probability and statistics with programming

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Probability and Statistics with Programming Course Code 009959 (Spring 2018) Prof. S. M. Riazul Islam, Dept. of Computer Engineering, Sejong University, Korea Probability Basics: Problem Solving E-mail: [email protected]

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Probability and Statistics with Programming

Course Code 009959 (Spring 2018)

Prof. S. M. Riazul Islam, Dept. of Computer Engineering, Sejong University, Korea

Probability Basics: Problem Solving

E-mail: [email protected]

Basic Concepts of Probability

Tree Diagram vs. Sample Space

Problem Solving: Conditional Probability

Problem Solving: Independence

Review: Law of Total Probability

Basic Concepts of Probability

Tree Diagram vs. Sample Space

Sample Space for a Double Coin-Toss Experiment

S= {HH, HT, TH, TT}

Tree Diagram

H

H

H

T

T

T

HH

HT

TH

TT

P(H)

P(T)

P(H/H)

P(T/H)

P(H/T)

P(T/T)

Basic Concepts of Probability

Problem 1

Book: Probability and Statistics for Engineering and the Sciences (Jay L. Devore)

Basic Concepts of Probability

Problem 2

Book: Probability and Statistics for Engineering and the Sciences (Jay L. Devore)

Solution 2

Basic Concepts of Probability

Solution 2

P A ∩ B ∩ C = 0.54

P B ∩ C = 0.54 + 0.14 = 0.68

P(C)=0.54+0.045+0.14+0.015=0.74

P A B ∩ C =P(A∩B∩C)

P(B∩C)=0.729

Basic Concepts of Probability

Problem 3

Book: Probability and Statistics for Engineering and the Sciences (Jay L. Devore)

Basic Concepts of Probability

Solution 3

P(both phenotypes are O)

=P O, O = P O × P O = 0.45 × 0.45 = 0.2025

P(Phenotypes of the selected individuals are matched)

=P A ∩ A) or (B ∩ B) or (AB ∩ AB) or (O ∩ O

= P A ∩ A + P B ∩ B + P AB ∩ AB + P O ∩ O= 0.16 + 0.0121 + 0.0016 + 0.2025

= 0.3762

Basic Concepts of Probability

Probability Review

Q&A