counting techniques and probability concepts

8/3/2019 Counting Techniques and Probability Concepts

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Counting Techniques

Probability Concepts

Antic, EricaBuquel, Marvilyn

Gepila, Gianina Moira

Macaraig, Romeo

Tajima, Sachi

Ysla, Kevin

Business Statistics

TF 2:30-4:00pm

H-358


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PERMUTATIONS

COMBINATIONS

AND

COUNTING TECHNIQUES


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Computer Science, Statistics and Probability all involvecounting techniques which are a branch of mathematicscalled combinatorics (ways to combine things). We'll be

introducing this topic in this section.For dinner you have the following choices:

soup salad chicken hamburgerprawns

icecream

ENTREES MAINS

DESSERTS

How many different combinations of mealscould you make?

We'll build a tree diagram to show all of

the choices.
http://www.jledu.com.cn/kbsc/graph/enghish/lt2/unit5/salad.jpg


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prawns

prawns

Now to get all possible choices wefollow each path.

soup, chicken, ice cream

soup, chicken,

soup, prawns, ice cream

soup, prawns,

soup, hamburger, ice cream

soup, hamburger,

salad, chicken, ice cream

salad, chicken,

salad, prawns, ice cream

salad, prawns,

salad, hamburger, ice cream

salad, hamburger,

Notice the number of choices at each branch

2

choices

3

choices

2

choices

We ended up with 12possibilities

2 3 2 = 12


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Multiplication Principle of CountingIf a task consists of a sequence of choices in which there are p

selections for the first choice, qselections for the second choice, rselections for the third choice, and so on, then the task of makingthese selections can be done in different ways.

pqr

If we have 6 different shirts, 4 different pants, 5different pairs of socks and 3 different pairs of

shoes, how many different outfits could wewear?

6 4 5 3 =360


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How many 4 digit can be formed onthe 26 letters of the english alphabet?

a. No repetition is allowed

b. Repitition is allowed

a. no. of ways = 26x25x24x23

= 358,800

b. no. of ways = 26x26x26x26= 456,976


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Factorial

The factorial of any integer n isdenoted by n! and is defined as:

n!=n(n-1)(n-2), 1

O! is defined = 1

The value of n! which may be readfactorial n or the factorial of n, is

obtained by multiplying all the integersfrom 1 to n.


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7! = 7.6.5.4.3.2.1

= 5040

2! 7! =(2.1)(7.6.5.4.3.2.1)

=(2)(5040)

10!+9!+9! =(10.9.8.7.6.5.4.3.2.1) +(9.8.7.6.5.4.3.2.1 ) +(9.8.7.6.5.4.3.2.1)

=3,628,800 + 362,880 +362,880

=4,354,560


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A permutation is an ordered arrangementof robjects chosen from nobjects.

For combinations order does not matter but for permutations it does.

There are three types of permutations.The first is distinct with repetition. This means there are ndistinct

objects but in choosing rof them

you can repeat an object.this means different

Let's look at a 3combination lock with

numbers 0 through 9

There are 10 choices for the first number

There are 10 choices for the second numberand you can repeat the first number

There are 10 choices for the third numberand you can repeat

By the multiplication principle there are 10 10 10 = 1000 choices


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Permutations: Distinct Objects with Repetition

The number of ordered arrangements of robjectschosen from nobjects, in which the nobjects aredistinct and repetition is allowed, is nr

This can be generalized as:

What if the lock had four choices fornumbers instead of three?

104 = 10 000choices


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The second type of permutation is distinct, without repetition.

Let's say four people have a race. Let's look at the possibilities of how they

could place. Once a person has been listed in a place, you can't use thatperson again (no repetition).

First place would be

choosing someone fromamong 4 people.

Now there are only 3 tochoose from for secondplace.

1st2nd

Now there are only 2to choose from forthird place. 3

rd4th

Only one possibility for

fourth place.

Based on the multiplication principle: 4 3 2 1 = 24 choices


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nPr , means the number of ordered arrangements of robjects chosen fromndistinct objects and repetition is not allowed.

In the last example:

0! =1

If you have 10 people racing andonly 1st, 2nd and 3rd place how

many possible outcomes arethere?

!!

rn

nP

r

n

24

!0

1234

!44

!44

4

P

720

!7

!78910

!310

!10

3

10

P


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A combination is an arrangement of robjects chosen from nobjects regardlessof order.

nCr, means the number combinations of robjects chosen from ndistinctobjects and repetition is not allowed.

Order doesn't matter here so the combination 1, 2, 3 is not different than 3, 2,

1 because they both contain the same numbers.

rn

rrnnC

r

n or!!

!


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You need 2 people on your committee and you have 5 to choose from. Youcan see that this is without repetition because you can only choose a persononce, and order doesnt matter. You need 2 committee members but it

doesn't matter who is chosen first. How many combinations are there?

10

2!3

!345

!2!25

!5

2

5

C

Th hi d f i i i l i bj h di i


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The third type of permutation is involving nobjects that are not distinct.

How many different combinations of letters in specific order (but notnecessarily English words) can be formed using ALL the letters in the wordREARRANGE?

The "words" we form will have 9 letters so we need 9 spots to place the letters.Notice some of the letters repeat. We need to use R 3 times, A 2 times, E 2 timesand N and G once.

First we choosepositions for theR's. There are 9positions and we'll

choose 3, orderdoesn't matter

9C3

That leaves6 positionsfor 2 A's.

R R R

6C2

A A

That leaves4 positionsfor 2 E's.

4C2

That leaves2 positionsfor the N.

E E

2C1

That leaves1 position

for the G.

N G

1C1

84 15 6 2 1 = 15 120 possible "words"

Not Examinable.. Just for Fun


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This can be generalized into the following:

Permutations Involving nObjects

That Are Not Distinct

1 2

1 2

1 2

The number of permutations of objects

of which are of one kind, are of a

second kind, . . ., and are of a th kind

is given by

!! ! !

where

k

k

k

n

n n

n k

n

n n n

n n n n


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Concepts of Probability

Probability is a mathematical conceptthat is used to measure the certaintyor uncertainty of occurrence of

statistical phenomenon. Sample Point each possible

outcomes of the experiment from the

sample space The total number of sample point in

the given sample space S is

symbolically denoted by n(S).


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Find the sample space of two coin in theevent of getting two tail and two heads in

tossing a coin.a. Let n1=number of possible outcomes ofthe first coin

n2= number of possible outcomes of

the second coinb. The event of getting two tails

A = {TT}

n(A)=1c. The event of getting two heads

A = {HH}

n(A)=1


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A box contains 7 blue, 5 red, 2 yellowballs. If 2 balls are drawn from the box,

determine the number of sample pointsof the following:

a. Sample spaceb. The event of getting all blue

c. The event of getting 1 red and 1

yellow


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a. Let S = event of drawing 3 balls fromthe box

n(S)= 14C2 = 14! / 2!(14-2)! = 91

b. Let A = event of getting all blue balls

n(A) = 7C2 = 7! / 2!(7-2)! = 21

c. Let B = event of getting 1 red & 1yellow

n(B) = 5C1 . 2C1

=5! / 1!(15-1)! . 2! / 1!(2-1)!

=5 . 2

= 10


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Probability of RelativeFrequency

P(A) = number of times A occurred /

number of times experiment wasrepeated


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Matthew decides to try to estimate theprobability that toast lands butter-side-

down when dropped.He drops a pieceof buttered toast 50 times andobserves that it lands butter-side-

down 30 times. He wants to estimatethe probability that the toast landsbutter-side-down.

P(toast land butter-side-down) = 30 / 50= 0.6 = 60%


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Sarah tosses a coin 200 times. Shegets 108 heads and 92 tails. Find the

probability of getting heads.

P(getting heads) = 108 / 200 = 0.54 =

54%


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Classical Probability

P(A) = n(A) / n(S)

Where:

n(A) = represents the number of samplepoints in event A

n(S) = represents the number of sample

points in the sample space


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If three coins are tossed find the eventof getting:

a. Two tailsb. Three heads

c. One tail and two head

{HHH, HHT, HTH, THH}

{TTT, TTH, THT, HTT}

n(S) = 2x2x2 = 8


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a. n(A) = 3 {TTH, THT, HTT}

P(A) = 3 / 8 = .375 = 37.50%

b. n(B) = 1 {HHH}

P(B) = 1 / 8 = 0.125 = 12.50%

c. n(C) = 3 {HHT, HTH, THH}

P(C) = 3 / 8 = .375 = 37.50 %


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In a deck of 52 cards, 5 card will bedrawn. What is the probability of getting:

a. 3 face card and 2 ace

b. 1 club and 4 spade

c. All hearts

n(S) =52

C5

= 2,598,960


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a. n(A) = 12C3 . 4C2 = 220 . 6 = 1,320

P(A) = 1,320 / 2,598,960 = 11/21658 =

0.05%

b. n(B) = 12C1 . 12C4 = 12 . 495 = 5,940

P(B) = 5,940 / 2,598,960 = 0.23%

c. n(C) =12

C5

= 792

P(C) = 792 / 2,598,960 = 0.03%


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Addition Rule

Two or more events are mutuallyexclusive if they cannot occur

simultaneously. This means that only

one of two or more events can occur

P(A B) = P(A) + P(B)

orP(A B) = n(A) / n(S) + n(B) / n(S)


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A spinner has 4 equal sectors coloredyellow, blue, green, and red. What is the

probability of landing on red or blue afterspinning this spinner?

P(A) = P(B) =

P(A B) = P(A) + P(B)

= +

= 2/4 = 1/2


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A glass jar contains 1 red, 3 green, 2blue, and 4 yellow marbles. If a single

marble is chosen at random from the jar,what is the probability that it is yellow orgreen?

P(A) = 4/10 P(B) = 3/10

P(A B) = P(A) + P(B)= 4/10+ 3/10

= 7/10


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Addition Rule

Events are non-mutually exclusivewhen two or more events can occur.

P(A B) = P(A) + P(B) - P(A B)

or

P(A B) = n(A) / n(S) + n(B) / n(S)

P(A B) / n(S)


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A single card is chosen at random froma standard deck of 52 playing cards.

What is the probability of choosing aking or a club?

P(A) = 4/52 P(B) = 13/52P(A B) = 1/52

P(A B) = P(A) + P(B) - P(A B)

= 4/52 + 13/52 1/52

= 16/52 = 4/13


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In a math class of 30 students, 17 areboys and 13 are girls. On a unit test, 4

boys and 5 girls made an A grade. If astudent is chosen at random from theclass, what is the probability of choosing agirl or an A student?

P(A) = 13/30 P(B) = 9/30

P(A B) = 5/30

P(A B) = P(A) + P(B) - P(A B)

= 13/30 + 9/30 5/30

= 17/30

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