MATH 105: Finite Mathematics 7-3: Probability from...
Transcript of MATH 105: Finite Mathematics 7-3: Probability from...
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