MATH 105: Finite Mathematics 7-3: Probability from Countingmath.wallawalla.edu/~duncjo/courses/math105/winter06/slides/finite... · Probability from Counting Examples (Lots and Lots

Probability from Counting Examples (Lots and Lots of Them!) Conclusion

MATH 105: Finite Mathematics7-3: Probability from Counting

Prof. Jonathan Duncan

Walla Walla College

Winter Quarter, 2006


Outline

1 Probability from Counting

2 Examples (Lots and Lots of Them!)

3 Conclusion


Outline



3 Conclusion


Counting and Probability

We have seen the following probability formula used quite often inthe last two sections.

Probability of Equally Likely Outcomes

if E is an event in a sample space S and outcomes in S are allequally likely, then

Pr [E ] =c(E )

c(S)

Counting Rules

We can use counting rules such as P(n, r) and C (n, r) to find c(E )and c(S).






Pr [E ] =c(E )

c(S)

Counting Rules







Pr [E ] =c(E )

c(S)

Counting Rules



Outline



3 Conclusion


Selecting a Subgroup of People

Example

A group of 6 women and 5 men wish to select 3 people to performsome task. They decide to draw names out of a hat.

1 What is the probability that all 3 are women?

2 What is the probability of 2 women and 1 man?

3 What is the probability of more women than men?

4 What is the probability of at least one man?



Example








Example



C (6, 3)

C (11, 3)=

20

165≈ 0.121






Example


1 What is the probability that all 3 are women? ≈ 0.121






Example




C (6, 2)C (5, 1)

C (11, 3)=

75

165≈ 0.455





Example



2 What is the probability of 2 women and 1 man? ≈ 0.455





Example





C (6, 3)C (5, 0) + C (6, 2)C (5, 1)

C (11, 3)=

95

165≈ 0.576




Example




3 What is the probability of more women than men? ≈ 0.576




Example




3 What is the probability of more women than men? ≈ 0.576


1− C (6, 3)

C (11, 3)= 1− 20

165≈ 0.879


License Plates

Example

A license plate is composed of 3 letters followed by 3 digits. If aplate is randomly produced, what is the probability that it containsat least one repeated character?

Let E be the event that the license has no repeatsIt is easier to count E than E

c(S) = 263 · 103 = 17, 576, 000

c(E ) = 26 · 25 · 24 · 10 · 9 · 8 = 11, 232, 000

Pr [E ] = 1− Pr [E ] = 1− 11232000

17576000≈ 0.361


License Plates

Example



c(S) = 263 · 103 = 17, 576, 000

c(E ) = 26 · 25 · 24 · 10 · 9 · 8 = 11, 232, 000

Pr [E ] = 1− Pr [E ] = 1− 11232000

17576000≈ 0.361


License Plates

Example



c(S) = 263 · 103 = 17, 576, 000

c(E ) = 26 · 25 · 24 · 10 · 9 · 8 = 11, 232, 000

Pr [E ] = 1− Pr [E ] = 1− 11232000

17576000≈ 0.361


License Plates

Example



c(S) = 263 · 103 = 17, 576, 000

c(E ) = 26 · 25 · 24 · 10 · 9 · 8 = 11, 232, 000

Pr [E ] = 1− Pr [E ] = 1− 11232000

17576000≈ 0.361


License Plates

Example



c(S) = 263 · 103 = 17, 576, 000

c(E ) = 26 · 25 · 24 · 10 · 9 · 8 = 11, 232, 000

Pr [E ] = 1− Pr [E ] = 1− 11232000

17576000≈ 0.361


License Plates

Example



c(S) = 263 · 103 = 17, 576, 000

c(E ) = 26 · 25 · 24 · 10 · 9 · 8 = 11, 232, 000

Pr [E ] = 1− Pr [E ] = 1− 11232000

17576000≈ 0.361


Casting a Play

Example

A play requires 3 male and 2 female rules, including that of“mother”. If there are 5 men and 4 women, including Daisy,auditioning for these parts, and the parts are chosen at random,find each probability.

1 the probability that Daisy gets a part

2 the probability that Daisy get the part of “mother”


Casting a Play

Example





Casting a Play

Example



C (5, 3)C (3, 1)

C (5, 3)C (4, 2)=

C (3, 1)

C (4, 2)=

3

6= 0.500



Casting a Play

Example



C (5, 3)C (3, 1)

C (5, 3)C (4, 2)=

C (3, 1)

C (4, 2)=

3

6= 0.500



Casting a Play

Example



C (5, 3)C (3, 1)

C (5, 3)C (4, 2)=

C (3, 1)

C (4, 2)=

3

6= 0.500


P(5, 3)P(3, 1)

P(5, 3)P(4, 2)=

P(3, 1)

P(4, 2)=

3

8= 0.375


Standing in a Row

Example

A family photo of a six-person family is to be taken. If the familymembers line up randomly in a straight line, what is the probabilitythat the mother and father stand next to each other?

Use the “combined-person” concept with 5 people including F-M

Don’t forget to count both F-M and M-F


Standing in a Row

Example





Standing in a Row

Example



P(5, 5)

P(6, 6)=

1

6≈ 0.166



Standing in a Row

Example



2 · P(5, 5)

P(6, 6)=

2

6≈ 0.333



Playing Cards

Example

Find the probability of each poker hand.

1 A flush

2 A full house

3 four of a kind

4 three of a kind

5 two pair

6 a pair


Playing Cards

Example


1 A flush

2 A full house

3 four of a kind

4 three of a kind

5 two pair

6 a pair


Playing Cards

Example


1 A flush2 · C (26, 4)

C (52, 5)=

29900

2598960≈ 0.0115

2 A full house

3 four of a kind

4 three of a kind

5 two pair

6 a pair


Playing Cards

Example


1 A flush ≈ 0.0115

2 A full house

3 four of a kind

4 three of a kind

5 two pair

6 a pair


Playing Cards

Example


1 A flush ≈ 0.0115

2 A full house

13 · C (4, 3) · 12 · C (4, 2)

C (52, 5)=

3744

2598960≈ 0.0014

3 four of a kind

4 three of a kind

5 two pair

6 a pair


Playing Cards

Example


1 A flush ≈ 0.0115

2 A full house ≈ 0.0014

3 four of a kind

4 three of a kind

5 two pair

6 a pair


Playing Cards

Example


1 A flush ≈ 0.0115


3 four of a kind

13 · C (4, 4) · C (48, 1)

C (52, 5)=

624

2598960≈ 0.00024

4 three of a kind

5 two pair

6 a pair


Playing Cards

Example


1 A flush ≈ 0.0115


3 four of a kind ≈ 0.0002

4 three of a kind

5 two pair

6 a pair


Playing Cards

Example


1 A flush ≈ 0.0115



4 three of a kind

13 · C (4, 3) · C (12, 2) · C (4, 1) · C (4, 1)

C (52, 5)=

54912

2598960≈ .0211

5 two pair

6 a pair


Playing Cards

Example


1 A flush ≈ 0.0115



4 three of a kind ≈ 0.0211

5 two pair

6 a pair


Playing Cards

Example


1 A flush ≈ 0.0115




5 two pair

C (13, 2)C (4, 2)C (4, 2)C (48, 1)

C (52, 5)=

134784

2598960≈ 0.0519

6 a pair


Playing Cards

Example


1 A flush ≈ 0.0115




5 two pair ≈ 0.0519

6 a pair


Playing Cards

Example


1 A flush ≈ 0.0115




5 two pair ≈ 0.0519

6 a pair

C (13, 1)C (4, 2)C (12, 3)C (4, 1)C (4, 1)C (4, 1)

C (52, 5)≈ 0.4226


Flipping a Coin

Example

A fair coin is tossed six times.

1 Find the probability exactly two tails appear.

2 Find the probability no more than two tails appear.


Flipping a Coin

Example





Flipping a Coin

Example



C (6, 2)

26=

15

64≈ 0.2344



Flipping a Coin

Example


1 Find the probability exactly two tails appear. ≈ 0.2344



Flipping a Coin

Example


1 Find the probability exactly two tails appear. ≈ 0.2344


C (6, 0) + C (6, 1) + C (6, 2)

26=

22

65≈ 0.344


Outline



3 Conclusion


Important Concepts

Things to Remember from Section 7-3

1 When dealing with equally likely events, remember:

Pr [E ] =c(E )

c(S)

2 Use Permutations and Combinations to find c(E ) and c(S).

3 Always ask yourself:1 Does order matter? (Yes: P, No: C)2 Am I done producing an event? (Yes: Add, No: Multiply)


Important Concepts



Pr [E ] =c(E )

c(S)




Important Concepts



Pr [E ] =c(E )

c(S)




Important Concepts



Pr [E ] =c(E )

c(S)




Important Concepts



Pr [E ] =c(E )

c(S)




Important Concepts



Pr [E ] =c(E )

c(S)




Next Time. . .

Now that we have used the tools learned in chapter 6 to computebasic probabilities in chapter 7, it is a good time to review whatwe’ve covered and assess how much you’ve learned.

For next time

Review Sections 6-1 through 7-3 (omit 6-6)

Prepare for Exam on Friday


Next Time. . .

Now that we have used the tools learned in chapter 6 to computebasic probabilities in chapter 7, it is a good time to review whatwe’ve covered and assess how much you’ve learned.

For next time

Review Sections 6-1 through 7-3 (omit 6-6)

Prepare for Exam on Friday

Documents

MATH 105: Finite Mathematics 7-3: Probability from Countingmath.wallawalla.edu/~duncjo/courses/math105/winter06/slides/finite... · Probability from Counting Examples (Lots and Lots