Download presentation

Presentation is loading. Please wait.

Published byKyree Moxham Modified over 2 years ago

1
MEGN 537 – Probabilistic Biomechanics Ch.2 – Mathematics of Probability Anthony J Petrella, PhD

3
Probability Relations Experimental outcomes can be represented by set theory relationships Union: A1 A3, elements belong to A1 or A3 or both P(A1 A3) = P(A1) + P(A3) - P(A1 A3) = A1+A3-A2 Intersection: A1 A3, elements belong to A1 and A3 P(A1 A3) = P(A3|A1) * P(A1) = A2 (multiplication rule) Complement: A’, elements that do not belong to A P(A’) = 1 – P(A) S

4
Important Concepts (review) Conditional Probability P(A B) = P(A|B) * P(B) = P(B|A) * P(A) Statistical independence P(A B) = P(A) * P(B) = P(B) * P(A) Mutual Exclusivity P(A B) = 0

5
Conditional Probability The likelihood that event B will occur if event A has already occurred P(A B) = P(B|A) * P(A) P(B|A) = P(A B) / P(A) Requires that P(A) ≠ 0 Remember P(A B) = P(A|B) * P(B) = P(B|A) * P(A) S AB

6
Theorem of Total Probability For collectively exhaustive and mutually exclusive events (E1, E2, …, En): P(A) = P(A|E1)*P(E1)+P(A|E2)*P(E2) + … + P(A|En)*P(En) A S E1 E2 E3 E4 En

7
Bayes’ Theorem Building on the idea of total probability, what is the probability that the event A was “caused” by the event E k (inverse probability) If E i are mutually exclusive events Theorem of Total Probability Conditional Probability For collectively exhaustive and mutually exclusive events A E1 E2 E3 For a specific event E k : S E4 En

8
Example Given a truss structure: If the likelihood of failure of any member is 10 -5, what is the likelihood of truss failure. The truss is statically determinant – failure of one member implies failure of the truss. w 1 2 3 4 5

9
Answer Application of Total Probability Let E = member failure F = truss failure P(E i ) = 10 -5 P(F|E i ) = 1 P(F) = P(F|E 1 )P(E 1 )+…P(F|E 5 )P(E 5 ) = 1*(10 -5 )+…+1*(10 -5 ) = 5*(1*(10 -5 )) = 0.00005

10
Answer Application of DeMorgan’s Rule Let E = member failure; F = truss failure P(E i ) = 10 -5 P(E i ’) = 1 - 10 -5 P(F)= P(E 1 E 2 E 3 E 4 E 5 ) = 1 - P((E 1 E 2 E 3 E 4 E 5 )’) = 1 - P(E 1 ’E 2 ’E 3 ’E 4 ’E 5 ’) [statistically independent] = 1 - P(E 1 ’) P(E 2 ’) P(E 3 ’) P(E 4 ’) P(E 5 ’) = 1 - (1 - 10 -5 ) 5 = 0.00005

11
Example A firm that manufactures syringes tested 1000 samples and finds that 10% are defective. The company concludes that a randomly selected shipment from the production process has a probability of 0.1 of being defective, P(D)=0.1. The firm also finds that the inspectors typically had a 5% error rate on both defective and non-defective parts. If “A” is the event of accepting the syringe as good P(A|D)=0.05, P(A|D’)=0.95 1.What is the probability of shipping a defective syringe at the current operating conditions? 2.What is the probability of shipping a non-defective syringe at the current operating conditions? 3.What is the probability of shipping a defective syringe if the company reduces its defect production rate to 0.01?

12
Answer: Using Bayes’ Theorem P(D) = 0.1 P(D) = 0.01 P(D|A) = P(A|D)*P(D) P(A|D)*P(D)+P(A|D’)*P(D’) P(D|A) = (0.05*0.1) = 0.0058 0.05*0.1+0.95*0.9 P(D|A) = (0.05*0.01) = 0.000531 0.05*0.01+0.95*0.99 P(D’|A) = P(A|D’)*P(D’) = 0.9942 P(A|D)*P(D)+P(A|D’)*P(D’)

Similar presentations

Presentation is loading. Please wait....

OK

LECTURE#2 Refreshing Concepts on Probability

LECTURE#2 Refreshing Concepts on Probability

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on new zealand culture and customs Ppt on solid dielectrics in electric fields Ppt on natural resources and conservation degrees Ppt on energy giving food example Ppt on addition for class 2 Heart anatomy and physiology ppt on cells Download ppt on fdi in retail in india Ppt on hard gelatin capsule market Ppt on aerobics music Ppt on game theory raleigh