chapter 5 possibilities and probability counting permutations combinations probability

Chapter 5 Possibilities and Chapter 5 Possibilities and ProbabilityProbability

CountingPermutationsCombinationsProbability

5.1 Counting5.1 Counting

Example 5.1: Ice cream cones

Flavor: chocolate , vanilla, strawberry

Cones: sugar, regular

How many different varieties?

Tree diagramTree diagram

sugar choc reg sugar van reg sugar str reg

3*2=6 choices

Rule 1: Multiplication RuleRule 1: Multiplication Rule

A choice consists of 2 distinct steps1st step can be made in m different waysFor each of these, 2nd step can be made in n

different ways

Then the whole choice can be made in m*n ways

Example 5.2Example 5.2

4 horses in a race How many ways can we pick a first

and second place finisher?

Answer: 43=12

You can get it from a tree diagram as well

Choice of 1st place choices of 2nd place

4*3=12 ways

A

B

C

DA

B

C

ABD

AC

D

B

C

D

Rule 2: Generalized Multiplication Rule 2: Generalized Multiplication RuleRule

A choice consists of k steps;

Step 1 can be made in n1 ways;

Step 2 can be made in n2 ways;

… …

Step k can be made in nk ways ;

Then whole choice can be made in n1n2…nk ways

Example 5.3Example 5.3

5 horses in a raceHow many ways can we pick a 1st , 2nd, and

3rd place finisher?

Example 5.4Example 5.4

A multiple choice exam has 5 questionsEach question has 4 possible answers

What is the number of ways to answer the exam?

Example 5.5Example 5.5

How many license plates can be formed with 3 letters followed by 3 numbers?

Example 5.6Example 5.6

How many ways can we write the letters

O W L ?

5.2 Permutations5.2 Permutations

The number of ways to order r of n objects

nPr=n(n-1)(n-2)…(n-r+1)

(application of multiplication rule)

Example 5.7Example 5.7

How many ways can we put 3 cards from a deck of cards in order?

52 51 50

(52)(51)(50)=132,600

Notation: n!=(1)(2)Notation: n!=(1)(2)……(n-1)(n)(n-1)(n)

The number of ways to order (permute) n of n objects is

nPn = n(n-1)(n-2)…(1) = n!

n! is called n-factorial

3!=(1)(2)(3)=6

4!=(1)(2)(3)(4)=3!(4)=24

5!=4!(5)=120

(0!=1)

Example 5.8Example 5.8

How many ways can we order 5 horses?

5!=120

Example 5.9Example 5.9

How many ways can we arrange the letters OLWS?

The Number of PermutationsThe Number of Permutations

)!(

!)1()1(

rn

nrnnnPrn

60)3)(4)(5()1)(2(

)1)(2)(3)(4)(5(

)!35(

!535

P

ExerciseExercise

How many different ways can we inject 3 of 15 mice with 3 different doses of a serum?

Exercise 5.21Exercise 5.21

In optics kits there are 5 concave lenses, 5 convex lenses, 2 prisms, and 3 mirrors. how many different ways can a person choose 1 of each kind?

Exercise 5.27Exercise 5.27

In how many different ways can a television director schedule a sponsor’s six different commercials during a telecast?

5.3 Combinations5.3 Combinations

How many ways can we choose 3 of 5 candidates to be in the final election?

Here the order of choosing the 3 finalists doesn’t matter!

Think this way Think this way …… ……

When we did worry about the order, there are (5)(4)(3)=60 ways

ABC ACB BAC BCA CAB CBA

ABD ADB BAD BDA DAB DBA

BCD BDC ...

… …

andand

Each choice of 3 candidates has 3!=6 ordered versions ABC ABC ACB BAC BCA CAB CBA

ThereforeTherefore

ABC ACB BAC BCA CAB CBAABC

ABD ADB BAD BDA DAB DBAABD

BCD BDC ... BCD

… …

The options are decreased by a factor of 6 compared to the ordered options

The number of ways to pick 3 out of 5The number of ways to pick 3 out of 5

10)1)(2(

)4)(5(

!3)!35(

!5

33

35

P

P

The Number of CombinationsThe Number of Combinations

The number of unordered ways to choose r of n objects is

(n choose r)

)!(!

!

rnr

n

r

nCrn

PascalPascal’’s Triangle s Triangle

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

Example 5.10Example 5.10

Choose 5 cards from a deck

!5552

)1)(2)(3)(4)(5(

)48)(49)(50)(51)(52(

)!47)(1)(2)(3)(4)(5(

)!47)(48)(49)(50)(51)(52(

)!552(!5

!52552

P

Example 5.11Example 5.11

5 flavors of ice cream. Choose 2. Order doesn’t matter.

Example 5.12Example 5.12

6 candidates in a primary election. Choose 2 for a final election

ExerciseExercise

Calculate the number of ways in which a chain of ice cream stores can choose 2 of 12 locations for new franchises.

ExerciseExercise

A computer store carries 15 kinds of monitors. Calculate the number of ways in which we can purchase 3 different ones.

ExerciseExercise

In planning a garden we have 5 kinds of bushes to choose from and 10 kinds of flowers. How many ways can we choose 2 kinds of bushes and 4 kinds of flowers?

5.4 Probability5.4 Probability

In a deck of cards what is the probability of picking an ace?

What is meant by “Probability”?

Frequency interpretation: The probability of an event happening is the proportion of times that event would occur in the long run.

FactsFacts

Each card is equally likely to be chosen for a shuffled deck of cards

If there are “n” equally likely possibilities and “s” of these are a “success” , then the probability of a success is s/n

P(ace)= 4/52=1/13

P(red)=26/52=1/2

Example 5.13Example 5.13

Draw 2 cards from a deck. What is the probability that we get 2 aces?

Ideas Ideas

Each card is equally likely to be selected if one card is selected

All pairs of cards are equally likely to be selected if only two cards are selected

Any three cards are equally likely to be selected if only three cards are selected

… … (we call these randomness)

Solution to Example 5.13Solution to Example 5.13

# of possible ways to pick 2 cards

# of ways to pick 2 aces

Probability P(2 aces)

13262

)51)(52(

2

52

n

62

)3)(4(

2

4

s

0045.01326

6

Example 5.14Example 5.14

Pick up 3 cards, what is the probability of getting 2 aces and 1 king?

Let’s work it out!

Step 1

Step 2

Step 3 Probability P(2 aces and 1 king)=s/n= 24/22100

241

4

2

4

s

22100)1)(2)(3(

)50)(51)(52(

3

52

n

# of ways to pick 2 aces out of 4

# of ways to pick 1 king out of 4

Example 5.15Example 5.15

Pick up 5 cards. What is the probability of getting 3 aces?

Step 1

Step 2

Step 3 Probability

P(3 aces in 5 cards )=s/n=

4 48

3 2s

52

5n

# of ways to pick 3 aces out of 4

# of ways to pick other 2 cards

5

52

2

48

3

4

Example 5.16Example 5.16 Roll a red die and a white die. Find the probability that sum=3. red die: 1, 2, 3, 4, 5, 6 white die: 1, 2, 3, 4, 5, 6

n = # of outcomes= (ways for red to land)*(ways for white to land)= 6×6=36 pairs

s = # of ways sum=3(R=1, W=2), (R=2, W=1)

=2 Probability=2/36=1/18