 5 4 3 2 As the players choose there weapons, fewer and

3 on Gun 4
2 on Gun 5

in a row.

6 5 4 3 2 1 As the players choose there weapons, fewer and

3 on Gun 4
2 on Gun 5

...and 1 on Gun 6

in a row.

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

in a row.


...multiplying them will give us...

in a row.



Therefore there are 720 ways that they can
be arranged!

in a row.

6 x 5 x 4 x3 2
x 1x be positioned after picking up there guns, we will
have to multiply them.
Another way of solving this problem is by using Factorial Notation.

= there are 720 ways that they can be arranged!

a row.

6 x 5 x 4 x3 2

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

a row.

6 x 5 x 4 x3 2

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

a row.

6 x 5 x 4 x3 2

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

a row.

6 x 5 x 4 x3 2

6! = 6 x 5 x 4 x 3 x 2 x 1 = 720
The quot;!quot; 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]

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.

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?

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.

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.

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.

H = Heads
T = Tails

H

T

For example, this tree represents 3 coins.

Diagram.


H = Heads
H
T = Tails

H T

H
T
T


Diagram.

H

H = Heads T
H
T = Tails
H
H T
T
H
H
T T
T H
T

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

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.
H
H = Heads H T
T = Tails
H
1st coin H T
T
H
H
T T
T H
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.

Diagram.

H
H = Heads T
2nd coin H
T = Tails
H
1st coin H T
T
H
H
T T
T H
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.

Diagram.

3rd coin H
H = Heads H T
2nd coin
T = Tails
H
1st coin H T
T
H
H
T T
T H
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.


event. HHH
H
T HHT
H = Heads H
T = Tails HTH
H
H T HTT
T
H THH
H
T T THT
T H
TTH
T
TTT

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


event. HHH
H
T HHT
H = Heads H
T = Tails HTH
H
H T HTT
T
H THH
H
T T THT
T H
TTH
T
TTT


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


event. HHH
H
T HHT
H = Heads H
T = Tails HTH
H
H T HTT
T
H THH
H
T T THT
T H
TTH
T
TTT


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


event. HHH
H
T HHT
H = Heads H
T = Tails HTH
H
H T HTT
T
H THH
H
T T THT
T H
TTH
T
TTT

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 quot;Hquot;s on them.


event. HHH
H
HHT
H = Heads
T = Tails
H T
HTH
3/8
H
H T HTT
T
H THH
H
T T THT
T H
TTH
T
TTT

Once we have the number of them with 2 quot;Hquot;s, we then divide it by the number of all the outcomes.


event. HHH
H
HHT
H = Heads
T = Tails
H T
HTH
3/8
H
H T HTT
T

H
H THH = .375
T T THT
H
T
T
TTH =37.5%
TTT

Once we have the number of them with 2 quot;Hquot;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.

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

Kyle
Allan
Kyle
Allan
Kyle
Rob
Kyle
Rob
Rob Allan
Rob
Allan Allan Rob
Kyle Kyle
Allan
Kyle Allan
Jason Jason Kyle
Kyle Jason
Allan
Kyle
Allan Jason
Jason Jason
Allan
Rob Kyle
Rob
Allan Kyle Rob
Kyle Jason Kyle
Jason
Rob Kyle
Jason
Kyle Rob Jason
Jason
Rob
Rob Allan
Jason Allan
Rob
Rob Jason
Allan
Allan Allan
Rob
Jason
Jason Jason
Rob

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 Kyle
second. Allan
Kyle
Allan
Kyle
Rob
Kyle
Rob
Rob Allan
Rob
Allan Allan Rob
Kyle Kyle
Allan
Kyle Allan
Jason Jason Kyle
Kyle Jason
Allan
Kyle
Allan Jason
Jason Jason
Allan
Rob Kyle
Rob
Allan Kyle Rob
Kyle Jason Kyle
Jason
Rob Kyle
Jason
Kyle Rob Jason
Jason
Rob
Rob Allan
Jason Allan Rob
Rob Jason Allan
Allan Allan
Rob
Jason
Jason Jason
Rob

chosen peoples names on the first column or/and Kyle
Allan
a.)
Allan
second. Kyle
Rob
Kyle
Rob
2/24
Kyle
Rob Allan = .08
Rob
Allan Allan Rob = 8%
Kyle Kyle
Allan
Kyle Allan
Jason Jason Kyle
Kyle Jason
Allan
Kyle
Allan Jason
Jason Jason
Allan
Rob Kyle
Rob
Allan Kyle Rob
Kyle Jason
Jason Kyle
Rob Kyle
Jason
Kyle Rob
Jason
Jason
Rob
Jason
Rob Allan
Allan
Rob Rob
Jason
Allan Allan
Allan
Rob Jason
Jason Jason
Rob

chosen peoples names on the first column or/and Kyle a.)
second. Allan Allan
Kyle

Rob
Kyle
Rob
2/24
Kyle
Rob Allan = .08
Rob
Allan Allan Rob = 8%
Kyle Kyle
Allan
Kyle Allan
Jason Jason Kyle
Kyle Jason
Allan
Kyle
Allan Jason
Jason Jason
Allan
Rob Kyle
Rob
Allan Kyle Rob
Kyle Jason Kyle
Jason
Rob Kyle b.)
Jason
Kyle Rob
Jason
Jason
Rob
1/24
Rob Allan = .04
Jason Allan
Rob = 4%
Rob Jason
Allan
Allan Allan
Rob
Jason
Jason Jason
Rob

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.

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?

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

• P(S) = p • P(F) = q = 1 p

• 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 a
defective pistol.

A pistol factory produces 42 pistols per month. Two percent of all
So how do we
the pistols are defective. What is the probability of getting four
answer now? I
need help...
defective pistols in one week?

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

• P(S) = p • P(F) = q = 1 p

• 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 a
defective pistol.


Binomial PD.
Data: Variable
X = 4
Num.Trial=42
P=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: Variable binompdf(42,0.06,5)
X = 4 .0083110781
Num.Trial=42
P=0.02

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


Binomial PD.
Data: Variable binompdf(42,0.06,5)
X = 4 .0083110781
Num.Trial=42
P=0.02

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

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

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.

FINAL CHALLENGE:(BOMB DEFUSING)

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.

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

2 8 7 4 3

Wilred & Alvin's Math Project '08

Wilred & Alvin's Math Project '08

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