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

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

Outline

1 Probability from Counting

2 Examples (Lots and Lots of Them!)

3 Conclusion

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

Outline

1 Probability from Counting

2 Examples (Lots and Lots of Them!)

3 Conclusion

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

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

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

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

Outline

1 Probability from Counting

2 Examples (Lots and Lots of Them!)

3 Conclusion

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

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

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

C (6, 3)

C (11, 3)=

20

165≈ 0.121

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?

Probability from Counting Examples (Lots and Lots of Them!) 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? ≈ 0.121

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?

Probability from Counting Examples (Lots and Lots of Them!) 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? ≈ 0.121

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

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

C (11, 3)=

75

165≈ 0.455

3 What is the probability of more women than men?

4 What is the probability of at least one man?

Probability from Counting Examples (Lots and Lots of Them!) 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? ≈ 0.121

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

3 What is the probability of more women than men?

4 What is the probability of at least one man?

Probability from Counting Examples (Lots and Lots of Them!) 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? ≈ 0.121

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

3 What is the probability of more women than men?

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

C (11, 3)=

95

165≈ 0.576

4 What is the probability of at least one man?

Probability from Counting Examples (Lots and Lots of Them!) 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? ≈ 0.121

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

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

4 What is the probability of at least one man?

Probability from Counting Examples (Lots and Lots of Them!) 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? ≈ 0.121

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

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

4 What is the probability of at least one man?

1− C (6, 3)

C (11, 3)= 1− 20

165≈ 0.879

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

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

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

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

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

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

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

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

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

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

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

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

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

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”

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

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”

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

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

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

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

C (3, 1)

C (4, 2)=

3

6= 0.500

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

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

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

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

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

C (3, 1)

C (4, 2)=

3

6= 0.500

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

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

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

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

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

C (3, 1)

C (4, 2)=

3

6= 0.500

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

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

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

P(3, 1)

P(4, 2)=

3

8= 0.375

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

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

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

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

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

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

P(5, 5)

P(6, 6)=

1

6≈ 0.166

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

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

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

2 · P(5, 5)

P(6, 6)=

2

6≈ 0.333

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

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

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

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

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

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

Playing Cards

Example

Find the probability of each poker hand.

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

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

Playing Cards

Example

Find the probability of each poker hand.

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

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

Playing Cards

Example

Find the probability of each poker hand.

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

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

Playing Cards

Example

Find the probability of each poker hand.

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

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

Playing Cards

Example

Find the probability of each poker hand.

1 A flush ≈ 0.0115

2 A full house ≈ 0.0014

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

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

Playing Cards

Example

Find the probability of each poker hand.

1 A flush ≈ 0.0115

2 A full house ≈ 0.0014

3 four of a kind ≈ 0.0002

4 three of a kind

5 two pair

6 a pair

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

Playing Cards

Example

Find the probability of each poker hand.

1 A flush ≈ 0.0115

2 A full house ≈ 0.0014

3 four of a kind ≈ 0.0002

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

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

Playing Cards

Example

Find the probability of each poker hand.

1 A flush ≈ 0.0115

2 A full house ≈ 0.0014

3 four of a kind ≈ 0.0002

4 three of a kind ≈ 0.0211

5 two pair

6 a pair

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

Playing Cards

Example

Find the probability of each poker hand.

1 A flush ≈ 0.0115

2 A full house ≈ 0.0014

3 four of a kind ≈ 0.0002

4 three of a kind ≈ 0.0211

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

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

Playing Cards

Example

Find the probability of each poker hand.

1 A flush ≈ 0.0115

2 A full house ≈ 0.0014

3 four of a kind ≈ 0.0002

4 three of a kind ≈ 0.0211

5 two pair ≈ 0.0519

6 a pair

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

Playing Cards

Example

Find the probability of each poker hand.

1 A flush ≈ 0.0115

2 A full house ≈ 0.0014

3 four of a kind ≈ 0.0002

4 three of a kind ≈ 0.0211

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

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

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.

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

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.

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

Flipping a Coin

Example

A fair coin is tossed six times.

1 Find the probability exactly two tails appear.

C (6, 2)

26=

15

64≈ 0.2344

2 Find the probability no more than two tails appear.

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

Flipping a Coin

Example

A fair coin is tossed six times.

1 Find the probability exactly two tails appear. ≈ 0.2344

2 Find the probability no more than two tails appear.

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

Flipping a Coin

Example

A fair coin is tossed six times.

1 Find the probability exactly two tails appear. ≈ 0.2344

2 Find the probability no more than two tails appear.

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

26=

22

65≈ 0.344

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

Outline

1 Probability from Counting

2 Examples (Lots and Lots of Them!)

3 Conclusion

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

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

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

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

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

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

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

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

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

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