lecture 7: bayes' rule, revision - stankova · beamer-tu-logo preliminaries bayes0 rulea lot...
TRANSCRIPT
![Page 1: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/1.jpg)
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Preliminaries Bayes′ rule A lot of computing . . .
Lecture 7: Bayes′ Rule, Revision
Katerina Stanková
Statistics (MAT1003)
April 26, 2012
![Page 2: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/2.jpg)
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Preliminaries Bayes′ rule A lot of computing . . .
Outline
1 PreliminariesPartition of a sample space
2 Bayes′ ruleTheoryExamples
3 A lot of computing . . .
![Page 3: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/3.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
And now . . .
1 PreliminariesPartition of a sample space
2 Bayes′ ruleTheoryExamples
3 A lot of computing . . .
![Page 4: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/4.jpg)
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Preliminaries Bayes′ rule A lot of computing . . .
Partition of a sample space
Partition of a sample space
Let S be a sample space. A set of subsets B1, B2, . . . , Bk of Sis called a partition of S if
k⋃i=1
Bi = B1 ∪ B2 ∪ . . .Bk = S
Bi ∩ Bj = ∅ for all i , j ∈ {1, . . . , k}, i 6= j
![Page 5: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/5.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Partition of a sample space
Partition of a sample spaceLet S be a sample space. A set of subsets B1, B2, . . . , Bk of Sis called a partition of S if
k⋃i=1
Bi = B1 ∪ B2 ∪ . . .Bk = S
Bi ∩ Bj = ∅ for all i , j ∈ {1, . . . , k}, i 6= j
![Page 6: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/6.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Partition of a sample space
Partition of a sample spaceLet S be a sample space. A set of subsets B1, B2, . . . , Bk of Sis called a partition of S if
k⋃i=1
Bi = B1 ∪ B2 ∪ . . .Bk = S
Bi ∩ Bj = ∅ for all i , j ∈ {1, . . . , k}, i 6= j
![Page 7: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/7.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Partition of a sample space
Partition of a sample spaceLet S be a sample space. A set of subsets B1, B2, . . . , Bk of Sis called a partition of S if
k⋃i=1
Bi = B1 ∪ B2 ∪ . . .Bk = S
Bi ∩ Bj = ∅ for all i , j ∈ {1, . . . , k}, i 6= j
![Page 8: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/8.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Partition of a sample space
Partition of a sample spaceLet S be a sample space. A set of subsets B1, B2, . . . , Bk of Sis called a partition of S if
k⋃i=1
Bi = B1 ∪ B2 ∪ . . .Bk = S
Bi ∩ Bj = ∅ for all i , j ∈ {1, . . . , k}, i 6= j
![Page 9: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/9.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Partition of a sample space
Partition of a sample spaceLet S be a sample space. A set of subsets B1, B2, . . . , Bk of Sis called a partition of S if
k⋃i=1
Bi = B1 ∪ B2 ∪ . . .Bk = S
Bi ∩ Bj = ∅ for all i , j ∈ {1, . . . , k}, i 6= j
![Page 10: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/10.jpg)
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Preliminaries Bayes′ rule A lot of computing . . .
Partition of a sample space
Example 1S = {1,2,3,4,5,6}
Partition {B1,B2,B3} with B1 = {1,3}, B2 = {5}, B3 = {2,4,6}
Example 2
S = [0,1]Partition {I1, I2} with I1 =
[0, 1
2
), I2 =
[12 ,1]
Example 3
Any sample space SPartition {B,B′} for any B ⊆ S
![Page 11: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/11.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Partition of a sample space
Example 1S = {1,2,3,4,5,6}Partition {B1,B2,B3} with B1 = {1,3}, B2 = {5}, B3 = {2,4,6}
Example 2
S = [0,1]Partition {I1, I2} with I1 =
[0, 1
2
), I2 =
[12 ,1]
Example 3
Any sample space SPartition {B,B′} for any B ⊆ S
![Page 12: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/12.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Partition of a sample space
Example 1S = {1,2,3,4,5,6}Partition {B1,B2,B3} with B1 = {1,3}, B2 = {5}, B3 = {2,4,6}
Example 2
S = [0,1]Partition {I1, I2} with I1 =
[0, 1
2
), I2 =
[12 ,1]
Example 3
Any sample space SPartition {B,B′} for any B ⊆ S
![Page 13: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/13.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Partition of a sample space
Example 1S = {1,2,3,4,5,6}Partition {B1,B2,B3} with B1 = {1,3}, B2 = {5}, B3 = {2,4,6}
Example 2S = [0,1]
Partition {I1, I2} with I1 =[0, 1
2
), I2 =
[12 ,1]
Example 3
Any sample space SPartition {B,B′} for any B ⊆ S
![Page 14: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/14.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Partition of a sample space
Example 1S = {1,2,3,4,5,6}Partition {B1,B2,B3} with B1 = {1,3}, B2 = {5}, B3 = {2,4,6}
Example 2S = [0,1]Partition {I1, I2} with I1 =
[0, 1
2
), I2 =
[12 ,1]
Example 3
Any sample space SPartition {B,B′} for any B ⊆ S
![Page 15: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/15.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Partition of a sample space
Example 1S = {1,2,3,4,5,6}Partition {B1,B2,B3} with B1 = {1,3}, B2 = {5}, B3 = {2,4,6}
Example 2S = [0,1]Partition {I1, I2} with I1 =
[0, 1
2
), I2 =
[12 ,1]
Example 3
Any sample space SPartition {B,B′} for any B ⊆ S
![Page 16: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/16.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Partition of a sample space
Example 1S = {1,2,3,4,5,6}Partition {B1,B2,B3} with B1 = {1,3}, B2 = {5}, B3 = {2,4,6}
Example 2S = [0,1]Partition {I1, I2} with I1 =
[0, 1
2
), I2 =
[12 ,1]
Example 3Any sample space S
Partition {B,B′} for any B ⊆ S
![Page 17: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/17.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Partition of a sample space
Example 1S = {1,2,3,4,5,6}Partition {B1,B2,B3} with B1 = {1,3}, B2 = {5}, B3 = {2,4,6}
Example 2S = [0,1]Partition {I1, I2} with I1 =
[0, 1
2
), I2 =
[12 ,1]
Example 3Any sample space SPartition {B,B′} for any B ⊆ S
![Page 18: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/18.jpg)
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Preliminaries Bayes′ rule A lot of computing . . .
Partition of a sample space
Observation
Let B1, . . . , Bk be a partition of S. Then for any A ⊆ S :
A = A ∩ S = A ∩ (B1 ∪ B2 ∪ . . . ∪ Bk )
= (A ∩ B1)︸ ︷︷ ︸A1
∪ (A ∩ B2)︸ ︷︷ ︸A2
∪ . . . ∪ (A ∩ Bk )︸ ︷︷ ︸Ak
P(Ai) = P(A ∩ Bi) = P(Bi) ·P(A|Bi)A1,. . . , Ak disjoint sets
P(A)=k∑
i=1
P(A ∩ Bi)
=k∑
i=1
P(Bi) · P(A|Bi)
![Page 19: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/19.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Partition of a sample space
ObservationLet B1, . . . , Bk be a partition of S. Then for any A ⊆ S :
A = A ∩ S = A ∩ (B1 ∪ B2 ∪ . . . ∪ Bk )
= (A ∩ B1)︸ ︷︷ ︸A1
∪ (A ∩ B2)︸ ︷︷ ︸A2
∪ . . . ∪ (A ∩ Bk )︸ ︷︷ ︸Ak
P(Ai) = P(A ∩ Bi) = P(Bi) ·P(A|Bi)A1,. . . , Ak disjoint sets
P(A)=k∑
i=1
P(A ∩ Bi)
=k∑
i=1
P(Bi) · P(A|Bi)
![Page 20: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/20.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Partition of a sample space
ObservationLet B1, . . . , Bk be a partition of S. Then for any A ⊆ S :
A = A ∩ S = A ∩ (B1 ∪ B2 ∪ . . . ∪ Bk )
= (A ∩ B1)︸ ︷︷ ︸A1
∪ (A ∩ B2)︸ ︷︷ ︸A2
∪ . . . ∪ (A ∩ Bk )︸ ︷︷ ︸Ak
P(Ai) = P(A ∩ Bi) = P(Bi) ·P(A|Bi)A1,. . . , Ak disjoint sets
P(A)=k∑
i=1
P(A ∩ Bi)
=k∑
i=1
P(Bi) · P(A|Bi)
![Page 21: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/21.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Partition of a sample space
ObservationLet B1, . . . , Bk be a partition of S. Then for any A ⊆ S :
A = A ∩ S = A ∩ (B1 ∪ B2 ∪ . . . ∪ Bk )
= (A ∩ B1)︸ ︷︷ ︸A1
∪ (A ∩ B2)︸ ︷︷ ︸A2
∪ . . . ∪ (A ∩ Bk )︸ ︷︷ ︸Ak
P(Ai) = P(A ∩ Bi) = P(Bi) ·P(A|Bi)A1,. . . , Ak disjoint sets
P(A)=k∑
i=1
P(A ∩ Bi)
=k∑
i=1
P(Bi) · P(A|Bi)
![Page 22: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/22.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Partition of a sample space
ObservationLet B1, . . . , Bk be a partition of S. Then for any A ⊆ S :
A = A ∩ S = A ∩ (B1 ∪ B2 ∪ . . . ∪ Bk )
= (A ∩ B1)︸ ︷︷ ︸A1
∪ (A ∩ B2)︸ ︷︷ ︸A2
∪ . . . ∪ (A ∩ Bk )︸ ︷︷ ︸Ak
P(Ai) = P(A ∩ Bi) = P(Bi) ·P(A|Bi)A1,. . . , Ak disjoint sets
P(A)=k∑
i=1
P(A ∩ Bi)
=k∑
i=1
P(Bi) · P(A|Bi)
![Page 23: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/23.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
And now . . .
1 PreliminariesPartition of a sample space
2 Bayes′ ruleTheoryExamples
3 A lot of computing . . .
![Page 24: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/24.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Theory
Bayes′ rule
If P(Bi) and P(A|Bi) are given for all i , we can calculate P(Bi |A)as follows:
P(Bi |A)=P(Bi ∩ A)
P(A)=
P(Bi) · P(A|Bi)
P(A)
where P(A) =k∑
i=1P(Bi) · P(A|Bi)
![Page 25: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/25.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Theory
Bayes′ ruleIf P(Bi) and P(A|Bi) are given for all i , we can calculate P(Bi |A)as follows:
P(Bi |A)=P(Bi ∩ A)
P(A)=
P(Bi) · P(A|Bi)
P(A)
where P(A) =k∑
i=1P(Bi) · P(A|Bi)
![Page 26: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/26.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Theory
Bayes′ ruleIf P(Bi) and P(A|Bi) are given for all i , we can calculate P(Bi |A)as follows:
P(Bi |A)=P(Bi ∩ A)
P(A)=
P(Bi) · P(A|Bi)
P(A)
where P(A) =k∑
i=1P(Bi) · P(A|Bi)
![Page 27: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/27.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Theory
Bayes′ ruleIf P(Bi) and P(A|Bi) are given for all i , we can calculate P(Bi |A)as follows:
P(Bi |A)=P(Bi ∩ A)
P(A)=
P(Bi) · P(A|Bi)
P(A)
where P(A) =k∑
i=1P(Bi) · P(A|Bi)
![Page 28: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/28.jpg)
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Preliminaries Bayes′ rule A lot of computing . . .
Examples
Exercise 2.95 + forget about 40
Events:
C : A randomly chosen person has cancerD : A randomly chosen person is diagnosed as having cancer
Complementary events: C′ and D′
Data:
P(C) = 0.05 (P(C′) = 0.95)P(D|C) = 0.78 (P(D′|C) = 0.22)P(D|C′) = 0.06 (P(D′|C′) = 0.94)
Question 1:P(D) =?
Solution:
P(D) =
P(D ∩ C) + P(D ∩ C′)
= P(C) · P(D|C) + P(C′) · P(D|C′)
= 0.05 · 0.78 + 0.95 · 0.06 = 0.039 + 0.054 = 0.093
![Page 29: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/29.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Examples
Exercise 2.95 + forget about 40Events:
C : A randomly chosen person has cancerD : A randomly chosen person is diagnosed as having cancer
Complementary events: C′ and D′
Data:
P(C) = 0.05 (P(C′) = 0.95)P(D|C) = 0.78 (P(D′|C) = 0.22)P(D|C′) = 0.06 (P(D′|C′) = 0.94)
Question 1:P(D) =?
Solution:
P(D) =
P(D ∩ C) + P(D ∩ C′)
= P(C) · P(D|C) + P(C′) · P(D|C′)
= 0.05 · 0.78 + 0.95 · 0.06 = 0.039 + 0.054 = 0.093
![Page 30: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/30.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Examples
Exercise 2.95 + forget about 40Events:C : A randomly chosen person has cancer
D : A randomly chosen person is diagnosed as having cancer
Complementary events: C′ and D′
Data:
P(C) = 0.05 (P(C′) = 0.95)P(D|C) = 0.78 (P(D′|C) = 0.22)P(D|C′) = 0.06 (P(D′|C′) = 0.94)
Question 1:P(D) =?
Solution:
P(D) =
P(D ∩ C) + P(D ∩ C′)
= P(C) · P(D|C) + P(C′) · P(D|C′)
= 0.05 · 0.78 + 0.95 · 0.06 = 0.039 + 0.054 = 0.093
![Page 31: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/31.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Examples
Exercise 2.95 + forget about 40Events:C : A randomly chosen person has cancerD : A randomly chosen person is diagnosed as having cancer
Complementary events: C′ and D′
Data:
P(C) = 0.05 (P(C′) = 0.95)P(D|C) = 0.78 (P(D′|C) = 0.22)P(D|C′) = 0.06 (P(D′|C′) = 0.94)
Question 1:P(D) =?
Solution:
P(D) =
P(D ∩ C) + P(D ∩ C′)
= P(C) · P(D|C) + P(C′) · P(D|C′)
= 0.05 · 0.78 + 0.95 · 0.06 = 0.039 + 0.054 = 0.093
![Page 32: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/32.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Examples
Exercise 2.95 + forget about 40Events:C : A randomly chosen person has cancerD : A randomly chosen person is diagnosed as having cancer
Complementary events: C′ and D′
Data:
P(C) = 0.05 (P(C′) = 0.95)P(D|C) = 0.78 (P(D′|C) = 0.22)P(D|C′) = 0.06 (P(D′|C′) = 0.94)
Question 1:P(D) =?
Solution:
P(D) =
P(D ∩ C) + P(D ∩ C′)
= P(C) · P(D|C) + P(C′) · P(D|C′)
= 0.05 · 0.78 + 0.95 · 0.06 = 0.039 + 0.054 = 0.093
![Page 33: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/33.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Examples
Exercise 2.95 + forget about 40Events:C : A randomly chosen person has cancerD : A randomly chosen person is diagnosed as having cancer
Complementary events: C′ and D′
Data:
P(C) = 0.05 (P(C′) = 0.95)P(D|C) = 0.78 (P(D′|C) = 0.22)P(D|C′) = 0.06 (P(D′|C′) = 0.94)
Question 1:P(D) =?
Solution:
P(D) =
P(D ∩ C) + P(D ∩ C′)
= P(C) · P(D|C) + P(C′) · P(D|C′)
= 0.05 · 0.78 + 0.95 · 0.06 = 0.039 + 0.054 = 0.093
![Page 34: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/34.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Examples
Exercise 2.95 + forget about 40Events:C : A randomly chosen person has cancerD : A randomly chosen person is diagnosed as having cancer
Complementary events: C′ and D′
Data:P(C) = 0.05 (P(C′) = 0.95)
P(D|C) = 0.78 (P(D′|C) = 0.22)P(D|C′) = 0.06 (P(D′|C′) = 0.94)
Question 1:P(D) =?
Solution:
P(D) =
P(D ∩ C) + P(D ∩ C′)
= P(C) · P(D|C) + P(C′) · P(D|C′)
= 0.05 · 0.78 + 0.95 · 0.06 = 0.039 + 0.054 = 0.093
![Page 35: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/35.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Examples
Exercise 2.95 + forget about 40Events:C : A randomly chosen person has cancerD : A randomly chosen person is diagnosed as having cancer
Complementary events: C′ and D′
Data:P(C) = 0.05 (P(C′) = 0.95)P(D|C) = 0.78 (P(D′|C) = 0.22)
P(D|C′) = 0.06 (P(D′|C′) = 0.94)
Question 1:P(D) =?
Solution:
P(D) =
P(D ∩ C) + P(D ∩ C′)
= P(C) · P(D|C) + P(C′) · P(D|C′)
= 0.05 · 0.78 + 0.95 · 0.06 = 0.039 + 0.054 = 0.093
![Page 36: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/36.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Examples
Exercise 2.95 + forget about 40Events:C : A randomly chosen person has cancerD : A randomly chosen person is diagnosed as having cancer
Complementary events: C′ and D′
Data:P(C) = 0.05 (P(C′) = 0.95)P(D|C) = 0.78 (P(D′|C) = 0.22)P(D|C′) = 0.06 (P(D′|C′) = 0.94)
Question 1:P(D) =?
Solution:
P(D) =
P(D ∩ C) + P(D ∩ C′)
= P(C) · P(D|C) + P(C′) · P(D|C′)
= 0.05 · 0.78 + 0.95 · 0.06 = 0.039 + 0.054 = 0.093
![Page 37: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/37.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Examples
Exercise 2.95 + forget about 40Events:C : A randomly chosen person has cancerD : A randomly chosen person is diagnosed as having cancer
Complementary events: C′ and D′
Data:P(C) = 0.05 (P(C′) = 0.95)P(D|C) = 0.78 (P(D′|C) = 0.22)P(D|C′) = 0.06 (P(D′|C′) = 0.94)
Question 1:P(D) =?
Solution:
P(D) =
P(D ∩ C) + P(D ∩ C′)
= P(C) · P(D|C) + P(C′) · P(D|C′)
= 0.05 · 0.78 + 0.95 · 0.06 = 0.039 + 0.054 = 0.093
![Page 38: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/38.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Examples
Exercise 2.95 + forget about 40Events:C : A randomly chosen person has cancerD : A randomly chosen person is diagnosed as having cancer
Complementary events: C′ and D′
Data:P(C) = 0.05 (P(C′) = 0.95)P(D|C) = 0.78 (P(D′|C) = 0.22)P(D|C′) = 0.06 (P(D′|C′) = 0.94)
Question 1:P(D) =?
Solution:
P(D) =
P(D ∩ C) + P(D ∩ C′)
= P(C) · P(D|C) + P(C′) · P(D|C′)
= 0.05 · 0.78 + 0.95 · 0.06 = 0.039 + 0.054 = 0.093
![Page 39: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/39.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Examples
Exercise 2.95 + forget about 40Events:C : A randomly chosen person has cancerD : A randomly chosen person is diagnosed as having cancer
Complementary events: C′ and D′
Data:P(C) = 0.05 (P(C′) = 0.95)P(D|C) = 0.78 (P(D′|C) = 0.22)P(D|C′) = 0.06 (P(D′|C′) = 0.94)
Question 1:P(D) =?
Solution:
P(D) = P(D ∩ C) + P(D ∩ C′)
= P(C) · P(D|C) + P(C′) · P(D|C′)
= 0.05 · 0.78 + 0.95 · 0.06 = 0.039 + 0.054 = 0.093
![Page 40: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/40.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Examples
Exercise 2.95 + forget about 40Events:C : A randomly chosen person has cancerD : A randomly chosen person is diagnosed as having cancer
Complementary events: C′ and D′
Data:P(C) = 0.05 (P(C′) = 0.95)P(D|C) = 0.78 (P(D′|C) = 0.22)P(D|C′) = 0.06 (P(D′|C′) = 0.94)
Question 1:P(D) =?
Solution:
P(D) = P(D ∩ C) + P(D ∩ C′)
= P(C) · P(D|C) + P(C′) · P(D|C′)
= 0.05 · 0.78 + 0.95 · 0.06 = 0.039 + 0.054 = 0.093
![Page 41: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/41.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Examples
Exercise 2.95 + forget about 40Events:C : A randomly chosen person has cancerD : A randomly chosen person is diagnosed as having cancer
Complementary events: C′ and D′
Data:P(C) = 0.05 (P(C′) = 0.95)P(D|C) = 0.78 (P(D′|C) = 0.22)P(D|C′) = 0.06 (P(D′|C′) = 0.94)
Question 1:P(D) =?
Solution:
P(D) = P(D ∩ C) + P(D ∩ C′)
= P(C) · P(D|C) + P(C′) · P(D|C′)
= 0.05 · 0.78 + 0.95 · 0.06 = 0.039 + 0.054 = 0.093
![Page 42: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/42.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Examples
Exercise 2.95Events:C : A randomly chosen person has cancerD : A randomly chosen person is diagnosed as having cancer
Complementary events: C′ and D′
Data:P(C) = 0.05 (P(C′) = 0.95)P(D|C) = 0.78 (P(D′|C) = 0.22)P(D|C′) = 0.06 (P(D′|C′) = 0.94)P(D) = 0.093 (P(D′) = 1− 0.093 = 0.907)
Question 2: What is the probability that someonediagnosed with cancer actually has the disease?
P(C|D) =?
Solution:
P(C|D) =
P(C ∩ D)
P(D)=
P(D|C) · P(C)
P(D)=
0.78 · 0.050.093
=0.0390.093
=1331
![Page 43: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/43.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Examples
Exercise 2.95Events:C : A randomly chosen person has cancerD : A randomly chosen person is diagnosed as having cancer
Complementary events: C′ and D′
Data:P(C) = 0.05 (P(C′) = 0.95)P(D|C) = 0.78 (P(D′|C) = 0.22)P(D|C′) = 0.06 (P(D′|C′) = 0.94)P(D) = 0.093 (P(D′) = 1− 0.093 = 0.907)
Question 2: What is the probability that someonediagnosed with cancer actually has the disease?
P(C|D) =?
Solution:
P(C|D) =
P(C ∩ D)
P(D)=
P(D|C) · P(C)
P(D)=
0.78 · 0.050.093
=0.0390.093
=1331
![Page 44: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/44.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Examples
Exercise 2.95Events:C : A randomly chosen person has cancerD : A randomly chosen person is diagnosed as having cancer
Complementary events: C′ and D′
Data:P(C) = 0.05 (P(C′) = 0.95)P(D|C) = 0.78 (P(D′|C) = 0.22)P(D|C′) = 0.06 (P(D′|C′) = 0.94)P(D) = 0.093 (P(D′) = 1− 0.093 = 0.907)
Question 2: What is the probability that someonediagnosed with cancer actually has the disease?P(C|D) =?
Solution:
P(C|D) =
P(C ∩ D)
P(D)=
P(D|C) · P(C)
P(D)=
0.78 · 0.050.093
=0.0390.093
=1331
![Page 45: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/45.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Examples
Exercise 2.95Events:C : A randomly chosen person has cancerD : A randomly chosen person is diagnosed as having cancer
Complementary events: C′ and D′
Data:P(C) = 0.05 (P(C′) = 0.95)P(D|C) = 0.78 (P(D′|C) = 0.22)P(D|C′) = 0.06 (P(D′|C′) = 0.94)P(D) = 0.093 (P(D′) = 1− 0.093 = 0.907)
Question 2: What is the probability that someonediagnosed with cancer actually has the disease?P(C|D) =?
Solution:
P(C|D) =
P(C ∩ D)
P(D)=
P(D|C) · P(C)
P(D)=
0.78 · 0.050.093
=0.0390.093
=1331
![Page 46: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/46.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Examples
Exercise 2.95Events:C : A randomly chosen person has cancerD : A randomly chosen person is diagnosed as having cancer
Complementary events: C′ and D′
Data:P(C) = 0.05 (P(C′) = 0.95)P(D|C) = 0.78 (P(D′|C) = 0.22)P(D|C′) = 0.06 (P(D′|C′) = 0.94)P(D) = 0.093 (P(D′) = 1− 0.093 = 0.907)
Question 2: What is the probability that someonediagnosed with cancer actually has the disease?P(C|D) =?
Solution:
P(C|D) =P(C ∩ D)
P(D)
=P(D|C) · P(C)
P(D)=
0.78 · 0.050.093
=0.0390.093
=1331
![Page 47: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/47.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Examples
Exercise 2.95Events:C : A randomly chosen person has cancerD : A randomly chosen person is diagnosed as having cancer
Complementary events: C′ and D′
Data:P(C) = 0.05 (P(C′) = 0.95)P(D|C) = 0.78 (P(D′|C) = 0.22)P(D|C′) = 0.06 (P(D′|C′) = 0.94)P(D) = 0.093 (P(D′) = 1− 0.093 = 0.907)
Question 2: What is the probability that someonediagnosed with cancer actually has the disease?P(C|D) =?
Solution:
P(C|D) =P(C ∩ D)
P(D)=
P(D|C) · P(C)
P(D)
=0.78 · 0.05
0.093=
0.0390.093
=1331
![Page 48: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/48.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Examples
Exercise 2.95Events:C : A randomly chosen person has cancerD : A randomly chosen person is diagnosed as having cancer
Complementary events: C′ and D′
Data:P(C) = 0.05 (P(C′) = 0.95)P(D|C) = 0.78 (P(D′|C) = 0.22)P(D|C′) = 0.06 (P(D′|C′) = 0.94)P(D) = 0.093 (P(D′) = 1− 0.093 = 0.907)
Question 2: What is the probability that someonediagnosed with cancer actually has the disease?P(C|D) =?
Solution:
P(C|D) =P(C ∩ D)
P(D)=
P(D|C) · P(C)
P(D)=
0.78 · 0.050.093
=0.0390.093
=1331
![Page 49: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/49.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Examples
Exercise 2.95 - Question 3Events:C : A randomly chosen person has cancerD : A randomly chosen person is diagnosed as having cancer
Data:P(C) = 0.05 (P(C′) = 0.95)P(D|C) = 0.78 (P(D′|C) = 0.22)P(D|C′) = 0.06 (P(D′|C′) = 0.94)P(D) = 0.093 (P(D′) = 1− 0.093 = 0.907)
Question 3: What is the probability that someone who isdiagnosed as not having cancer actually has the disease?P(C|D′) =?
Solution:
P(C|D′) =
P(C ∩ D′)
P(D′)=
P(D′|C) · P(C)
P(D′)
=0.05 · 0.22
0.907=
0.00110.907
≈ 0.012
![Page 50: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/50.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Examples
Exercise 2.95 - Question 3Events:C : A randomly chosen person has cancerD : A randomly chosen person is diagnosed as having cancer
Data:P(C) = 0.05 (P(C′) = 0.95)P(D|C) = 0.78 (P(D′|C) = 0.22)P(D|C′) = 0.06 (P(D′|C′) = 0.94)P(D) = 0.093 (P(D′) = 1− 0.093 = 0.907)
Question 3: What is the probability that someone who isdiagnosed as not having cancer actually has the disease?
P(C|D′) =?
Solution:
P(C|D′) =
P(C ∩ D′)
P(D′)=
P(D′|C) · P(C)
P(D′)
=0.05 · 0.22
0.907=
0.00110.907
≈ 0.012
![Page 51: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/51.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Examples
Exercise 2.95 - Question 3Events:C : A randomly chosen person has cancerD : A randomly chosen person is diagnosed as having cancer
Data:P(C) = 0.05 (P(C′) = 0.95)P(D|C) = 0.78 (P(D′|C) = 0.22)P(D|C′) = 0.06 (P(D′|C′) = 0.94)P(D) = 0.093 (P(D′) = 1− 0.093 = 0.907)
Question 3: What is the probability that someone who isdiagnosed as not having cancer actually has the disease?P(C|D′) =?
Solution:
P(C|D′) =
P(C ∩ D′)
P(D′)=
P(D′|C) · P(C)
P(D′)
=0.05 · 0.22
0.907=
0.00110.907
≈ 0.012
![Page 52: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/52.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Examples
Exercise 2.95 - Question 3Events:C : A randomly chosen person has cancerD : A randomly chosen person is diagnosed as having cancer
Data:P(C) = 0.05 (P(C′) = 0.95)P(D|C) = 0.78 (P(D′|C) = 0.22)P(D|C′) = 0.06 (P(D′|C′) = 0.94)P(D) = 0.093 (P(D′) = 1− 0.093 = 0.907)
Question 3: What is the probability that someone who isdiagnosed as not having cancer actually has the disease?P(C|D′) =?
Solution:
P(C|D′) =
P(C ∩ D′)
P(D′)=
P(D′|C) · P(C)
P(D′)
=0.05 · 0.22
0.907=
0.00110.907
≈ 0.012
![Page 53: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/53.jpg)
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Preliminaries Bayes′ rule A lot of computing . . .
Examples
Exercise 2.95 - Question 3Events:C : A randomly chosen person has cancerD : A randomly chosen person is diagnosed as having cancer
Data:P(C) = 0.05 (P(C′) = 0.95)P(D|C) = 0.78 (P(D′|C) = 0.22)P(D|C′) = 0.06 (P(D′|C′) = 0.94)P(D) = 0.093 (P(D′) = 1− 0.093 = 0.907)
Question 3: What is the probability that someone who isdiagnosed as not having cancer actually has the disease?P(C|D′) =?
Solution:
P(C|D′) =P(C ∩ D′)
P(D′)
=P(D′|C) · P(C)
P(D′)
=0.05 · 0.22
0.907=
0.00110.907
≈ 0.012
![Page 54: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/54.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Examples
Exercise 2.95 - Question 3Events:C : A randomly chosen person has cancerD : A randomly chosen person is diagnosed as having cancer
Data:P(C) = 0.05 (P(C′) = 0.95)P(D|C) = 0.78 (P(D′|C) = 0.22)P(D|C′) = 0.06 (P(D′|C′) = 0.94)P(D) = 0.093 (P(D′) = 1− 0.093 = 0.907)
Question 3: What is the probability that someone who isdiagnosed as not having cancer actually has the disease?P(C|D′) =?
Solution:
P(C|D′) =P(C ∩ D′)
P(D′)=
P(D′|C) · P(C)
P(D′)
=0.05 · 0.22
0.907=
0.00110.907
≈ 0.012
![Page 55: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/55.jpg)
beamer-tu-logo
Preliminaries Bayes′ rule A lot of computing . . .
Examples
Exercise 2.95 - Question 3Events:C : A randomly chosen person has cancerD : A randomly chosen person is diagnosed as having cancer
Data:P(C) = 0.05 (P(C′) = 0.95)P(D|C) = 0.78 (P(D′|C) = 0.22)P(D|C′) = 0.06 (P(D′|C′) = 0.94)P(D) = 0.093 (P(D′) = 1− 0.093 = 0.907)
Question 3: What is the probability that someone who isdiagnosed as not having cancer actually has the disease?P(C|D′) =?
Solution:
P(C|D′) =P(C ∩ D′)
P(D′)=
P(D′|C) · P(C)
P(D′)
=0.05 · 0.22
0.907=
0.00110.907
≈ 0.012
![Page 56: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/56.jpg)
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Preliminaries Bayes′ rule A lot of computing . . .
And now . . .
1 PreliminariesPartition of a sample space
2 Bayes′ ruleTheoryExamples
3 A lot of computing . . .
![Page 57: Lecture 7: Bayes' Rule, Revision - Stankova · beamer-tu-logo Preliminaries Bayes0 ruleA lot of computing ... Lecture 7: Bayes0Rule, Revision Kateˇrina Sta nkovᡠStatistics (MAT1003)](https://reader033.vdocuments.us/reader033/viewer/2022042400/5f0f26e27e708231d442be53/html5/thumbnails/57.jpg)
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Preliminaries Bayes′ rule A lot of computing . . .
Try all odd exercises from Section 2.7 (pp. 76–77),Exercise 2.99 was done in the classAlso, you should be able to compute all Review exercisesfrom pp. 77-79, check it!