1 / 4

Probability Problems: Permutations and Bayes' Rule

Solve permutation and probability calculations involving boys, girls, and Bayes' rule in various scenarios. Understand how to compute probabilities based on given conditions and apply them to determine outcomes accurately.

luther
Download Presentation

Probability Problems: Permutations and Bayes' Rule

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Homework #2 Chapter 3

  2. Problem 18 • Permutation between 5 boys and 10 girls • The 4th position is a boy • (14!/15!) X 5 =1/3 • The 12th position is a boy • The same reasoning= 1/3 • The 3rd position is a particular boy • (14!)/(15!)=1/15

  3. Problem 34 • If P(HPSA/C)=0.268 (i.e., P(LPSA/C)=0.732), P(HPSA/NC)=0.135 (i.e., P(LPSA/NC)=0.865), and, P(C)=0.7, then P(C/HPSA)=? P(C/LPSA)=? • By using the Bayes’ rule, P(C/HPSA)= P(C∩HPSA) / P(HPSA) =[P(C)P(HPSA/C)]/ {[P(C)P(HPSA/C)]+[P(NC)P(HPSA/NC)]} =0.82 • P(C/LPSA)= P(C∩LPSA) / P(LPSA) =[P(C)P(LPSA/C)]/ {[P(C)P(LPSA/C)]+[P(NC)P(LPSA/NC)]}=0.69 • In the same testing results, if P(C)=0.3, then P(C/HPSA)=? P(C/LPSA)=? • By the same approach, P(C/HPSA)=0.46 • P(C/LPSA)=0.266

  4. Problem 37 • P{(1-2-3)U(4-5)}=P(1-2-3)+P(4-5)-P(1-2-3-4-5)=p1p2p3+p4p5-p1p2p3p4p5 , (it is also equal to 1 -P[(1-2-3)c]P[(4-5)c]=1-(1-p1p2p3)(1-p4p5) • P{[(1-2)U(3-4)]∩5}= p5X [1-(1-p1p2)(1-p3p4)] • If the circuit 3 is open, then the probability will be the same as a): [1-(1-p1p4)(1-p2p5)]X(1-p3); • In the other condition, if the circuit 3 is closing, then the probability will be [1-(1-p1)(1-p2)] X[1-(1-p4)(1-p5)] X p3 • The summation will be the probability of current flowing from A to B

More Related