100AHW3 - STAT 100A HWIII Solution Problem 1: Prove the...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
STAT 100A HWIII Solution Problem 1: Prove the following two identities ( C stands for cause, E stands for effect), where { C i ,i = 1 ,...,n } partition the whole sample space. (1) Rule of total probability: P ( E ) = n i =1 P ( C i ) P ( E | C i ). A: P ( E ) = P ( n i =1 ( E C i )) = n i =1 P ( E C i ) = n i =1 P ( C i ) P ( E | C i ). (2) Bayes rule: P ( C j | E ) = P ( C j ) P ( E | C j ) / n i =1 P ( C i ) P ( E | C i ). A: P ( C j | E ) = P ( C j E ) /P ( E ) = P ( C j ) P ( E | C j ) / n i =1 P ( C i ) P ( E | C i ). Problem 2: Suppose 1% of the population is inflicted with a particular disease. For a medical test, if a person has the disease, then 95% chance the person will be tested positive. If a person does not have the disease, then 90% chance the person will be tested negative. Using precise notation, calculate (1) The probability that a randomly selected person will be tested positive. A: Let
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.
Ask a homework question - tutors are online