Probability from Counting Examples (Lots and Lots of Them!) Conclusion MATH 105: Finite Mathematics 7-3: Probability from Counting Prof. Jonathan Duncan Walla Walla College Winter Quarter, 2006
Probability from Counting Examples (Lots and Lots of Them!) Conclusion Outline Probability from Counting 1 Examples (Lots and Lots of Them!) 2 Conclusion 3
Probability from Counting Examples (Lots and Lots of Them!) Conclusion Outline Probability from Counting 1 Examples (Lots and Lots of Them!) 2 Conclusion 3
Probability from Counting Examples (Lots and Lots of Them!) Conclusion Counting and Probability We have seen the following probability formula used quite often in the last two sections. Probability of Equally Likely Outcomes if E is an event in a sample space S and outcomes in S are all equally 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 in the last two sections. Probability of Equally Likely Outcomes if E is an event in a sample space S and outcomes in S are all equally 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 in the last two sections. Probability of Equally Likely Outcomes if E is an event in a sample space S and outcomes in S are all equally 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 Probability from Counting 1 Examples (Lots and Lots of Them!) 2 Conclusion 3
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 perform some 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 perform some 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 perform some task. They decide to draw names out of a hat. 1 What is the probability that all 3 are women? C (11 , 3) = 20 C (6 , 3) 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 perform some 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 perform some 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) = 75 165 ≈ 0 . 455 C (11 , 3) 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 perform some 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 perform some 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) = 95 165 ≈ 0 . 576 C (11 , 3) 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 perform some 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 perform some 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 a plate is randomly produced, what is the probability that it contains at least one repeated character? Let E be the event that the license has no repeats It is easier to count E than E c ( S ) = 26 3 · 10 3 = 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 a plate is randomly produced, what is the probability that it contains at least one repeated character? Let E be the event that the license has no repeats It is easier to count E than E c ( S ) = 26 3 · 10 3 = 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 a plate is randomly produced, what is the probability that it contains at least one repeated character? Let E be the event that the license has no repeats It is easier to count E than E c ( S ) = 26 3 · 10 3 = 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 a plate is randomly produced, what is the probability that it contains at least one repeated character? Let E be the event that the license has no repeats It is easier to count E than E c ( S ) = 26 3 · 10 3 = 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 a plate is randomly produced, what is the probability that it contains at least one repeated character? Let E be the event that the license has no repeats It is easier to count E than E c ( S ) = 26 3 · 10 3 = 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 a plate is randomly produced, what is the probability that it contains at least one repeated character? Let E be the event that the license has no repeats It is easier to count E than E c ( S ) = 26 3 · 10 3 = 17 , 576 , 000 c ( E ) = 26 · 25 · 24 · 10 · 9 · 8 = 11 , 232 , 000 Pr [ E ] = 1 − Pr [ E ] = 1 − 11232000 17576000 ≈ 0 . 361
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